WBCHSE Class 11 Some Basic Concepts of Chemistry Notes

Some Basic Concepts of Chemistry

Chemistry is the study of the materials that make up the universe. It deals with the composition, structure, properties, and interaction of substances. The universe is made up of a large variety of materials. Chemists determine how chemical transformations occur among these substances.

Chemistry finds application in diverse areas. For instance, chemical fertilisers have helped increase crop yield to a large extent. Chemical industries also manufacture acids, alkalis, salts, drugs, dyes, soaps, detergents, alloys, polymers, etc.

  • Chemistry has provided many life-saving drugs. Cisplatin and Taxol are very effective in cancer therapy, and AZT (Azidothymidine) is used to treat AIDS victims.
  • In recent years chemistry has given us many new materials with specific magnetic, electric, and optical properties. These types of materials include conducting polymers, superconducting ceramics, and optical fibers.
  • Environmentally hazardous refrigerants like CFCs (chlorofluorocarbons), which are responsible for the depletion of the ozone layer, have been replaced by alternatives.
  • Like the study of all other sciences, the study of chemistry is based on the careful observation of various phenomena (like a reaction between substances) under controlled conditions.
  • In other words, experiments form the foundation of chemistry. The observations made during an experiment may be qualitative or quantitative in nature. Quantitative observations involve the measurement of one or more quantities and are discussed later in the chapter.

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Classification Of Matter

We are often struck by the diversity of the things that surround us. The diversity of plant and animal life, the diversity of climate, and the diversity of geographical features. And yet, what is perhaps more wonderful is that all that surrounds us, in fact, everything in the entire universe, of which our surroundings comprise an infinitesimal part, is made of two things—matter and energy. Energy is something that we cannot hold in our hands or feel directly.

WBCHSE Class 11 Some Basic Concepts of Chemistry Notes.

  • We can experience its effects like when we feel the warmth of sunlight on our bodies.
  • Matter, on the other hand, is something we can feel and see directly. Expressed in a more formal way, matter is anything that occupies space, has mass, offers resistance, and can be perceived of directly by our senses.
  • Matter can be classified according to its physical state or according to its chemical composition. Physically, matter can exist in three states, viz., solid, liquid, and gaseous.

Chemical classification of matter: On the basis of chemical composition, matter can be broadly divided into pure substances and mixtures. Pure substances can be of two types, viz., elements and compounds, while mixtures can be homogeneous or heterogeneous. An element can be a metal, a nonmetal, or a metalloid, while a compound can be either inorganic or organic in nature. The following chart should make you more familiar with this classification of matter.

Basic Chemistry Class 11 Chapter 1 Some Basic Concepts of Chemistry Chemicall Classification Flowchart

Elements, Compounds And Mixtures

The substances that we see around us are either elements (like gold), compounds (like sugar), or mixtures (like air). Molecules are the units of matter. They may be made up of two or more atoms of the same element or of different elements combined in a definite ratio (compounds).

A mixture is a combination of elements, compounds, or element(s) and compound(s) in any ratio. Elements and compounds are pure substances that have a particular set of properties. Early chemists found it very difficult to distinguish an element from a compound. It was Lavoisier who made a breakthrough and showed how elements can be distinguished from compounds.

1. Elements: An element is the simplest form of a pure substance, which can neither be decomposed into nor built from simpler substances by ordinary physical or chemical methods. This definition of an element arose out of Lavoisier’s work, about which you will leam later in this chapter.

  • Of course, there were difficulties related to defining an element this way. Until a substance could be disintegrated into simpler substances, it was considered to be an element, and when someone showed that it could be broken up, it was no longer considered to be an element. For example, water was considered to be an element until Sir Humphrey Davy showed that it could be decomposed by passing an electric current through it.
  • A more exact definition of an element came up after the discovery of radioactivity, which showed that elements can be broken up and also that they can be synthesised from simpler substances. The modem definition, thus, is that an element is a pure substance that contains only one kind of atoms. Mono-, di- or polyatomic molecules with the same kinds of atoms constitute elements, for example, He, O2, N2, and S8.
  • We know of 111 elements so far. Only 90 of these occur in nature and just 20 elements make up 99% of the Earth’s crust. Some of the most abundant elements on the Earth’s crust are oxygen, silicon, aluminum, iron, carbon, calcium, sodium, and potassium. An element can be a metal, a nonmetal, or a metalloid.

Metals With the exception of mercury, metals are generally solid at room temperature. They are malleable (can be beaten into thin sheets) and ductile (can be drawn into wires), and are good conductors of heat and electricity. Most metals have a high tensile strength and are lustrous. About 75-80 percent of the known elements are metals, gold, silver, copper, iron, and aluminum being some of the common ones.

Nonmetals These elements are poor conductors of electricity and heat. They are generally brittle (if solid), and without luster except iodine which shows a metallic luster. Carbon, sulfur, hydrogen, oxygen, and nitrogen are some common nonmetals

Metalloids These elements have some properties of metals and some of nonmetals. Some common examples of metalloids are silicon, arsenic, antimony, and selenium.

2. Compounds: A compound is a pure substance which is formed by the union of two or more elements in a definite proportion by mass and which can be decomposed into its constituent elements by suitable chemical methods. The composition of a compound is always the same regardless of the source from which it is obtained or the method by which it is produced.

  • For example, water obtained from different sources (example seas, lakes, rivers, or wells) is always made of hydrogen and oxygen chemically combined in the ratio of 1: 8 by mass.
  • A compound is homogeneous (uniform throughout) and has fixed properties. For example, sodium chloride is a compound of sodium and chlorine and has a fixed melting point, density, and solubility in water.
  • The properties of a compound are completely different from those of its constituent elements. For example, hydrogen and oxygen are gases at room temperature, also hydrogen is combustible and oxygen is a supporter of combustion.
  • These two gases combine to form water, which is liquid at room temperature and is used to extinguish fire.

The basic differences between elements and compounds are listed in Table, while some compounds and the ratios of their constituents are mentioned in Table.

Differences between elements and compounds

Basic Chemistry Class 11 Chapter 1 Some Basic Concepts of Chemistry Differences Between Elements And Compounds

Some compounds and their constituents

Basic Chemistry Class 11 Chapter 1 Some Basic Concepts of Chemistry Some Compounds And Their Constituents

Compounds can be of two types, inorganic and organic.

Inorganic compounds These compounds are derived from nonliving sources like rocks and minerals. Common salt, washing soda, and lime are some inorganic compounds we use in everyday life.

Organic compounds These compounds are derived from living sources like plants and animals, or their remains buried under the earth (for example, petroleum). All organic compounds, like carbohydrates, fats, oils, waxes, and proteins contain carbon.

3. Mixtures: Mixtures, like compounds, contain more than one element, but the constituents of a mixture are not present in any particular ratio. Also, the components of a mixture do not lose their identity, so the properties of a mixture are not completely different from those of its constituents.

For example, when we dissolve sugar in water, it still tastes sweet. The components of a mixture do not combine chemically. They can be separated by some physical means. Mixtures can be homogeneous or heterogeneous.

Homogeneous mixtures A mixture that has the same composition throughout is called a homogeneous mixture. Air is a homogeneous mixture of CO2, O2, N2, etc. A solution of sugar in water, natural gas, and alloys, like brass, bronze, and steel, are all homogeneous mixtures. Obviously, homogeneous mixtures can be solids, liquids, or gaseous. Such mixtures are also called solutions. The components of a homogeneous mixture cannot be distinguished even under a microscope.

Heterogeneous mixtures Such mixtures do not have the same composition throughout. In other words, the components of a heterogeneous mixture are not distributed evenly. These components (called phases) can be distinguished with the naked eye, for example, a mixture of sand and sugar, a mixture of salt and pepper, or clay and water. Some mixtures look homogeneous to the naked eye, but are actually heterogeneous, for example, milk. Seen under a microscope, milk looks like a clear liquid in which droplets of fat arc suspended.

Differences between mixtures and compounds

Basic Chemistry Class 11 Chapter 1 Some Basic Concepts of Chemistry Differences Between Mixtures And Compounds

Some of the characteristics of mixtures mentioned in Table do not apply to homogeneous mixtures like bronze. Bronze is an alloy (a solution in the solid state) of copper and zinc in a definite proportion (3:2). The properties of bronze are different from those of copper or zinc. Besides, bronze cannot be separated into copper and zinc by ordinary physical methods. And yet, bronze is a mixture and not a compound because the constituents do not combine chemically.

Properties of Matter: The characteristics of matter are called properties of matter. Such characteristics may include solubility, melting point, odour and chemical behaviour. Properties of matter can be classified as physical or chemical depending on whether the property involves a change in the chemical make-up of a substance.

  • The physical properties are those characteristics that do not involve a chemical reaction. Examples are colour, odour, boiling point, solubility and electrical conductivity. The chemical properties, in contrast, describe the chemical reactions a substance will undergo. For example, sodium reacts violently with water to form sodium hydroxide solution and hydrogen gas. This is a chemical property of sodium.
  • Properties can also be classified as either intensive or extensive depending on whether their values change with the size/amount of the sample. Properties like melting point and boiling point are intensive—they do not change with the change in amount of a substance. As you know, the boiling point of 1 g or 1 kg water is the same—373 K. Length and volume are extensive properties. The length and volume of an ice cube are certainly much smaller than those of an iceberg.

Measurements

When any quantity has to be measured, it is compared with a fixed standard, known as the unit of measurement. As an illustration, if we say that the height of a person is 150 centimeters (usually written as cm), it means that this height is 150 times the unit of measurement, which in this case is 1 cm. The height is measured using a scale with centimetre and millimetre markings. Thus the result of any measurement has two parts—a number (150) and a unit (cm). One does not make sense without the other.

  • Different types of units have been used for the measurement of physical quantities around the world. Even in the first half of the twentieth century people used different units to measure the same physical quantity. For example, weight was measured in pounds and ounces, and seers; distance was measured in feet and yards, and furlongs and miles.
  • The units used varied according to the profession, too. For example, jewellers used tolas and raties to measure weight. The use of different units to measure the same physical quantity was confusing and caused complications. Besides, these systems of units were rather cumbersome. For example, 1 stone = 14 pounds j and 1 pound —16 ounces.
  • To get around these difficulties, the French Academy of Science devised the metric system in 1791. In this simple system, the different units of a physical quantity are related to each other by powers of 10, and these powers are indicated by prefixes used with the unit for the particular physical quantity. The unit of length, for example, is k the metre. The kilometer, a larger unit, is 103 meters, while the centimeter, a smaller unit, is 10-2 metre. Soon, scientists across the world adopted the metric system, and today, the whole world uses it.

The Measurement International System of Units (SI): Though almost the whole world was using the metric system by the middle of the twentieth century, scientists noticed that different metric units were being used for the same physical quantity.

  • In 1960, the General Conference of Weights and Measures adopted a set of units to be used by scientists all over the world to measure six basic physical quantities.
  • Mole, as a unit of amount of substance, was added in the year 1971. This system, known as the International System of Units, is popularly referred to as the SI units (after the French Sysfeme International d’ Unites).

Basic Chemistry Class 11 Chapter 1 Some Basic Concepts of Chemistry Seven Basic SI Units

The seven basic units from which all other units are derived are given in Table Some derived units are listed in Table, while the standard prefixes used to reduce or enlarge the basic units are mentioned in Table.

Basic Chemistry Class 11 Chapter 1 Some Basic Concepts of Chemistry Some Derived Units

Measuring quantities: Some of the quantities which are often used in experimentation in chemistry at this level have been discussed below.
Mass The mass of a substance is the quantity of matter possessed by it.

  • Though the SI unit of mass is the kilogram, it is too large to be used for many purposes in a chemical laboratory. The milligram (1 mg = 0.001 g = 10-6 kg) and the microgram (1 μg = 0.001 mg = 10-6 g = 10-9 kg), are more commonly used in the laboratory.
  • Mass is measured during chemical transformations. To measure with high accuracy, one needs an analytical balance. The standard kilogram is defined as the mass of a cylindrical bar of a platinum-sodium alloy stored in a vault in a suburb of Paris. Here it is worthwhile to mention that mnss is a physical property measuring the amount of matter in a substance whereas weight is the measure of the pull of gravity on that object by the Earth (or any other celestial body).
  • Volume It is a derived quantity, defined as the amount of space occupied by an object. The volume is measured in ST units in cubic metres (m3). But for convenience, measurements in chemistry are usually made in cubic decimetres (1 dm3 = 0.001 m3) and cubic centimetres (1 cm3 = 0.001dm3 = 10-6 m3). The litre is a common unit of volume in everyday life (1L = 1000 mL = 1000 cm3 = 1 dm3). In the laboratory, mostly a graduated cylinder, a volumetric flask, a burette, and pipette are used for measuring volumes of liquids.
  • Density It is an intensive physical property of a substance and relates the mass of an object to its volume. The density of a substance is its mass per unit volume. The SI-derived unit of density is kg m-3, which is too large to measure densities such as that of water (1.00 g cm-3). Chemists often use smaller units—g cm-3 for solids and g/mL for liquids.
    • The densities of substances change with change in temperature (since volume of a substance varies with temperature). Therefore, it is better to specify the temperature while stating the density of a substance. Temperature The three common scales to measure temperature are °C (degree Celsius), °F (degree Fahrenheit) and K (kelvin).

The temperature on the Celsius and the Fahrenheit scales are related to each other as:

∴ C =5(F-32)/9.

The magnitude of the kelvin is equal to that of the degree Celsius but the two are related as:

∴ K=°C + 273.15.

Practically speaking, the lowest temperature permitted in nature is -273.15 °C (0 K). This temperature is known as absolute zero.

Chemists measure time as it helps to know how long it takes for a chemical process or reaction to occur. Some reactions like the formation of fossil fuels take millions of years. On the other hand, when hydrochloric add is added to a solution of silver nitrate, silver chloride is precipitated in a fraction of a second.

Measurement Dimensional analysis: All derived quantities, like area, volume, density, and speed, can be derived from the seven basic quantities. The units of the derived quantities can be obtained from the seven basic units, as shown in Table. For example, if the side of a square is expressed in metres (m), the area is expressed in square metres (m2). Similarly, if the distance covered is expressed in metres and the time taken in seconds, the velocity is expressed in m s-1.

  • Quite often it becomes necessary to convert one set of units into another. This can be done quite simply by dimensional analysis. The dimensions of a derived quantity are the powers to which the basic quantities have to be raised in a product defining the quantity.
  • Dimensional analysis involves calculations based on the fact that if two quantities have to be equated, they must have the same dimensions or the same units. Some calculations involving the dimensions of physical quantities follow.

Measurement Conversion of one unit into another

1. Suppose you want to convert 10 minutes into seconds.

First find the unit conversion factor.

1 min = 60 s.

∴ 1 = \(\frac{60 \mathrm{~s}}{1 \mathrm{~min}}\)

Thus, 60 s/1 min is the conversion factor. Note that the quantities dimensionless, since time is divided by time on the right-hand side.

Let us convert 10 minutes into seconds.

10 min = 10 min x \(\frac{60 \mathrm{~s}}{1 \mathrm{~min}}\) = 600 s. ….(1)

Dimensionally both sides of the Equation (1) are the same. Also, multiplying 10 minutes by (60 s/1 min) does not change its value, since (60 s/1 min) is equal to unity. A great advantage of using this method is that any mistake made in writing the conversion factor can be spotted easily.

In Equation (1) the unit minutes in the numerator and denominator of the right-hand side cancel, leaving behind the unit seconds. This would not have been so if the wrong conversion factor had been used and the mistake would have been noticed immediately. For example, if 1 min/60 s had been used as the conversion factor

10 min = 10 min \(\frac{1 \mathrm{~min}}{60 \mathrm{~s}}=\frac{1 \mathrm{~min}^2}{6 \mathrm{~s}}\)

2. Similarly, to convert 170 pounds (lb) into kilograms (kg), first find the unit conversion factor.

1 kg = 2205 lb.

∴ 1 = \(\frac{2.205 \mathrm{lb}}{1 \mathrm{~kg}}=\frac{1 \mathrm{~kg}}{2.205 \mathrm{lb}}\)

Mass(kg) = 170 lb x \(\frac{1 \mathrm{~kg}}{2.205 \mathrm{Ib}}\) = 77.09 kg ……..(2)

If the wrong conversion factor had been used, the units on the two sides of Equation would not have been the same.

Measurement Validity of an equation: Dimensional analysis is also used to check the validity of an equation. If the dimensions on both sides of the equation do not match, the equation cannot be correct. Let us take an example.

∴ E = mc2 …….(3)

Using SI units, E is in joules, m in kg and c in m s-1. Let us see if the units on both sides of Equation (3) match.

J = kg (ms-1)2 or

or, Nm = kg m2 s-2

or kg m2 s-2 = kg m2 s-2.

Uncertainty in measurement: You can be very certain about the measurement of a quantity when it involves the counting of objects. For example, when you count the number of chairs in your class you can be certain about the exactness of your answer. You can also be sure that if someone else were to count the number of chairs, one would get the same result.

However, can you be as certain about the result you get when you measure your friend’s height with a measuring tape? You may read it as 150.1 cm and another friend may measure it as 150.3 cm. At the most you and your friend can reach the conclusion that his height is greater than 150 cm and less than 151 cm.

  • There is always a certain amount of uncertainty in such measurements. This uncertainty arises not because the height of your friend (or any other measured quantity) is uncertain, but because of the way the measuring instrument is calibrated (accuracy of the scale) and because two people may record the same reading differently (human error).
  • The accuracy of a measurement thus, depends on the accuracy of the measuring instrument and the skill of the person making the measurement. The difference between the two kinds of measurements that we have just considered (counting chairs and measuring a person’s height) is that in one case, we used a discrete variable and in the other we used a continuous variable.
  • This means that while the number of chairs in your class can be 30 or 31 but not between 30 and 31 (discrete value), the height of your friend can be anything between 150 cm and 151 cm (continuous variable).

Measurement Precision and accuracy: If you repeat a particular measurement, you may not obtain precisely the same result (experimental error). Precision refers to the closeness of a set of values obtained from two or more measurements of the same quantity.

Accuracy, on the other hand, refers to the closeness of a single measurement to its true value. For example, three students A, B, and C separately determine the weight of the same person which is actually 50.45 kg. The results obtained are given below. (The measurements are made in kilograms.)

Basic Chemistry Class 11 Chapter 1 Some Basic Concepts of Chemistry Precision And Accuracy

The values obtained by student A differ widely from one another and the average value is incorrect. Thus the data is neither precise nor accurate. The values determined by student 13 are more precise as they deviate little from one another. The average weight is still not accurate. The data obtained by student C arc both precise and accurate.

Measurement Significant figures: How does one express the uncertainty about the accuracy of a measurement while writing the result of an observation? The convention is to include all the digits which are certain and a last digit that is uncertain while writing the result of a measurement.

  • The total number of digits is called the number of significant figures. It represents the accuracy and precision with which a quantity has been measured. Remember that the number of significant figures is the number of digits in the result, including the last digit which is uncertain.
  • Suppose the length of a piece of wood is reported by three students as 123 cm, 123.0 cm, and 123.00 cm. Thu results may seem equivalent to you but the scientific significance of the three results is very different. The number of significant figures in the three results is 3, 4, and 5 respectively.
  • In other words, in the first case (123 cm), the digits 1 and 2 are certain but not 3—it is only the best estimate. This generally means that the actual value lies between 122 and 124 or that the length is 123 ± 1 cm. When the result of a measurement is expressed in this fashion, it implies that the measurement has been made with a crude scale.
  • The second result has four significant numbers, which means that the length lies between 122.9 cm and 123.1 cm. In other words, in this case, the third digit is also certain, so obviously the scale used for this measurement (123.0 cm) is more precise than the one used for the first measurement.
  • The last result shows that the length lies between 122.99 cm and 123.01 cm (5 significant figures) and implies that the scale used is the most precise of the three.
  • While reporting the result of a measurement, one must be very careful about indicating the precision with which the measurement has been made. To write more significant figures than the scale of measurement allows would be wrong, while writing fewer significant figures than the situation allows for would be holding back information that could be useful. It would be useful to remember the following rules in this context.
  1. All digits are significant except zeros in the beginning of a number. For example, 132 cm, 0.132 cm, and 0. 0132 cm have three significant figures each. Zeros to the left of the first nonzero digit are not significant, even if such zeros follow the decimal point.
  2. Zeros to the right of the decimal point are significant. Thus, 151 g, 151.0 g, and 151.00 g have 3, 4, and 5 significant figures respectively.

The following examples should help you understand these rules better.

  • 678 has 3 significant figures
  • 0. 42 has 2 significant figures
  • 40.5 has 3 significant figures
  • 2004 has 4 significant figures
  • 0. 02 has one significant figure
  • 0. 0022 has two significant figures
  • 2.20 has three significant figures
  • 0. 0200 has three significant figures

In numbers that do not contain a decimal point, trailing zeros may or may not be significant. In case of a very large number of significant figures, for instance the value of n (n = 3.1415926) one may choose tlie number of digits according to the calculation. In other words, all the numeral values involved in a calculation should be so expressed that they have almost the same number of significant figures.

Measurement Calculations involving significant figures: Several calculations may have to be made in the course of an experiment and these calculations may involve the addition, subtraction, multiplication or division of different measured quantities. The measurement of all the quantities (used to arrive at the final result of the experiment) may not be made with the same precision.

The precision of the final result depends on the precision of the least accurate of the measurements. In other words, the final result cannot be more precise than the least precise of the quantities involved in the calculations. Remembering a couple of rules will be helpful while making calculations involving quantities of different precisions.

Rule 1: In additions or subtractions the final result should be reported to the same number of decimal places as that of the term with the least number of decimal places. The number of significant figures of the different numbers have no role to play.

Example 1. The three numbers to be added have 3,1 and 2 decimal places respectively, while the significant figures in the three numbers are 4,2 and 3 respectively.
Solution:

⇒ \(\begin{aligned}
2.512 \\
2.2 \\
5.23 \\
\hline 9.942
\end{aligned}\)

∴ Actual sum 9.942

∴ Reported sum 9.9

The reported sum is 9.9 because the term with the least number of decimal places is 2.2, which has only one decimal place.

Example 2. Each number has three decimal places, so the answer is to be reported up to three decimal places. The significant figures in each of the terms is 4 but the result has 5 significant figures.
Solution:

⇒ \(\begin{array}{r}
6.612 \\
5.234 \\
2.020 \\
\hline 13.866
\end{array}\)

∴ Actual sum 13.866

∴ Reported sum 13.866

Example 3. The first number has 2 decimal places and the second has 4 decimal places, so the answer is reported up to 2 decimal places.
Solution: 

⇒ \(\begin{array}{r}
19.26 \\
(-)11.02534 \\
\hline 8.23466
\end{array}\)

∴ Actual difference 8.23466

∴ Reported difference 8.23

Rule 2: In multiplications and divisions, the result is reported to the same number of significant figures as the least precise term or the term with the least number of significant figures.

Example 1: The first term has 5 significant figures, while the second term has 3 significant figures, so the result can have only 3 significant figures.
Solution:

⇒ \(\begin{aligned}
41.012 \\
\times 1.21 \\
\hline 49.62452
\end{aligned}\)

∴ Actual product 49.62452

∴ Reported product 49.6

Example 2: In this division, 0.41 has 2 significant figures, while 15-724 has 5 significant figures. The result can have only two significant figures.
Solution:

⇒ \(\begin{aligned}
& 15.724 \\
& +0.41 \\
& \hline 0.0260747
\end{aligned}\)

∴ Actual result 0.0260747

∴ Reported result 0.026

Rule 3: If one of the term is in an expression is an exact number, the result should have the same number of significant figures as the least precise term other than the exact number. The presence of exact numberfs) in an expression atfect the number of significant figures that should be included in the answer because an exact number is supposed to have an infinite number of significant figures.

Example 1: All the terms other than the two exact numbers have 3 significant figures. So the answer should also have 3 significant figures.
Solution:

⇒ \(\frac{4.23 \times 0.141 \times 3}{0.0214 \times 2}\)

∴ Actual result 41.8058

∴ Reported reult 41.8

  • Rounding off As seen in the above examples, the final result often has more figures than die number of significant figures in the least precise quantity involved in the calculations. lVhile reporting the result, only the significant figures are retained. The others are dropped. This procedure is called rounding off. The convention followed while rounding off a number can be summarised as follow’s.
  • If the digit which follows the last digit to be reported is less than 5, the last digit is left unchanged and all the digits to its right are dropped. If, however, the digit following the last digit is equal to or more than 5, the last significant figure is increased by one. Thus 1.234,1277 and 12252 become 122,13 and 13 respectively, if the fire result is to be reported up to the first decimal place.
  • While solving a problem remember to take up all significant digits used in calculations and round off only the final result to the desirable number of significant digits.

Scientific (exponential notation): Consider the quantity 10600 g. Written as such, it implies that there are 5 significant figures. Suppose the measurement was made in such a way that the number is precise to only 3 significant figures? Writing the result as 10600 g would then not be correct because it would not indicate the precision of the measurement To remove such ambiguity, results are expressed in the scientific notation.

Thus, 10600 g may be expressed in one of the following ways, depending on the precision of the measurement

1.06 x 104 g (3 significant figures)

1.060 x 104 g (4 significant figures)

1.0600 x 104 g (5 significant figures)

  • You are familiar with the Avogadro number, which is 6.022 x 1023. If you write it in the ordinary way and not in terms of 10 raised to a power, it will be 602,213,700 000 000 000 000 000. It is difficult to write such numbers in an ordinary way and errors may creep in while doing so. The scientific or exponential notation is a better way to represent such numbers.
  • In scientific notation, all numbers, large or small, are expressed as a number between 1.000 and 9.999 multiplied or divided by 10 an appropriate number of times. For example, the number 2484.32 may be expressed as 2.48432 x 10 x 10 x 10 or 24832 x 103. Here 3 is the power or exponent, to which 10 is raised. In general in scientific notation a number is represented as N x 10n,
  • where N is a number between 1.000 and 9.999 and n is a number (not necessarily a single digit) called exponent. To express a number smaller than 1.000 in scientific notation, the decimal point is moved to the right until there is only one nonzero digit before the decimal point.
  • The number is then divided by 10 an appropriate number of times. For example, to express 0.00032481 in scientific notation, we move the decimal point to the right. The number is divided 4 times by 10, as the decimal has shifted four places. In this example, the exponent n = -4.
  • Take another example of a number larger than 9.999. To express it in scientific notation the decimal point is moved to the left until there is only one nonzero digit before the decimal point. Suppose the number is 14872.5. When this number is expressed in scientific notation, the decimal point is moved to the left and the number thus obtained is multiplied by 10.

⇒ 00032481 = \(\frac{3.2481}{10 \times 10 \times 10 \times 10}\) =3.2481 x 10-4

⇒ 14872.5 = \(1.48725 \times 10 \times 10 \times 10 \times 10=1.48725 \times 10^4\)

∴ The exponent n = 4.

To subtract or add numbers in scientific notation the exponent n or power of 10 should be the same in all the numbers. If the exponent is not the same, it has to be made the same by shifting the decimal place in any of the numbers before adding or subtracting. For example, suppose we have to add 6.426 x 103 and 2.045 x 104. We can transform 6.426 x 103 to 0.6426 x 104 and then add,

∴ 2.045 x 104 + 0.6426 x 104 = (2.045 + 0.6426) x 104 = 2.6876 x 104.

To multiply two numbers in scientific notation, we make use of the relation, (10)x x(10)y =10(x+y).

In this case, the exponent need not be the same. For example, (2.0456 x 104) (4.132 x 10-7) = 8.452 x 10[4+(-7)] = 8.452 x 10-3.

To divide two numbers in scientific notation, we make use of the relation

∴ \(\frac{10^x}{10^y}=10^{x-y}\)

The exponent need not be the same here too. For example,

∴ \(\frac{2.45 \times 10^{14}}{9.24 \times 10^{24}}=\frac{2.45}{9.24} \times 10^{14-24}=0.265 \times 10^{-10}\)

In scientific notation, the number is expressed as 2.65 x 10-11. In similar example,

∴ \(\frac{4.65 \times 10^{-4}}{2.92 \times 10^{-10}}=\frac{4.65}{2.92} \times 10^{[-4-(-10)]}=1.59 \times 10^{(-4+10)}=1.59 \times 10^6\)

Historical Approach To The Particulate Nature Of Matter

The concept that matter is made up of tiny bits of material—particles—originated in Greek natural philosophy. The Greek philosopher Democritus in the fifth century BC proposed the “particulate nature of matter” He believed that matter is composed of indivisible particles called atoms (from the Greek atoms, meaning indivisible).

However, now we know that atom is divisible into smaller subatomic particles like protons, neutrons, and electrons. Nevertheless, the word “atom” as indivisible still makes sense as once split, an atom loses its identity. Also, it is the smallest particle of an element that takes part in a chemical combination.

  • The atomic theory of Democritus was not appreciated for many years. Aristotle’s theory that matter is composed of four elements—air, wind, fire, and water—was widely accepted. However, after a series of experiments in more recent times, the particulate nature of matter was established.
  • The French chemist Antoine Laurent Lavoisier, in the late 1700s, discovered that during a chemical reaction, the components involved change in terms of appearance and form only, and the total mass remained the same. In other words, the total mass of the reactants is equal to that of the products (law of conservation of mass).
  • This prompted him to conclude that some basic part of the components which was not visible, did not change. Later, in 1800, scientists working on chemical reactions found that something ruled the behaviour of matter that even their microscopes could not perceive.
  • The law of conservation of mass helped establish the law of constant composition or the law of definite proportions. These laws were a result of Joseph Proust’s extension of Lavoisier’s work.
  • In 1803, John Dalton proposed the atomic theory of matter on the basis of the laws of chemical combination and other related chemical observations. The theory stated that all matter is made of indivisible and indestructible ultimate particles called atoms. Thus the start of the nineteenth century brought back Democritus’ view to the forefront of science.

Percentage Composition And Molecular Formula

To study a chemical compound, mainly its chemical properties, it is essential to know its chemical formula. The chemical formula of a compound can be determined by analyzing the compound for the amount of elements (moles) in a given mass of the compound. The result may also be expressed in percentage composition.

The mass percentage composition of a compound is defined as the number of grams of different elements present in 100 g of the compound. The mass percentage of an element may be calculated by dividing the mass of the element in one mole of the compound by the molar mass of the compound and multiplying the result by 100.
Let us consider CO2.1 mol of carbon dioxide will always contain 1 mol of carbon and 2 mol of oxygen atoms.

The molar mass of CO2 is 12 + (2 x 16) = 44.0 g.

Now 44.0 g of CO2 contains 12.0 g of C and 32.0 g of O.

Mass percentage of C in CO2 = \(\frac{\text { mass of } \mathrm{C} \text { in } 1 \mathrm{~mole} \mathrm{CO}_2}{\text { molar mass of } \mathrm{CO}_2} \times 100\)

= \(\frac{12.0 \mathrm{~g}}{44.0 \mathrm{~g}} \times 100=27.27\)

Similarly, we may calculate the mass percentage of O in 1 mole of CO2.

Mass percentage of O in CO2 = \(\frac{32.0 \mathrm{~g}}{44.0 \mathrm{~g}}\) x 100 = 72.73.

The molecular composition of a compound can be expressed in any of the following ways.

  1. A chemical formula giving the number of atoms of each type per molecule, i.e., CO2.
  2. The number of moles of each element per mole of a compound.
  3. The mass of each element per 100 g of a compound.

We may conclude that if the formula of a compound is known we can determine its percentage composition and vice versa.

Empirical Formula And Molecular Formula

Tire empirical formula of a compound represents the simplest whole-number ratio of the atoms of the various elements present in a molecule. For example, the empirical formula of benzene is CH. This indicates that carbon and hydrogen are present in benzene in a the ratio 1 :1, The empirical formula of ethanoic acid is CH2O which means that that a molecule of ethanoic acid contains carbon, hydrogen, and oxygen atoms in the ratio 1:2:1

The molecular formula of a compound, on the other hand, shows the actual number of atoms of the various elements present in one molecule of the compound. For example, the molecular formula of benzene is C6H6, which means that one molecule of benzene contains 6 atoms of carbon and 6 atoms of hydrogen. Similarly, the molecular formula of ethanoic acid is C2H4O2, which indicates that 1 molecule of ethanoic acid contains 2 atoms of carbon, 4 atoms of hydrogen, and 2 atoms of oxygen.

The molecular formula of a compound is a simple whole-number multiple of its empirical formula. This can be expressed mathematically as follows.

Molecular formula = n x empirical formula

∴ where n is an integer. The value of n can be determined from the relation

∴n = \(\frac{\text { molecular mass }}{\text { empirical formula mass }}\)

For example, the molecular mass of glucose (C6H12O6) is 180 and its empirical formula (CH2O) mass is 30.

∴ n = \(\frac{180}{30}\)

When the value of n = 1, the empirical formula and molecular formula are the same.

Determination of empirical formula: The empirical formula of a compound can be determined from the percentage of the different elements present in it and their atomic masses. Such calculations involve the following steps.

  1. Divide the percentage of each element by its atomic mass. This will tell you the relative number of moles of the elements present in a molecule of the compound.
  2. Divide the quotients obtained by the smallest in value to get a simple ratio of moles of various elements present.
  3. If the results obtained in step 2 are not whole numbers, raise the values to the nearest whole numbers or multiply all of them by a suitable integer to obtain whole numbers.
  4. Write the symbols of the elements side by side and insert the corresponding numerical values you have obtained in step 3 at the lower right-hand comer of each symbol. This is the empirical formula of the compound.

Determination of molecular formula: The determination of the molecular formula of a compound involves the following steps.

  1. Determine the empirical formula as described above.
  2. Find the empirical formula mass by adding the atomic masses of the atoms in the empirical formula.
  3. Divide the molecular mass of the compound (determined by a suitable method) by the empirical formula mass to obtain n.
  4. Multiply the empirical formula by n to obtain the molecular formula.

Example 1. The percentages of carbon and hydrogen in an organic compound are 92.5 and 7.5. Determine the molecular formula of the compound if its molecular mass is 78.
Solution:

Basic Chemistry Class 11 Chapter 1 Some Basic Concepts of Chemistry Percentages Of Carbon And Hydrogen In Organic Compound

Thus the empirical formula of the compound is CH.

Therefore, its empirical formula mass = 12×1 + 1×1 = 13

Given that its molecular mass is 78.

∴ n = \(\frac{\text { molecular mass }}{\text { empirical formula mass }}=\frac{78}{13}=6\)

Therefore, the molecular formula of the compound is 6 x (CH) = C2H6.

Example 2. Calculate the molecular formula of a compound on the basis of the following data.

  1. 4.24 mg of an organic compound yielded 8.45 mg of CO2 and 3.46 mg of H2O on combustion.
  2. The moleadar mass of the compound is 88 g.

Solution :

1. Calculate the percentage of carbon in the compound as follows.

44 mg of CO2 contains 12 mg of C.

Therefore, 8.45 mg of CO2 contains 12/44 x 8.45 mg of carbon.

∴ percentage of carbon =\(\frac{\text { weight of carbon }}{\text { weight of compound }}\) x 100

= \(\frac{12}{44}\) x \(\frac{8.45}{4.24}\) x 100 = 543.

2. Calculate the percentage of hydrogen in the compound as follows.

18 mg of H2O contains 2 mg of hydrogen.

Therefore, 3.46 mg of H2O contains 2/18 x 3.46 mg of hydrogen.

∴ percentage of hydrogen = \(=\frac{\text { weight of hydrogen }}{\text { weight of compound }}\) x 100

= \(\frac{2}{18}\) x \(\frac{3.46}{4.24}\) x 100 = 9.0

3. Percentage of oxygen = 100 – (54.3 + 9.0) = 36.7.

4. Determine the empirical formula as usual.

Basic Chemistry Class 11 Chapter 1 Some Basic Concepts of Chemistry Empirical Formula

Thus the empirical formula of the compound is C2H4O.

∴ empirical formula mass = 2 x 12 + 4 x 1 +16 = 44 g.

Given that the molecular mass of the compound is 88 g.

∴ n = \(\frac{88}{44}\)= 2.

∴ molecular formula = 2(C2H4O) = C4H8O2.

WBCHSE Class 11 Chemistry Notes For Periodic Properties Of Elements and Their Functions

Classification of Elements and Periodicity of Properties

Man has been fascinated by the chemical properties of the natural substances in his environment from almost the dawn of civilization. Initially, his curiosity was fuelled entirely by his need to improve his living conditions. He learned to harden clay by baking it at a high-temperature thousands of years ago.

  • Then, quite by accident, he discovered metals, which changed his life completely and led to the study of alchemy. Alchemy was not really a scientific study of metals.
  • It was based on this belief in the existence of a magical substance (Philosopher’s Stone) that could transform base metals into gold.

However, modern chemistry owes a lot to alchemists, who in their search for the Philosopher’s Stone, tested all the known substances and increased man’s knowledge of the properties of these substances.

Read and Learn More WBCHSE For Class11 Basic Chemistry Notes

  • Early scientists, who inherited the knowledge passed on by the alchemists, could afford to study the properties of the few known elements individually. But as knowledge grew, and more and more elements came to be known, this approach proved to be cumbersome.
  • Scientists started looking for a way to group together similar elements so that it would be possible to study their properties and the properties of their compounds more systematically.
  • After years of effort and several attempts at classifying all the known elements, they succeeded in coming up with an arrangement in which similar elements, or elements which have similar properties, are grouped together.
  • This arrangement, called the periodic table, has systematized the study of elements and their compounds. It has provided a framework for organizing the vast amount of information available on the chemical behavior of elements.

WBCHSE Class 11 Chemistry Notes For Periodic Properties Of Elements and Their Functions

History Of The Periodic Table

All the attempts made to classify elements, initially, were based on their atomic weights. This was because Dalton had proposed, earlier, that different elements have different types of atoms, characterized by their atomic weight.

Dobereiner’s triads: J W Dobereiner, a German chemist, classified similar elements into groups of three, or triads, such that the atomic weight of the central element was approximately the arithmetic mean of the atomic weights of the other two. He also pointed out that the properties of the middle element were in between those of the other two members.

These groups were called Dobereiner’s triads, for example, lithium (7), sodium (23) and potassium (39), calcium (40), strontium (88) and barium (137), and chlorine (35), bromine (80) and iodine (127). The major drawback of this classification was that it could be applied to only a limited number of elements.

Newlands’ law of octaves: An English chemist, John Newlands, came up with this arrangement in 1865, based on the observation that when the lighter elements are arranged in order of increasing atomic weight, every element is similar to the element eight placed from it in the list.

Hus was called Newlnnds’ law of octaves because it was like a musical scale in which the first note is the same as the eighth. Thus, the properties of lithium are similar to those of the element eight places from it, i.e., sodium.

Basic Chemistry Class 11 Chapter 3 Classification Of Elements And Periodicity Of Properties Prt Of Newland's Octaves

This arrangement too was discarded because it could not be applied to elements with atomic weights greater than 40 u. Also, the discovery of noble gases meant that the eighth element was no longer similar to the first. However, for his work Newlands was awarded the Davy Medal in 1887 by the Royal Society, London.

Lothar Meyer’s arrangement: Julius Lothar Meyer (1830-1895) was a German chemist who studied the atomic volumes, melting points, boiling points, and other physical properties of various elements.

  • He found that if a graph is plotted between the atomic volumes and atomic weights of the various elements, similar elements occupy similar positions on the curve.
  • The most strongly electropositive alkali metals (Li, Na, K, Rb, Cs) occupy the peaks on the curve, the less electropositive metals (Be, Mg, Ca, Sr, Ba) occupy descending positions on the curve and the most electronegative elements (F, Cl, Br, I) occupy ascending positions on the curve.

Basic Chemistry Class 11 Chapter 3 Classification Of Elements And Periodicity Of Properties Lothae Meyer's Atomic Volume Curve

On the basis of his observations, Meyer proposed in 1869 that the physical properties of the elements are a periodic function of their atomic weights.

Mendeleev’s periodic law: Around the same time, DI Mendeleev (1834-1907), a Russian chemist, was busy trying to arrange the elements in some order. He found that if they are arranged in increasing order of atomic weights, their chemical properties vary in a regular pattern. He proposed that the chemical properties of the elements are a periodic function of their atomic weights.

  • When Mendeleev came to know of Lothar Meyer’s conclusion, he combined his own proposal with Meyer’s, and the Jaw came to be known as Lothar Meyer-Mendeleev law or Mendeleev’s periodic law.
  • According to Mendeleev: The properties of the elements, as well as the formulae and properties of their compounds depend on the atomic weights of the elements. This means that when the elements are arranged in increasing order of atomic weight, – elements with similar properties occur at regular intervals in the list.

Mendeleev’s Periodic Table

Mendeleev’s arrangement of elements in increasing order of atomic weights is called Mendeleev’s periodic table. He grouped together similar elements and was insightful and courageous enough to leave blank spaces (on his table) for elements he believed would be discovered later.

The elements predicted by him were discovered later, and remarkably enough, were found to possess the properties he had predicted they would have. For example, gallium and germanium, called eka-aluminum and eka-silicon by Mendeleev, were discovered after he proposed his table and are very similar to aluminum and silicon respectively, as he had foreseen.

Basic Chemistry Class 11 Chapter 3 Classification Of Elements And Periodicity Of Properties Mendeleev's Predictions For The Elements Germanium And Gallium

Mendeleev’s periodic table had eight vertical columns (groups). Later, when the inert gases were discovered, a new group called the zero group was added to the table.

Basic Chemistry Class 11 Chapter 3 Classification Of Elements And Periodicity Of Properties Modified Form Of Mendeleev's Periodic Tabl

Vertical Columns There are nine vertical columns in Mendeleev’s periodic table called groups. These groups are designated 1, 2, 3, 4, 5, 6, 7, 8 and zero. With the exception of the 8th and zero groups, every group is divided into two subgroups A and B. Group 8 contains nine elements in three sets of three each, and group zero comprises the inert gases.

Horizontal rows The seven horizontal rows are called periods. The first period contains only two elements, the second and third periods contain eight elements each and are called short periods. The fourth, fifth, and sixth periods contain 18 elements each, while the seventh is incomplete and contains only three elements.

Merits of Mendeleev’s periodic table: Mendeleev’s periodic table was a giant step forward in the study of chemistry.

  • Systematic study His classification of the elements made the study of their properties simpler and more systematic. Placing elements which have similar chemical properties in the same group is helpful in that if one knows the properties of one element of a group, one can predict those of the rest of the members of the group.
  • Prediction of new elements While arranging the 56 elements known then, Mendeleev left blank spaces for elements which had not been discovered at that time. He predicted the properties of these unknown elements on the basis of their positions in his table. He was proved right later, when these elements were discovered.
  • Determination of atomic weights Mendeleev’s periodic table has helped correct errors made in the calculation of tire atomic weights of some elements. For example, beryllium was initially assigned an atomic weight of 13.5 on the basis of its equivalent weight (4.5) and valency, which had been wrongly calculated as 3.

This would have placed it between C and N in the periodic table. But there was no such space in Mendeleev’s table. Besides, its properties showed that its correct position should be between lithium (7) and boron (11), so beryllium was assigned a valency of 2 (atomic weight = 2 x 4.5 = 9) and placed where it belonged.

Drawbacks of Mendeleev’s periodic table: So great was Mendeleev’s contribution to the systematic study of elements that he is usually given credit for the periodic table as we know it today. Nonetheless, the table had the following drawbacks.

  • Position of isotopes Isotopes are atoms of the same elements that have different atomic weights. As Mendeleev’s. classification is based on atomic weight, isotopes would have to be placed in different positions. For example, the isotopes of hydrogen, \({ }_1^1 \mathrm{H},{ }_1^2 \mathrm{H},{ }_1^3 \mathrm{H}\) would occupy different positions.
  • Position of hydrogen Though hydrogen is placed in Group IA (alkali metals), it resembles the elements of both Group IA and Group 7A (halogens). Thus, the position of hydrogen in the periodic table is controversial.
  • Anomalous positioning of elements Some elements with higher atomic weight are placed before those with lower atomic weight. For example, argon (atomic weight = 40) is placed before potassium (atomic weight = 39).

Also, some dissimilar elements are grouped together, while some similar elements are placed in different groups. For example, alkali metals, such as Li, Na, and K (LA) are grouped together with the coinage metals, Cu, Ag, and Au (IB) though their properties are different. On the other hand, some chemically similar elements like Cu (LA) and Hg (2B) are placed in different groups.

Elements of Group 8 The nine elements of Group 8 have not been placed in a proper order. They have been arranged in three triads without any justification.

Modern Periodic Law

Mendeleev’s arrangement of elements, though rational and systematic, was empirical. Several decades later, after the electron had been discovered and the modem theory of atomic structure had been developed, H G J Moseley studied the spectral lines emitted by heavy elements in the X-ray region of the electromagnetic spectrum.

  • He found that the X-rays radiated by each element have a characteristic frequency that differs according to a regular pattern. The frequencies were different not because of the change in atomic weight but due to the charge on the nucleus (atomic number).
  • Thus, he proposed that the elements should be classified on the basis of atomic number (Z) and not atomic weight. He said that the properties of an element depend on its atomic number, rather than its periodic functions of their atomic numbers.
  • This means that if the elements are arranged in order of increasing atomic number, elements that have similar properties will appear at regular intervals. This is known as periodicity.

Cause of periodicity: It is not difficult to understand periodicity if one considers atomic structures and what happens to an atom during a chemical reaction. Neither the nucleus, nor the electrons of the inner shells of an atom participate in a chemical reaction.

  • Only the valence electrons participate in chemical reactions, so it is these electrons that govern the chemical properties of atoms. In that case, atoms which have the same kind of arrangement of electrons in their outermost shells should have similar chemical properties.
  • Now consider the electronic configurations of any group of elements (for example, the alkali metals or Group 1 in the periodic table).

Basic Chemistry Class 11 Chapter 3 Classification Of Elements And Periodicity Of Properties Cause Of Periodicity

  • They have the same number of valence-shell electrons (Group 1 f elements have one electron in their valence shell). This explains why they have similar properties.
  • The atomic number of an element is equal to the number of protons it contains, which is again equal to the number of electrons it has. And an atom’s chemical behaviour is governed by the number of electrons it has.
  • That is why it makes more sense to arrange the elements in order of increasing atomic number, rather than in order of increasing atomic weight.

Long Form Of Periodic Table

Several periodic tables have been proposed after it was realised that the atomic number, and not the atomic weight V of elements, should be the basis for the arrangement of elements. The most commonly used one is called the long form of the periodic table. It has eighteen columns and seven rows.

The rows are called periods and the columns contain elements with similar valence-shell configuration. These columns are called groups and elements belonging to the same group (obviously with similar properties) constitute a family, for example, the halogen family (Group 17). In this form of the periodic table, 14 elements each of the sixth and seventh periods (i.e., lanthanides and actinides) are placed separately at the bottom.

Electronic configurations of the elements and the periodic table: As a result of elucidation of the structure of the atom, it is now recognised that the periodic law is essentially the consequence of the periodic variation in electronic configurations.

The electronic configuration of an element indeed determines the physical and chemical properties of the element and its compounds. The electronic configuration of elements can be best studied in terms of variation in periods and groups of the periodic table.

Periods: You already know that an electron in an atom is characterised by a set of four quantum numbers. The principal quantum number (M) denotes the main energy level or shell.

Basic Chemistry Class 11 Chapter 3 Classification Of Elements And Periodicity Of Properties Long Form Of Periodic Table Of The Elements With Their Atomic Numbers And Atomic

Basic Chemistry Class 11 Chapter 3 Classification Of Elements And Periodicity Of Properties Number Of Elemenets In Different Periods

Each period begins with the filling up of a new energy shell and all the elements of a period have the same number of electron shells, or the same principal quantum number (n) of the outermost shell. In fact, the number of a period is the same as the principal quantum number of the valence shell of the elements it contains. The number of elements in a period is equal to the number of electrons required to fill the orbitals of that shell.

First period This corresponds to the filling up of the first energy shell (n = 1). This shell has only one orbital, the Is orbital, which can accommodate only two electrons, so the first period has only ttvo elements.

Second period This is associated with filling up of the second energy shell (n = 2). This shell has one 2s and three
2p orbitals, which can accommodate eight electrons. The second period, thus, has eight elements.

Third period This corresponds to the filling up of the third shell (n = 3). This shell has one 3s, three 3p and five 3d orbitals, but the 3d orbitals have higher energy levels than the 4s orbitals. Consequently, the 3d orbitals are filled after the 4s orbitals and the third period involves the filling up of only four orbitals. Thus this period contains only eight elements, from sodium to argon.

Fourth period This corresponds to the filling up of the fourth energy level (n = 4). It starts with the filling up of the 4s orbital after which the five 3d orbitals are filled and then the three 4p orbitals. The 4d and 4f orbitals can be filled only after the 5s orbital. Therefore, this period involves the filling up of nine orbitals and contains eighteen elements, from potassium to krypton.

Fifth period The fifth period is associated with the filling up of the fifth energy level (n = 5). After the 5s orbital the five 4d and three 5p orbitals are filled. Neither the 4f, nor the 5d orbitals can be filled until the 6s orbital is filled. Thus, this period too has only eighteen elements, from rubidium to xenon.

Sixth period This corresponds to the filling up of the sixth energy level (n = 6). Starting with the 6s orbital, it involves the filling up of sixteen orbitals (6s, 4f, 5d and 6p), so it contains thirty-tivo elements, from caesium to radon. The fourteen elements from cerium to lutetium correspond to the filling up of the seven 4f orbitals. These are separated from the main table and placed below it to save space. They constitute the first f-transition series and are called lanthanides.

Seventh period This corresponds to the filling up of the seventh energy level (n = 7). Like the sixth period this should also have contained thirty-two elements, corresponding to the filling up of the 7s, 5f, 6d and 7p orbitals, but it is still incomplete and contains only twenty-five elements. It includes most of the man-made radioactive elements. The fourteen elements corresponding to the filling of the 5f orbitals are called actinides and constitute the second f-transition series. These too have been placed below the main table to save space and to allow similar elements to come under the same group.

The first three periods are called short periods and the next three are called long periods.

Division of elements into blocks: The long form of the periodic table has four blocks—s, p, d and f. Each block is named after the atomic orbital which receives the last electron during the filling up of orbitals in order of their increasing energies, s Block elements in which the last electron enters the s orbital of their respective valence shells are called s-block elements.

This subshell has only one orbital which can accommodate only two electrons, so there are only two groups of elements in this block—Groups 1 and 2. The general electronic configuration of the outermost shell of these elements is ns1-2 where n represents the value of the principal quantum number of the valence shell. Group 1 comprises alkali metals, while Group 2 contains the alkaline earth metals. These elements have the following general characteristics.

  1. They are soft metals.
  2. They have low ionisation enthalpies and a re highly electropositive.
  3. They are very reactive and form ionic compounds.
  4. The metallic character and reactivity of the elements increase down the group.
  5. They exhibit oxidation states of +1 and +2 as they have only one or two electrons in their valence shell.
  6. They are good reducing agents because they lose electrons easily.

p block This block includes elements in which the last electron enters the p subshell of the valence shell, helium (1s2) being the exception. The general electronic configuration of the outermost shell of these elements is ns2np6.

This block contains the groups 13 to 18. Group 18 comprises noble gases with completely filled orbitals (ns2np6). Thus, each period ends in a noble gas with a closed-shell configuration. The other two important groups are chalcogens (Group 16) and halogens (Group 17).

The p-block elements have the following general characteristics.

  1. Most of them are nonmetals and the character increases as we move from left to right across a period.
  2. Most of them are highly electronegative.
  3. They exhibit variable oxidation states.
  4. They form ionic and covalent compounds.
  5. Most of them form acidic oxides.

The elements of the s and p blocks together are called representative elements.

  • Two exceptions are observed in this classification. The electronic configuration of helium is 1s2 so that it should belong to the s block, but it has been placed in the p block with Group 18 elements (noble gases). Its position is justified because it has a completely filled valence shell and shows the characteristic properties of a noble gas. The other exception is hydrogen. The electronic configuration of hydrogen is 1s1, hence it can be placed in Group 1 (s block).
  • However, like halogens it is one electron short of noble-gas configuration. This controversy regarding position of hydrogen in the periodic table has been discussed in detail. It is placed at the top of the alkali metals in Period 1 of the periodic table.

d block Elements in which the last electron enters the d subshell of the penultimate (second last) energy level are called d-block elements. These include elements of groups 3 to 12. Their general valence shell configuration is (n – 1) d1-10 ns1-2 where n represents the outermost energy level. This block contains three rows of ten elements each. The fourth row is incomplete.

Tire rows which are complete are called the first, second and third transition series. They correspond to the filling of the 3d, 4d and 5d orbitals respectively. Members of this block are called transition elements because they represent a change in character from reactive metals on one side to nonmetals on the other. These elements have the following general characteristics.

  1. They are hard metals with high melting points.
  2. They exhibit variable valencies and oxidation states.
  3. They form ionic as well as covalent compounds.
  4. They form coloured complexes.
  5. Most exhibit paramagnetism.
  6. Most possess catalytic properties.

f block This block contains elements in which the last electron enters the f from the outermost) shell. The general electronic configuration of f-block elements is \((n-2) \mathrm{f}^{1-14}(n-1) \mathrm{d}^{0-1} n \mathrm{~s}^2\), where n is the principal quantum number of the valence shell. This block comprises two series of elements placed below the main body of the periodic table. Elements of the first series are called lanthanides, while elements of the second series are called actinides. Together they are called the inner transition elements.

These elements have the following general characteristics.

  1. They show variable oxidation states.
  2. They are metals with high densities and high melting points.
  3. They form coloured compounds.
  4. Most elements of the actinide series are radioactive.

Elements up to uranium (Z = 92) arc found in nature, except technetium (Z = 43) and promethium (Z = 61), which arc produced from the disintegration of radioactive elements. Elements beyond uranium are called synthetic or transuranic as they are produced synthetically. Many of these have been made only in nanogram quantities or even less by nuclear reactions and their chemistry is not fully known.

  • In addition to the classification of elements into s, p, d and f blocks another classification is done, based on their properties. The elements can be broadly classified into metals and nonmetals.
  • Elements on the left-hand side of the periodic table are metals. More than seventy-five per cent of all known elements are metals. Elements at the top right-hand side of the periodic table are nonmetals.
  • Metals are generally solids at room temperature (except mercury) with high melting and boiling points. They are malleable and ductile, and good conductors of heat and electricity. Nonmetals are generally solids or gases at room temperature with low melting and boiling points. They are brittle and poor conductors of heat and electricity. The change from metallic to nonmetallic character is gradual. Some elements like germanium, arsenic, silicon and antimony show characteristics of both metals and nonmetals. These elements are called semimetals or metalloids.

Predicting period, group number and block of an element: If you have to predict the group, period and block of an element, proceed step by step.

  1. Write the electronic configuration of the element according to the Aufbau principle. Do not change the configuration by considering the relative stability of half-filled and completely filled orbitals. Do not rearrange in order of increasing values of quantum number.
  2. The period of the element is the same as the value of the principal quantum number of the valence shell. For example, consider aluminium, whose atomic number is 13. Its electronic configuration is 1s2 2s2 2p6 3s2 Sp1. The value of the principal quantum number of its valence shell is 3, so it belongs to the third period.
  3. The block of an element is predicted on the basis of the subshell which receives the last electron. Consider copper, whose electronic configuration should be 1s2 2s2 2p6 3s2 3p6 4s2 3d9.
  4. However, due to a slight difference in the energies of 4s and 3d subshells, and stability of the completely filled and half-filled orbitals, the outer electronic configuration of Cu is 3d10 4s1. The last electron is received by the d subshell, so it belongs to the d block.
  5. The group number is predicted on the basis of the number of electrons in the outermost or the penultimate shell.
  • For an s-block element, the group number is the same as the number of electrons in the valence shell. For example, the electronic configuration of lithium is 1s2 2s1. It belongs to Group 1 as there is one electron in its valence shell.
  • For d-block elements, the group number is obtained by adding 1 to the number of electrons in the d subshell of the penultimate shell. For example, the electronic configuration of copper is 1s2 2s2 2p6 3s2 3p6 4s1 3d10. Its group number is (10 +1) = 11.
  • For p-block elements, the group number is obtained by adding 12 to the number of electrons in the p subshell of the valence shell. For example, the electronic configuration of gallium is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p1. It belongs to Group 13.

Nomenclature of the elements with atomic number more than 100: Enrico Fermi and his co-workers in 1934 made attempts to prepare the elements beyond uranium, The new elements were prepared by the bombardment of uranium with slow neutrons. As of now, elements with numbers up to 112 and 114 have been discovered.

Basic Chemistry Class 11 Chapter 3 Classification Of Elements And Periodicity Of Properties IUPAC Nomenclature Of Elements With Atomic Number Above 100

  • Earlier the naming of new elements was done y discoverer(s) traditionally. In recent years, disputes have arisen over the original discoveries of some of the elements of atomic number 104 and above, when two scientists from two different countries have common criteria for discovering the same element.
  • To avoid such discrepancies IUPAC has recommended that until the c a m or a newly discovered element is proved and its official name announced, a nomenclature is to be followed of name these new elements.
  • The nomenclature of new elements and elements which are yet to be discovered is based on the Latin words for the atomic number of the elements. The names are derived using the numerical roots for 0 and numbers 1 to 9.
  • The roots are combined together in the sequence of digits which make up the atomic number and (ium) is added at the end. The notation for IUPAC nomenclature of elements is shown in Table 3.7. For example, the IUPAC name and symbol for the element with atomic number 122 may be written from the table. The roots for 1 and 2 are un and bi respectively. Hence the name is un (1) + bi (2) + bi (2) + ium or unbibium and the symbol assigned is Ubb.

Basic Chemistry Class 11 Chapter 3 Classification Of Elements And Periodicity Of Properties Nation For IUPAC Nomenclature Of Elements

Example: Write the IUPAC names and symbols for the elements with atomic numbers 121 and 140. Also, identify the group where these elements would be placed (once discovered) by giving their electronic configuration.
Solution:

The IUPAC names for the elements with atomic numbers 121 and 140 are Unbiunium (Ubu) and Unquadnilium (Uqn) respectively.

  • The electronic configuration of elements which are yet to be discovered can be written on the basis of the periodicity in filling of the valence shell orbitals of elements in the periodic table.
  • Thus, the electronic configuration of the element with atomic number 121 is [Uuo]7d1 8s2. Therefore, it will be placed in Group 3.
  • The electronic configuration of the element with atomic number 140 is [Uuo]6f14 7d6 8s2. Therefore, it will be placed in Group 8.

 

Periodic Trends In Chemical Properties

We have learnt about the periodicity shown by atomic properties like atomic size, ionisation enthalpy, electron gain enthalpy and electronegativity. The discussion involving trends in chemical properties can be quite elaborate.

Here we shall restrict ourselves to the periodicity of valency and anomalous properties of second-period elements only. We have already discussed the periodicity of valency earlier in the chapter.

Example: Refer to the periodic table and predict the formulae of compounds which might be formed by the following pairs k, of elements:

  1. Carbon and chlorine, and
  2. Boron and oxygen.

Solution:

  1. Carbon is a Group 14 element with a valency of 4, while chlorine belongs to Group 17 with a valency of 1. Therefore, the formula of the compound formed will be CCl4.
  2. Boron is a Group 13 element with a valency of 3, while oxygen is a member of Group 16 and possesses a valency of 2. Hence, the formula of the compound formed would be B2O3.

Anomalous properties of second-period elements: The representative elements of the second period—lithium, beryllium, boron, etc., show anomalous characteristics, as compared to the other members in the respective groups.

  • This is attributed to the high ionisation enthalpy, high electronegativity, small size and large charge-to-radius ratio of the first members of the said groups.
  • Further these elements showing anomalous characteristics have a close chemical similarity to their diagonal neighbours in the next group of the third period. This is a relationship within the periodic table and is referred to as diagonal relationship. You will study about the anomalous characteristics and similarity in properties of the s block elements due to the diagonal relationship.
  • The p-block elements of the second period have a tendency to enter into 7t bonding (or pπ-pπ multiple bonding) with themselves, as in C=C, C≡C and N=N, O=O. They also form π bonds with the other second-period elements. Such bonds exist in oxides of nitrogen, (N=O), cyanide and carboxyl groups (C=N, C=O).
  • However, the subsequent members of the same group do not form σ bonds with other elements. For example, in silicon dioxide, there are only a bonds.

Periodic Trends And Chemical Reactivity

By now you are quite aware of the periodic trends in certain fundamental properties of elements. The ionisation enthalpy decreases down a group and increases across a period. The size (atomic and ionic radii) increases from J top to bottom in a group, and decreases from left to right in a period.

  • The electronegativity and electron affinity both increase from left to right in a period and decrease from top to bottom in a group. All these properties are related to electronic configuration. But how does this affect the chemical reactivity of elements in the periodic table?
  • The elements at the extreme left of the periodic table are very reactive due to their low ionisation enthalpies. The elements at the extreme right are also very reactive due to their high electronegativities or electron affinities.
  • The elements in the middle of the periodic table are relatively unreactive. Thus, transition elements are only moderately reactive since their ionisation enthalpies are relatively large and electronegativities or electron affinities are relatively small.
  • Tire elements at the extreme left of the periodic table are very reactive due to their tendency to lose an electron and form cations. Similarly, the elements at the extreme right of the periodic table readily accept an electron to form anions, and are, thus, very reactive. The metallic character decreases and nonmetallic character increases on moving from left to right across the periodic table.

The chemical reactivity of an element can be best shown by its reaction with oxygen. Elements at the two extremes of a period form oxides easily. The elements on extreme left form basic oxides (like Na2O) while elements in the extreme right form acidic oxides (like Cl2O7).

The elements in the centre of the periodic table form amphoteric oxides (like Al2O3, and As2O3) or neutral oxides (like CO, NO, N2O). Amphoteric oxides behave as acidic oxides with bases and as basic oxides with acids. However, neutral oxides have no acidic or basic properties.

Example: Show that Na2O is a basic oxide and Cl2O7 is an acidic oxide.
Solution:

On reaction with water, Na2O forms a strong base (NaOH) and so it is a basic oxide.

⇒ \(\mathrm{Na}_2 \mathrm{O}+\mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{NaOH}\)

On reaction with water, Cl2O7 forms a strong acid (HClO7) so it is an acidic oxide.

⇒ \(\mathrm{Cl}_2 \mathrm{O}_7+7 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{HClO}_7\)

The acidic or basic nature of the oxides can be easily tested by the action of litmus paper on their aqueous solutions.

Example: Predict the formulae of the stable compounds that ivottld be formed by the combination of the following pair of elements.

  1. Lithium and oxygen
  2. Magnesium and nitrogen
  3. Aluminium and iodine
  4. Silicon and oxygen
  5. Phosphorus and fluorine
  6. Element with atomic number 71 (Lu) and fluorine

Solution:

  1. Li2O
  2. Mg3N2
  3. All3
  4. SiO2
  5. PF3 or PF5
  6. LuF3

 

WBCHSE Class 11 Chemistry Notes For Atomic Structure – Discovery of Subatomic Particles

Atomic Structure

The atom as pictured by Dalton, Berzelius, Cannizzaro, and Avogadro was an indestructible particle, which could neither be created nor destroyed. It was thought to be indivisible, and the smallest building block of matter. This concept of the atom, as you have read in the previous chapter, could not explain how or why atoms combine, or how exactly atoms of different elements differ from each other.

  • The explanation had to wait until the end of the nineteenth century when some path-breaking experiments proved that the atom is made of still smaller particles—electrons, protons, and neutrons. These discoveries led to a clearer understanding of the chemical behaviour of atoms.
  • They also led to the formulation of quantum mechanics, since Newtonian mechanics could not describe the motion of the electron. The laws of quantum mechanics helped to explain the stability of the atom in terms of the arrangement of electrons in the atom and the characteristic radiation emitted by atoms.

Faraday’s Contribution

The earliest indication that scientists had regarding the existence of a relation between matter and electricity came from the observation that static electricity is generated when substances like glass (or ebonite) are rubbed against silk (or fur). Michael Faraday, an English physicist, performed some experiments in the 1830s, which made this connection more evident.

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Discovery Of Subatomic Particles

Faraday showed that when electricity is passed through an electrolyte, the electrolyte undergoes chemical changes. The phenomenon is called electrolysis and the quantitative results obtained by Faraday are expressed in Faraday’s laws of electrolysis.

According to these laws,

  1. The weights of the elementary substances produced (liberated) during electrolysis are proportional to the quantity of charge (current x time) passed through the electrolyte and
  2. The number of moles of various substances liberated by a fixed quantity of charge bears a simple, whole-number ratio with each other.
  • These results showed clearly that there is a relation between atoms and electricity. They were also very similar to the laws of chemical combination. Only years later, a scientist called Stoney claimed that Faraday’s laws on the interaction between matter and electricity were like the laws of chemical combination, which dealt with the interaction between different materials.
  • He claimed further that since the laws of chemical combination led to the formulation of the atomic theory of matter, Faraday’s laws should imply the discrete nature of electricity. In other words, that electricity must be made of discrete particles and he called these particles of electricity electrons.

WBCHSE Class 11 Chemistry Notes For Atomic Structure – Discovery of Subatomic Particles

Discovery Of The Electron

The electron was discovered during experiments involving the passage of electric current through a gas. Under ordinary conditions, gases do not conduct electricity.

Discovery Of Subatomic Particles

However, if a gas is filled in a sealed tube fitted with electrodes and the pressure inside the lube is lowered to about 10-2 atm, the gas starts conducting electricity when a high voltage (5000-10,000 V) is applied across the electrodes. Such a tube is called a gas-discharge tube. The neon signs we are so familiar with are actually gas-discharge tubes filled with neon.

Basic Chemistry Class 11 Chapter 2 Atomic Structure Cathode Ray Tube

Discovery Of The Electron Cathode rays: When the pressure inside the discharge tube is 1 atm, no electric current flows through the tube even on applying high voltage. The pressure of the gas inside the tube can be decreased by pumping out the gas with the help of a vacuum pump. As the pressure is reduced to about 10-2 atm, a purple light is emitted if the gas being used is air. The whole tube appears purple.

Discovery Of Subatomic Particles Class 11 Notes

Basic Chemistry Class 11 Chapter 2 Atomic Structure Cathode Rays Produce A Shadow Of An Object Placed In Their Way

  • The colour of the light depends on the gas being used. If the pressure is reduced much further to about 10-4 atm, the purple light disappears and the part of the glass tube directly opposite to the cathode begins to glow with a greenish light.
  • A British physicist, Sir William Crookes, who conducted this experiment in 1879, showed that the green glow (fluorescence) is caused by the bombardment of invisible rays emitted by the cathode. An object placed inside the tube, between the cathode and the opposite end of the tube, casts a sharp shadow on the wall of the tube.
  • Besides, the glow can then be seen only around the shadow. This proved that the glow was due to the emission of rays from the cathode and that the rays moved in straight lines. These rays are called cathode rays.

Discovery Of The Electron Properties of cathode rays: A series of experiments conducted by several scientists ultimately proved that cathode rays are actually a stream of electrons. Becquerel’s discovery of radioactivity and subsequent experiments conducted by other scientists to study the phenomenon of radioactivity acted as secondary proof to establish that the particles emitted by the cathode are electrons.

Discovery Of Subatomic Particles Class 11 Notes

Basic Chemistry Class 11 Chapter 2 Atomic Structure Cathode Rays Cause A Light Paddle To Rotate

  • But first it had to be proved that the emission from the cathode consisted of a stream of material particles. This was done by placing a light paddle wheel in the path of the cathode rays. It was seen that the rays rotate the paddle wheel, which could happen only if they comprised particles with mass and energy.
  • A metal foil placed in the path of the cathode rays gets heated if the rays are focussed on it. This, too, shows that emission from the cathode consists of fast-moving particles. Of course, light rays could also cause the foil to get heated. In fact, later experiments did show that cathode rays consist of electrons and light rays, but we will not concern ourselves with that at the moment.

Discovery Of Subatomic Particles Class 11 Notes

Basic Chemistry Class 11 Chapter 2 Atomic Structure Cathode Rays Are Reflected Towards The Positive Plate

  • The next thing that needed to be investigated was whether the particles emitted by the cathode were charged. An electric field was applied to the beam of cathode rays as shown in Figure and it was seen that the rays get deflected towards the positive plate. This proved that the particles of the cathode rays are negatively charged.
  • When a magnetic field was applied to the beam of cathode rays, the rays were deflected towards the south pole, as negatively charged particles should.

Basic Chemistry Class 11 Chapter 2 Atomic Structure Cathode Rays Are Reflected By A Magnetic Field

Other experiments conducted to investigate the properties of cathode rays showed that they ionise the gas through which they pass (as charged particles should) and that they produce X-rays when they are made to fall on metals, such as tungsten.

Discovery Of The Electron Charge To Mass Ratio (e/m): Sir J J Thomson, a British physicist, observed the deflection of cathode rays under the influence of an electric field and a magnetic field applied simultaneously and at right angles to the electric field in a specially designed discharge tube.

Basic Chemistry Class 11 Chapter 2 Atomic Structure Cathode Rays Are Defleted By Electric And Magnetic Fields

  • He then adjusted the two fields so that the cathode rays would strike a fluorescent screen placed at the end of the tube opposite the cathode at the original position as did before the two fields were applied. From his observations, he calculated the charge (e) to mass (m) ratio of the electron.
  • It turned out to be the same (1.76 x 108 C g-1) irrespective of the nature of the cathode or the gas taken in the discharge tube. This proved that electrons, the negatively charged subatomic particles, are universal constituents of matter.

Discovery Of The Electron Charge : In 1909, Robert Millikan, an American physicist, measured the charge carried by an electron by doing an experiment which has come to be known as the Millikan oil-drop experiment. He allowed a spray of oil droplets (produced by an atomiser) to enter the apparatus through a small hole and fall between two charged plates.

  • He observed the motion of the droplets of oil as they fell under the influence of the force of gravity through a telescope. He then passed X-rays into the chamber to ionise the air inside. The electrons thus produced imparted a negative charge to the oil droplets which then came under the influence of the electric field between the charged plates.
  • He adjusted the electric field until it balanced the gravitational field. (In a balanced field a droplet would either be stationary or move with constant velocity in accordance with the first law of motion.) The magnitude of the electric field required to balance the two fields would be a measure of the charge on the droplet of oil. The charge on a droplet, according to Millikan’s observations, was always a whole-number multiple of 16 x 10-19 C. He, therefore, concluded that the charge on an electron is 16 x 10-19 C.

Discovery Of Mass Of The Electron: The mass of an electron can be calculated quite simply from the results of Thomson’s and Millikan’s experiments.

Basic Chemistry Class 11 Chapter 2 Atomic Structure Millikan Oil Drop Experiment

According to Thomson,

⇒ \(\frac{e}{m}=1.76 \times 10^8 \mathrm{Cg}^{-1}\)

According to Millikan, e =1.6 x 10-19 C.

∴ mass of electron (m) = \(\frac{e}{e / m}\)

= \(\frac{16 \times 10^{-19}}{1.76 \times 10^8}=9.1 \times 10^{-28} \mathrm{~g}\)

= \(9.1 \times 10^{-31} \mathrm{~kg}\)

Anode Rays Or Canal Rays

After the discovery of the electron began the search for its positive counterpart. Since the atom as a whole is electrically neutral, scientists began to look for positively charged subatomic particles. In 1886, Eugen Goldstein, a German physicist, performed an experiment that led to the discovery of the proton.

  • He used a perforated cathode instead of the normal one used in the gas discharge tube.
  • When he passed an electric discharge through the tube, he observed a glow on the wall of the tube behind the perforated cathode. This obviously meant that a stream of particles (or light) was moving towards the cathode, passing through it and causing the tube to glow.
  • The rays were named anode rays or canal rays. These rays traveled in a direction opposite to that of the cathode rays.

Investigations followed, as in the case of cathode rays, and the following conclusions were reached.

  1. Anode rays travel in straight lines.
  2. They consist of material particles.
  3. They are positively charged since they are deflected towards the negatively charged plate in an electric field.
  4. They are deflected by a magnetic field.
  5. Their charge-to-mass (e/m) ratio depends upon the gas used in the discharge tube.
  6. The mass of the positively charged particles in the anode rays is equal to the atomic mass of the gas used.

Origin Of Cathode And Anode Rays: When a high voltage is applied across the gas-discharge tube, the atoms or molecules of the gas get ionised, resulting in the formation of positively charged ions and negatively charged particles. The negatively charged particles, we know, are electrons.

⇒ \(\underset{substack{\text { a neutral } \\ \text { gaseous atom }}}{\mathrm{A}} \stackrel{\text { ionised }}{\longrightarrow} \mathrm{A}^{+}+\mathrm{e}^{-}\)

Under the influence of the high voltage, the electrons move at a high velocity towards the anode. In the process, they ionize more neutral gas atoms, or produce more electrons and positive ions, The negatively charged particles or electrons constitute cathode rays, while the positively charged ions constitute anode rays.

Basic Chemistry Class 11 Chapter 2 Atomic Structure Canal Rays

Obviously, the e/m ratio of the cathode rays is independent of the gas used because these rays comprise electrons. The e/m ratio of the anode rays, on the other hand, depends on the gas used because the mass of the ion depends on the mass of the original atom.

Cathode And Anode Rays Discovery Of Proton: Observations showed that the positive particles (ions) produced by hydrogen had the highest e/m ratio. In other words, hydrogen produced the lightest positive particle. This particle was called proton.

Its charge-to-mass ratio was found to be 9.58 x 104 C g-1. The charge on the proton is equal and opposite to that on the electron, i.e., 16 x 10-19 C and its mass is 167 x 10-24 g. The proton is about 1837 times heavier than the electron and its mass is almost the same as that of a hydrogen atom.

Radioactivity

As mentioned earlier, Becquerel’s investigations into the phenomenon of radioactivity further confirmed that the atom is divisible and is made up of charged particles. The spontaneous emission of radiation by some substances like uranium is called radioactivity. Elements that have this property are called radioactive elements.

  • When the radiation emitted by a radioactive substance is allowed to pass through a strong electric field or magnetic field, it splits into three parts which are deflected in different directions.
  • The rays deflected towards the negative plate are called α (alpha) rays. These consist of positively charged He2+ ions.
  • The rays deflected towards the positive plate are called β (beta) rays and consist of negatively charged β particles or electrons.
  • The undeflected rays are high-energy electromagnetic radiations called γ (gamma) rays. You will learn more about this phenomenon later.

Basic Chemistry Class 11 Chapter 2 Atomic Structure Effect Of Electric Field On The Radiations Emiited Form A Radioactive Element

Thomson’s Model Of The Atom

Once it was discovered that the atom consisted of negatively and positively charged particles (this was before Chadwick’s discovery of the neutron), scientists wished to know how these particles were arranged inside the atom.

In 1898 Thomson proposed a model of the atom in which the positive charge was supposed to be spread over a sphere of radius 10’8 cm. He assumed that electrons were embedded in this sphere, more or less evenly, as plums in a pudding.

Basic Chemistry Class 11 Chapter 2 Atomic Structure Thomson's Model Of The Atom

Consequently, Thomson’s model of the atom came to be known as the plum pudding model. Thomson’s model was a pioneering effort to explain the atomic structure and he received the Nobel prize in physics in 1906.

 

WBBSE Solutions For Class 6 Maths Chapter 11 Geometrical Concepts Related To Regular Solids

Class 6 Math Solutions WBBSE Chapter 11 Geometrical Concepts Related To Regular Solids Exercise 11

Question 1. Let’s place a tetrahedron and place the face of a hemisphere on an exercise book and mark the boundary with a pencil. Let’s observe the shapes and write these names.
Solution:

WBBSE Solutions For Class 6 Maths Chapter 11 Geometrical Concepts Related To Regular Solids Tetrahedron ANd Hemisphere

Question 2. Let me try and find a solid which has only one surface.
Solution:

WBBSE Solutions For Class 6 Maths Chapter 11 Geometrical Concepts Related To Regular Solids Sphere Curved Surface

Name of solid: Football (sphere)

Surface: Curved surface

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Question 3. Let me write the name of such a soild that has one curved surface and two plane surfaces.
Solution:

WBBSE Solutions For Class 6 Maths Chapter 11 Geometrical Concepts Related To Regular Solids Cylinder One Curved Surface And Two Plane Surfaces

Name of solid: Glass(cylinder)

Surface: One curved surface and two plane surfaces

Class 6 WBBSE Math Solutions

Question 4. Let me draw such a solid which has one curved and one plane surface.
Solution:

WBBSE Solutions For Class 6 Maths Chapter 11 Geometrical Concepts Related To Regular Solids Cone One Plane And One Curved Surface

Name of solid: Cone

Surface: One plane and one curved surface

Question 5. With 6 plane surfaces, let me make a solid figure and write my observations for the solid made.
Solution:

WBBSE Solutions For Class 6 Maths Chapter 11 Geometrical Concepts Related To Regular Solids Cuboid Six Plane Surfaces

Name of solid: Box (cuboid)

Surface: Six-plane surface

Class 6 WBBSE Math Solutions Chapter 11 Geometrical Concepts Related To Regular Solids Exercise 11.1

Question 1. Fill in the blank spaces given in the chart below

WBBSE Solutions For Class 6 Maths Chapter 11 Geometrical Concepts Related To Regular Solids Name Of The Solids
Solution:

WBBSE Solutions For Class 6 Maths Chapter 11 Geometrical Concepts Related To Regular Solids Name Of The Solids with Bases

Question 2. Let us identify the solids and write their names: Pyramid, Tetrahedron, Cube, Cuboid

WBBSE Solutions For Class 6 Maths Chapter 11 Chapter 11 Geometrical Concepts Related To Regular Solids Identify The Solids

Solution:

WBBSE Solutions For Class 6 Maths Chapter 11 Geometrical Concepts Related To Regular Solids Solids And Their Names

Question 3. Let us choose the right option

1. Which of these is not cuboid?

  1. Brick
  2. Dice
  3. Book
  4. Bottle

Answer: 4. Bottle

2. Which of these is of cylindrical shape?

  1. Eraser
  2. Duster
  3. Milk powder can
  4. Ball

Answer: 3. Milk powder can

WBBSE Math Solutions Class 6

3. A solid having one surface is

  1. Dice
  2. Pencil
  3. Exercise Book
  4. Ball

Answer: 4. Ball

4. Observing the figures find out which one will not make a cube

WBBSE Solutions For Class 6 Maths Chapter 11 Geometrical Concepts Related To Regular Solids Wihich One Will Not Make A Cone
Answer: 2.

WBBSE Solutions For Class 6 Maths Chapter 11 Geometrical Concepts Related To Regular Solids Cube And Tetrahedron And Cuboid

 

 

WBBSE Solutions For Class 6 Maths Chapter 10 Recurring Decimal Numbers

Class 6 Math Solutions WBBSE Chapter 10 Recurring Decimal Numbers Exercise 10

Question 1. Identify which is a Terminating and Non-terminating decimal number

⇒ \(\frac{1}{3}, \frac{7}{9}, \frac{1}{6}, \frac{7}{11}, \frac{11}{12}, \frac{15}{37}, \frac{2}{15}, \frac{49}{63}, \frac{11}{37}, \frac{12}{70}, \frac{1}{2}, \frac{9}{45}, 11 \frac{10}{12}, \frac{6}{13}\)

Solution:

WBBSE Solutions For Class 6 Maths Chapter 10 Recurring Decimal Numbers Non Terminating Decimal Number

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These are non-terminating decimals

WBBSE Solutions For Class 6 Maths Chapter 10 Recurring Decimal Numbers Non Terminating Decimal Number 1

These are non-terminating decimals

WBBSE Solutions For Class 6 Maths Chapter 10 Recurring Decimal Numbers Terminating Decimal Number 2

These are terminating decimals

Question 2. Identify which is Mixed recurring and pure recurring decimal number.

⇒ \(\frac{5}{6}, \frac{34}{510}, \frac{1}{15} \frac{52}{41}, \frac{15}{13}, \frac{4}{7}, \frac{6}{7}, \frac{8}{9}, \frac{5}{11}, \frac{7}{11}, \frac{3}{13}, \frac{4}{15}, \frac{13}{15}\)

Class 6 Math Solutions WBBSE

Solution:

WBBSE Solutions For Class 6 Maths Chapter 10 Recurring Decimal Numbers Mixed Recurring Decimal Number

These are mixed recurring decimals

WBBSE Solutions For Class 6 Maths Chapter 10 Recurring Decimal Numbers Pure Recurring Decimal Number 2

These are mixed recurring decimals

Class 6 Math Solutions WBBSE

WBBSE Solutions For Class 6 Maths Chapter 10 Recurring Decimal Numbers Pure Recurring Decimal Number 3

These are mixed recurring decimals

Question 3. Let’s directly convert the following pure recurring decimal numbers to corresponding fractions:

1. \(0 . \dot{3}\)
Solution:

\(0 . \dot{3}\) = \(\frac{3}{9}\)

2. \(0 . \dot{8}\) = ______
Solution:

= \(\frac{8}{9}\)

3. \(0 . \ddot{1} \dot{8}\) = _______
Solution:

= \(\frac{18}{99} = \frac{2}{11}\)

4. \(0 . \dot{2} \dot{7}\) = _______
Solution:

= \(\frac{27}{99}=\frac{3}{11}\)

5. \(0 . \dot{1} 6 \dot{2}\) _____
Solution:

= \(\frac{162}{999}=\frac{18}{111}\)

6. \(0 . \dot{2} 9 \dot{7}\) = ______
Solution:

= \(\frac{297}{999}=\frac{33}{111}\)

7. \(0 . \dot{5} 64 \dot{3}\) = ________
Solution:

= \(\frac{5643}{9999}=\frac{627}{1111}\)

Question 4. Let’s express the following pure recurring decimal numbers in fractions

⇒ \(0 . \dot{5}, 0 . \ddot{4} \dot{5}, 0 . \ddot{5} \dot{3}, 0 . \ddot{1}, \quad 0 . \dot{5} \dot{1}\)

Solution:

⇒ \(0 . \dot{5}=\frac{5}{9} ; \quad 0 . \dot{4} \dot{5}=\frac{45}{99}=\frac{5}{11} ; \quad 0 . \dot{5} \dot{3}=\frac{53}{99}\)

⇒ \(0 . \ddot{12}=\frac{12}{99}=\frac{4}{33} ; \quad 0 . \dot{5} 1 \dot{2}=\frac{512}{999}\)

Class 6 Maths Solutions WBBSE

Question 5. Convert the following mixed recurring decimal numbers into fractions:

1. \(0.2 \dot{7}\)
Solution:

= \(\frac{27-2}{90}\)=\(\frac{25}{90}\) = \(\frac{5}{18}\)

2. \(0.08 \ddot{1}\)
Solution:

= \(\frac{81}{990}=\frac{9}{110}\)

3. \(2.8 \dot{2}\)
Solution:

= \(\frac{282-28}{90}=\frac{254}{90}\)

= \(\frac{127}{45}\)

4. \(0.2 \ddot{7} \dot{2}\)
Solution:

= \(\frac{272-2}{990}=\frac{270}{990} =\frac{3}{11}\)

5. \(3.4 \ddot{3} \dot{2}\)
Solution:

= \(\frac{3432-34}{990}=\frac{3398}{990} =\frac{1699}{495}\) .

Class 6 Maths Solutions WBBSE Chapter 10 Recurring Decimal Numbers Exercise 10.1

Question 1. By actual division let us find if the quotients are terminating decimal numbers or recurring decimal numbers

1. A 7 m long ribbon is divided into 8 equal pieces, let’s find the length of each piece.
Solution:

7m long ribbon is divided into 8 equal pieces.

∴ Length of each piece = \(\frac{7}{8}\) m.

WBBSE Solutions For Class 6 Maths Chapter 10 Recurring Decimal Numbers It Is A Terminating

= 0.875 m (T erminating)

2. Let’s find the weight of each packet of sugar when 11 kg sugar is divided equally in 12 packets.
Solution:

11 kg sugar is divided in 12 packets.

∴ Each packet contains \(\frac{11}{12}\)kg.

WBBSE Solutions For Class 6 Maths Chapter 10 Recurring Decimal Numbers It Is A Recurring

= 0.9166 kg

= 0.916 kg (Recurring)

3. Let’s calculate the amount of water each bottle can hold when 12 litres water is poured into 7 equal-sized bottles.
Solution:

The amount of water in each bottle = \(\frac{12}{7}\) litre.

WBBSE Solutions For Class 6 Maths Chapter 10 Recurring Decimal Numbers Calculate the Amonut Of Water Each Bottle

= 1.7142857

= 1.71428 5 litre (Recurring)

WBBSE Math Solutions Class 6

4. If 15 trees are planted along the side of a 24 m long road including its two ends, let’s find the distance between two consecutive trees.
Solution:

The distance between two consecutive trees

= \(\frac{24}{15}\) m= \(\frac{8}{5}\) m = 1.6 m. (Terminating)

Question 2. Express the following fractions in decimal fractions and identify the terminating and recurring decimal numbers:
Solution:

WBBSE Solutions For Class 6 Maths Chapter 10 Recurring Decimal Numbers Decimnal Fractions Identifying The Terminating And Recurring Decimal Number 1

WBBSE Solutions For Class 6 Maths Chapter 10 Recurring Decimal Numbers Decimnal Fractions Identifying The Terminating And Recurring Decimal Number 1

WBBSE Solutions For Class 6 Maths Chapter 10 Recurring Decimal Numbers Decimnal Fractions Identifying The Terminating And Recurring Decimal Number 3

WBBSE Solutions For Class 6 Maths Chapter 10 Recurring Decimal Numbers Decimnal Fractions Identifying The Terminating And Recurring Decimal Number 4

WBBSE Solutions For Class 6 Maths Chapter 10 Recurring Decimal Numbers Decimnal Fractions Identifying The Terminating And Recurring Decimal Number 5

Question 3. Let’s identify the pure and mixed recurring decimal numbers and convert them to fractions:

WBBSE Math Solutions Class 6

1. \(0 . \dot{5} \dot{4}\)
Solution:

= \(\frac{54}{99}=\frac{6}{11}\)

2. \(0 . \ddot{3} \dot{9}\)
Solution:

= \(\frac{39}{99}=\frac{13}{33}\)

3. \(0.0 \dot{2} \dot{4}\)
Solution:

= \(\frac{24}{990}=\frac{4}{165}\)

4. \(0 . \dot{6} \dot{9}\)
Solution:

= \(\frac{69}{99}=\frac{23}{33}\)

5. \(0 . \dot{9} \dot{3}\)
Solution:

= \(\frac{93}{99}=\frac{31}{33}\)

6. \(0.0 \ddot{8} \ddot{1}\)
Solution:

= \(\frac{81}{990}=\frac{9}{110}\)

7. \(0.27 \dot{2}\)
Solution:

= \(\frac{272-27}{900}=\frac{245}{900}\)

= \(\frac{49}{180}\)

WBBSE Math Solutions Class 6

8. \(0 . \dot{5} 1 \dot{3}\)
Solution:

= \(\frac{513}{999}=\frac{19}{37}\)

9. \(0 . \dot{14} \dot{4}\)
Solution:

= \(\frac{144}{999}\)

10. \(3: 4 \ddot{3} \dot{2}\)
Solution:

= \(\frac{3432-34}{990}=\frac{3398}{990}\)

= \(\frac{1699}{495}=3 \frac{214}{495}\)

11. \(7.0 \ddot{2} \dot{8}\)
Solution:

= \(\frac{7028-70}{990}=\frac{6958}{990}\)

= \(\frac{3479}{495}=7 \frac{14}{495}\)

12. \(0 . \dot{3} 7 \dot{5}\)
Solution:

= \(\frac{375}{999}\)=\(\frac{125}{333}\)

13. \(0 . \dot{29} \dot{1}\)
Solution:

= \(\frac{291}{999}=\frac{97}{333}\)

14. \(3 . \dot{200}\)
Solution:

= \(\frac{3205-3}{999}\)=\(\frac{3202}{999}\)

15. 0.375
Solution:

= \(\frac{375}{999}=\frac{125}{333}\)

16. \(0 . \dot{29 \dot{1}}\)
Solution:

= \(\frac{291}{999}=\frac{97}{333}\)

WBBSE Math Solutions Class 6

17. \(3 . \dot{20} \dot{5}\)
Solution:

= \(\frac{3205-3}{999}=\frac{3202}{999}\)

= \(3 \frac{205}{999}\)

18. \(0 . \dot{0} 12 \dot{1}\)
Solution:

= \(\frac{121}{9999}=\frac{11}{909}\)

Pure: 1, 2, 4, 5, 8, 9, 12, 13, 14, 15

Mixed: 3, 6, 7, 10, 11

Question 4. Let’s arrange the following in ascending order:

1. \(0 . \dot{3}, 0 . \dot{1}, 0 . \dot{1}\)
Solution:

⇒ \(0 . \dot{3}=0.333\)

⇒ \(0 . \dot{16}=0.166\)

⇒ \(0 . \dot{1}=0.111\)

∴ In ascending order: 0.111 ; 0.166; 0.333

i.e., \(0 . \dot{1}, 0 . \dot{1}, 0 . \dot{3}\).

2. \(0 . \dot{6} \dot{3} ; \frac{3}{4}, \frac{5}{6}\)
Solution:

⇒ \(0 . \ddot{6} \dot{3}=0.6363\)

⇒ \(\frac{5}{6}=0.8333\)

⇒ \(\frac{3}{4}=0.7500\)

∴ In ascending order: 0.6363; 0.7500; 0.8333

i.e., \(0 . \dot{6} 3 ; \frac{5}{6}, \frac{3}{4}\)

3. \(0.5 \dot{3} ; \frac{2}{25} ; \frac{16}{75}\)

⇒ \(0.5 \dot{3}=0.533\)

⇒ \(\frac{2}{25}=0.80\)

⇒ \(\frac{16}{75}=0.213\)

∴ In ascending order: 0.080; 0.213,0.533

i.e., \(\frac{2}{25}, \frac{16}{75}, 0.5 \dot{3}\)

4. \(0.91 \dot{6}, \frac{1}{121} ; \frac{3}{44}\)

⇒ \(0.916=0.916000\)

⇒ \(\frac{1}{121}\)=0.008264

⇒ \(\frac{3}{44}=0.068181\)

∴ In ascending order: 0.008264; 0.068181; 0.916000

i.e., \(\frac{1}{121} ; \frac{3}{44} ; 0.916\)

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers

Class 6 Math Solutions WBBSE Chapter 12 HCF And LCM Of Three Numbers Exercise 12

Question 1. Let’s find the greatest number which will divide both 564 and 630 to keep the remainder 3 in both cases.
Solution:

564 – 3 = 561

and 630-3 = 627

∴ The greatest number is the H.C.F. of 561 and 627.

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers HCF Of 561 And 627

∴ H.C.F. = 3 x 11 =33

Question 2. Let’s find the greatest number which divides 78,182 and 195 and keeps number remainder.
Solution: 78; 182; 195

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers HCF Of 78, 182 And 195

∴ Required H.C.F = 13

Read and Learn More WBBSE Solutions For Class 6 Maths

Question 3. Mill bought few exercise books for Rs. 80.50 p. Her brother bought few more for Rs. 57.50 paise. Let’s find the maximum price of an exercise book and total number of exercise books bought.

Class 6 Math Solutions WBBSE
Solution:

Rs. 80.50 = 8050 p

Rs. 57.50 = 5750 p

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers HCF Of 8050 And 5750

∴ H.C.F. =  5x 10 x 23 = 1150

∴ Maximum price of an exercise book = 1150 p.

Total Amount = (8050 + 5750) p = 13800 p.

∴ Number of exercise books = 13800/1150 = 12

Class 6 WBBSE Math Solutions Chapter 12 HCF And LCM Of Three Numbers Exercise 12.1

Question 1. Four bells ring at an interval of 45 min, 1 hour, 1 hr 15 min, and 1 hr 30 m. Let’s find bells ringing together at 12, when will these ring together again? Also, let’s find out how many times the bells will separately ring during these hours.
Solution:

Class 6 Math Solutions WBBSE

45 min = 45 mins

1 hour = 60 mins

1 hr 15 mins = 75 mins

1 hr 30 mins = 90 mins

L.C.M. of 45,60,75, 90

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers LCM Of 45, 60, 75 And 90

∴ Required L.C.M =2x3x5x3x2x5 = 900 mins = 15hrs.

The bells ring together at 12 room, then again the bells will ring together at 12 noon + 15 hrs. = 3 A.M.

Number of times 1st bell rings separately = 900 ÷ 45 = 20

Number of times 2nd bell rings separately = 900 ÷ 60 = 15

Number of times 3rd bell rings separately = 900 ÷ 75 = 12

Number of times 3rd bell rings separately = 900 ÷ 90 = 10

Class 6 WBBSE Math Solutions

Question 2. Let’s work to find the least number of four digits which will be divisible by 12,15, 20, and 35.
Solution:

L.C.M. of 12,15, 20 and 35

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers LCM Of 12, 15, 20 And 35

= 2x2x3x5x7 = 420

Least number of 4 digits, divisible by 12,15, 20, and 35 is = 420 x 3 = 1260.

Question 3. Let’s work out to find the greatest number of five digits which when divided by 16, 24, 30 and 36 will leave a remainder 10 in each case.
Solution:

L.C.M. of 16, 24, 30 and 36

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers LCM Of 16, 24 30, And 36

= 2x2x3x2x5x3x1 = 720

Greatest number of five digits = 99999

Class 6 WBBSE Math Solutions

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers Greatest Number Of 5 Digit Is 99999

∴ The required number = (99999 – 639) + 10

= 99360 + 10 = 99370.

WBBSE Math Solutions Class 6 Chapter 12 HCF And LCM Of Three Numbers Exercise 12.2

Question 1. There are three small tanks of capacity 35 liters, 56 liters, and 84 liters. Let’s find what will be the biggest capacity of a container that will measure the oil of the 3 tanks in exact whole numbers.
Solution:

To find the biggest capacity of a container we have to find the H.C.F. of 35 litres, 56 litres, and 84 litres

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers Division Of 35 And 7

∴ The biggest capacity of the container = 7 litres.

Question 2. The length and breadth of our school hall are 2000 cm and 1600 cm respectively. Let s find the length of the longest tape which can measure both length and breadth in exact whole numbers.
Solution:

WBBSE Math Solutions Class 6

To find the longest tape to measure 2000 cm and 1600 cm we have to find the H.C.F. of 2000 and 1600 cm.

2000 cm = 20 m.

1600 cm = 16 m.

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers Division Of 16

∴ Length of the tape = 4 m.

Question 3. There is a stock of 1071 dhotis, 595 sarees, and 357 dresses. Let us calculate the maximum number of families among which these can be distributed equally. How many of these things will each family receive?
Solution:

Number of dhotis = 1071

Number of sarees = 595

Number of dresses = 357

WBBSE Math Solutions Class 6

Now find H.C.F. of 1071,595,357 357,595, 357

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers HCF Of 1071, 595, 357

Number of Dhotis = \(\frac{1071}{119}=9\)

Number of Sarees = \(\frac{595}{119}=5\)

Number of Dresses = \(\frac{357}{119}=3\)

∴ Maximum number of families = 119.

Question 4. The perimeter of the front wheel of an engine is 1 m 4 dcm and the perimeter of its hind wheel is two and a half times more than the front wheel. Let’s find the least distance covered by the wheels when they will simultaneously take an exact number of complete revolutions.
Solution:

The perimeter of the front wheel = 1 m 4 dcm = 14 dcm

and the perimeter of its hind wheel = 14 x 2\(\frac{1}{2}\)

= 14 x \(\frac{5}{2}\) dcm.

= 35 dcm.

Now, H.C.F. of 14 dcm and 35 dcm = 7 dcm.

∴ Least distance covered by the wheels = 7 dcm.

WBBSE Class 6 Maths Solutions

Question 5. Let’s find the H.C.F. of the following numbers,

  1. 24, 36, 54
  2. 24, 30, 40, 48
  3. 296, 703, 814
  4. 160,165, 305
  5. 165,264,286
  6. 906,1510,1057

Solution:

1. 24, 30, 40, 48
Solution:

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers HCF Of 24, 36 And 54

∴ H.C.F. = 2×3 = 6

2. 24, 30, 40, 48
Solution:

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers HCF Of 24, 30, 40 And 48

H.C.F. of 24, 30,40, 48

24 = 2 x 2 x 2 x 3.

30 = 2 x 3 x 5

40 = 2 x 2 x 2 x 5

48 = 2 x 2 x 2 x 2 x 3

∴ H.C.F. = 2

WBBSE Class 6 Maths Solutions

3. 296,703,814
Solution:

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers HCF Of 296, 703 And 814

H.C.F. of 296, 703, 814

∴ Required H.C.F. = 37

4. 160,165, 305
Solution:

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers HCF Of 160, 165 And 305

H.C.F. of 160, 165, 305

∴ H.C.F. = 5

WBBSE Class 6 Maths Solutions

5. 165, 264, 286
Solution:

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers HCF Of 2165, 264 And 286

H.C.F. of 165, 264, 286

∴ H.C.F. = 11

6. 906, 1510, 1057
Solution:

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers HCF Of 906, 1510 And 1057

H.C.F. of 906, 1510, 1057

∴ Required H.C.F. = 151

Question 6. Let’s find the greatest number which divides 306,810 and 2214 and keeps the number remainder.
Solution:

Class 6 Math WBBSE Solutions

To find the required greatest number we have to find the H.C.F. of 306, 810, and 2214.

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers HCF Of 306, 810 And 2214

∴ H.C.F. = 18.

∴ The required number = 18.

Question 7. The traffic signal lights at three different crossings of a road change at every 16 seconds, 28 sec, and 40 sec respectively. If the signal lights changed together at 8 a.m in the morning, let’s find when the three signal lights will change together again.
Solution:

L.C.M. of 16 sec, 28 sec, and 40 sec

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers LCM Of 16, 28 And 40

∴ L.C.M = 2x2x2x2x7x5 = 560 sec = 9 mins 20 sec.

The three signal lights will change together 9 min 20 sec after 8 am, i.e., 8 hr 9 min 20 sec.

Question 8. There are three sticks in our house of lengths 45 cm, 50 cm, and 75 cm. Let’s find the least length of a tape which can be completely measured by each stick.
Solution:

To find the required length of the tape we have to find the LC.M. of 45,50, and 75.

Class 6 Math WBBSE Solutions

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers LCM Of 45, 50 And 75

L.C.M. = 5x5x 3x3x2 = 450

∴ The least length of the tape = 450 cm.

Question 9. Let’s find the least number which is divisible by 15,20,24 and 32.
Solution:

The least number is the L.C.M. of 15, 20, 24, and 32.

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers LCM Of 15, 20, 24 And 32

∴ L.C.M. = 2x2x2x3x5x4 = 480

Question 10. Let’s find the L.C.M. of the following:

  1. 36, 60, 72
  2. 24, 36, 45, 60
  3. 105,119,289
  4. 144,180, 348
  5. 110,165, 330
  6. 204, 408, 306

Solution:

Class 6 Math WBBSE Solutions

1. 36, 60,72
Solution:

L.C.M. of 36, 60, 72

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers LCM Of 36, 60 And 72

∴ L.C.M. = 2x2x3 x 3x5x2 = 360

2. 24, 36, 45, 60
Solution:

L.C.M. of 24, 36, 45, 60

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers LCM Of 24, 36, 45 And 60

∴ L.C.M. = 2x2x3x3x5x2 = 360

3. 105,119,289
Solution:

Class 6 Math WBBSE Solutions

L.C.M. of 105,119, 289

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers LCM Of 105, 119, 289

∴ L.C.M. = 7x 17 15 x 17 = 30345

4. 144,180, 348
Solution:

L.C.M of 144,180, 348

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers LCM Of 144, 180 And 348

∴ L.C.M. = 2x2x3x3x4x5x29 = 20880

5. 110, 165, 330
Solution:

L.C.M Of 110, 165, 330

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers LCM Of 144, 180 And 348

L.C.M = 2x2x3x3x4x5x29 = 20880

6. 204, 408, 306
Solution:

WBBSE Class 6 Maths Solutions

L.C.M of 204, 408, 306

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers LCM Of 204, 408 And 306

L.C.M = 2x2x3x17x2x3 = 1224

Question 11. Let’s find the H.C.F. and the L.C.M. of the following.

1. 6 Rs. 50 paise, 5 Rs. 20 paise and 7 Rs. 80 paise.
Solution:

Rs. 6, 50p = 650p

Rs. 5, 20p = 520p

Rs. 7, 80p = 780p

H.C.F.

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers HCF Of 650 And 780

H.C.F. = 130p = Rs. 1 30p

L.C.M.

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers LCM Of 520, 650 And 780

L.C.M. = 2x2x3x5x10x13 = 7800p = Rs. 78.

2. 2 m 28 cm, 3 m 42 cm, 4 m 56 cm.
Solution:

WBBSE Class 6 Maths Solutions

2 m 28 cm. = 228 cm.

3 m 42 cm = 342 cm.

56 cm = 456 cm.

H.C.F

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers HCF Of 342 And 456

H.C.M. = 114 cm

= 1 m 14 cm

L.C.M

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers LCM Of 228, 342 And 456

L.C.M = 2 x 2 x 2 x 3 x 3 x 1 9 = 1 368 cm

= 13 m 68 cm

3. 3 I 600 ml, 4 I 800 ml, 6 I.
Solution:

3 litre 600 ml = 3600 ml

4 litre 800 ml = 4800 ml

6 litre = 6000 ml

WBBSE Class 6 Maths Solutions

H.C.F

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers HCF Of 4800 And 6000

H.C.F. = 1200 ml.

= 1 litre 200 ml

L.C.M

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers LCM Of 3600, 4800 And 6000

L.C.M. = 2 x 2 x 3 x 3 x 3 x 4 x 5 x 100 = 72000 ml.

= 72 litre,

4. 6 hours 4 min 30 sec, 2 hours 42 min.
Solution:

6 hr, 4 min 30 sec – (6 x 3600 + 4 x 60 + 30) secting

= 21600 + 240 + 30 sec = 21870 sec

2 hr 42 min = (2 x 60 + 42) = 162 min

= 162 x 60 sec

WBBSE Class 6 Maths Solutions

H.C.F

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers HCF Of 9720 And 21870

H.C.F.= 9720 sec

L.C.M

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers LCM Of 9720, 21870

L.C.M = 9 x 9 x 3 x 4 x 9 x 10

= 87480 sec

= 1458 min

= 24 hr 18 min

WBBSE Class 6 Maths Solutions

Question 12. From the pairs of numbers given below, let’s find if the product of two numbers is equal to the product of their H.C.F and L.C.M

1. 87,145
Solution:

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers HCF And LCM Is 87 And 145

147 = 5×29

L.C.M = 3 x 5 x 29 = 435

∴ Product of the two numbers = 87 x 145 = 12615

Product of their H.C.F. x L.C.M. = 29 x 435 = 12615

∴ Product of the two numbers = Product of their H.C.F. x L.C.M.

2. 60, 75
Solution:

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers HCF And LCM Is 60 And 75

60 = 2x2x3x5

75 = 3 x 5 x 5

H.C.F = 3×5 = 15

L.C.M = 2x2x3x5x5 = 300

Product of the two numbers = 60 x 75 = 4500

Product of their H.C.F. x L.C.M. = 15 x 300 = 4500

∴ Product of the two numbers = Product of their H.C.F. x L.C.M.

3. 42, 63
Solution:

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers HCF And LCM Is 42 And 63

42 = 2 x 3 x 7

63 = 3x3x7

∴ H.C.F. = 3×7 = 21

L.C.M. = 2x3x3x7 = 126

Product of the two numbers = 42 x 63 = 2646

Class 6 WBBSE Math Solutions

Product of their H.C.F. and L.C.M. = 21 x 126 = 2646

Product of the two numbers = Product of their H.C.F. and L.C.M.

4. 186, 403
Solution:

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers HCF And LCM Is 186 And 403

186 = 2x3x31

403= 13×31

H.C.F. = 31

L.C.M. = 2x3x13x31 =2418

Product of the two numbers = 186 x 403 = 74958

Product of their H.C.F. and L.C.M. = 74958

∴ Product of the two numbers = Product of their H.C.F. and L.C.M.

Question 13. The L.C.M. and H.C.F. of the two numbers are 2175 and 145 respectively. If one number is 725, let’s find the other number.
Solution:

L.C.M. = 2175

H.C.F. = 145

∴ Product of the two numbers = Product of L.C.M. and H.C.F. = 2175 x 145

One number = 725

∴ Other number = \(\frac{2175 \times 145}{725}=435\)

Question 14. Let’s find the H.C.F. of 145 and 232. Using H.C.F. let us find L.C.M.
Solution:

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers HCF Of 145 And 232

145 = 5×29

232 = 2 x 2 x 2 x 29

∴ H.C.F. = 29

∴ H.C.F. x L.C.M. = Product of two numbers

∴ 29 x L.C.M. = 145×232

∴ L.C.M = \(\frac{145 \times 232}{29}=1160\)

∴ L.C.M. = 1160

Class 6 WBBSE Math Solutions

Question 15. Let’s find the L.C.M of 144 and 384. Using L.C.M let’s find their H.C.F.
Solution:

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers HCF Of 144 And 384

144 = 2x2x2x2x3x3

384 = 2x2x2x2x2x2x2x3

∴ L.C.M = 2x2x2x2x2x2x2x3x3 = 1152

H.C.F x L.C.M = Product of two numbers

∴ H.C.F x 1152 = 144×384

HCF = \(\frac{144 \times 384}{1152}=48\) = 48

∴ H.C.F = 48

Question 16. Find the least number that must be subtracted from 5834, so that the result is divisible by 20, 28, 32, and 35.
Solution:

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers LCM Of 20, 28, 32 And 35

L.C.M. = 2x2x5x7x8 = 1120

∴ The required least number = 234

Class 6 WBBSE Math Solutions

Question 17. Let’s find the greatest number which divides 2300 and 3500 to leave the remaining 32 and 56 respectively.
Solution:

The required number is the H.C.F. of (2300 – 32) = 2268 and (3500 – 56) = 3444

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers HCF Of 3444

H.C.F. = 84

∴ Required greatest number = 84

Question 18. Let’s find the greatest number that divides 650, 775 and 1250 to keep equal remainder in all cases.
Solution:

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers HCF Of 775

H.C.F. = 25

∴ Required greatest number = 25

Question 19. The sum of two numbers is 384 and their H.C.F. is 48, what may be the 2 possible numbers?
Solution:

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers HCF Of 384

8 = 1+7……..(1)

= 2 + 6 …….(2)

= 3 + 5 …….(3)

= 4 + 4 …….(4)

In 1st case, the numbers are (48 x 1) and (48 x 7) = 48, 336.

In 2nd case, the numbers are (48 x 2) and (48 x 6) = 96, 288.

But it is not possible as the H.C.F of 96 and 288 is 96.

In 3rd case, the numbers are (48 x 3) and (48 x 5) = 144, 240.

In 4th case, the numbers are (48 x 4) and (48 x 4) = 192, 192.

But it is not possible as the H.C.F of 192 and 192 is 192.

.-. Required numbers are (48, 336) and (144,240).

Question 20. The H.C.F. and L.C.M. of two numbers are 12 and 720. Let’s try to find how many pairs are possible and what may be those numbers.
Solution:

The factors of 12 are 1,2, 3, 4, 6, and 12.

Case 1 12×1= 12; 720-1 =720

H.C.F. =12, L.C.M. = 720

∴ (12, 720) is possible.

Case 2 12×2 = 24; 720 + 2 = 360

H.C.F. = 24, L.C.M. = 360

∴ (24,360) is not possible.

Case 3 12×3 = 36; 720 + 3 = 240

H.C.F. = 12, L.C.M. = 720,

∴ (36, 240) is possible.

Case 4 12 x 4 = 48; 720 + 4 = 180

∴ (48, 180) is possible.

H.C.F. = 48, L.C.M. = 720

Case 5 12×6 = 72; 720 + 6=120

H.C.F. =24, L.C.F. = 360

∴ (72,120) is not possible.

Case 6 12×12= 14; 720 + 2 = 60

H.C.F. = 12 L.C.M. = 60

∴ (60, 144) is possible.

∴ 4 pairs are (12, 720); (36,240); (48,180), and (60, 144)

Question 21. Let us find the least number from which if 4000 is subtracted, the result will be divisible by 7,11, and 13.
Solution:

L.C.M. of 7, 11 and 13 = 7×11 x 13 = 1001

∴ Required number = 4000 + 1001 = 5001

Question 22. Let’s find two pairs of numbers between 50 and 100, whose H.C.F. is 16.
Solution:

16 x 3 = 48

16×4 = 64

16×5 = 80

16×6 = 96

16×7 = 112

As the numbers are in between 50 and 100.

∴ The numbers are, (64, 80) and (80, 96).

Question 23. Let us find a number that is divisible by 28, 33, 42, and 77 and t nearest to 98765.
Solution:

L.C.M. of 28, 33, 42, and 77

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers LCF Of 28, 33, 42 And 77

L.C.M. = 2 x 3 x 7 x 11 x 2 = 924

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers HCF Of 98765

∴ The required number = 107 x 924 = 98868.

Question 24. Let us find the least number divisible by 13 such that when the number is divided by 8,12,16 and 20, it leaves 1 as the remainder of all the cases.
Solution:

L.C.M of 8, 12, 16 and 20

WBBSE Solutions For Class 6 Maths Chapter 12 HCF And LCM Of Three Numbers LCM Of 18, 12, 16 And 20

If it leaves 1 as the remainder the number is 240 + 1 = 241 which is not possible as 241 is not divisible by 13.

Again, the L.C.M is 240 x 2 = 480 and 480 + 1 = 481, it is divisible by 13.

∴ The required number = 481.

 

WBBSE Solutions For Class 6 Maths Chapter 9 Percentage

Class 6 Math Solutions WBBSE Chapter 9 Percentage Exercise 9

Question 1. Let’s convert the following percentages into proper fractions:

1. 10%
Solution:

10% = \(\frac{10}{100}=\frac{1}{10}\)

2. 70%
Solution:

70% = \(\frac{70}{100}=\frac{7}{10}\)

3. 15%
Solution:

15% = \(\frac{15}{100}=\frac{3}{20}\)

4. 257 %
Solution:

257 % = \(\frac{257}{100}\)

5. \(33 \frac{1}{3} \%\)
Solution:

= \(\frac{100}{3} \times \frac{1}{100}=\frac{1}{3}\)

Question 2. Let’s convert the following percentages into proper fractions:

Read and Learn More WBBSE Solutions For Class 6 Maths

1. 61%
Solution:

61% = \(\frac{61}{100}=0.61\)

2. 3%
Solution:

3% = \(\frac{3}{100}=0.03\)

Class 6 Math Solutions WBBSE

3. 105%
Solution:

105% = \(\frac{105}{100}=1.05\)

4. 1.26%
Solution:

1.26% = \(\frac{126}{100 \times 100}=0.0126\)

5. 0.07%
Solution: 

= \(\frac{7}{100} \times \frac{1}{100}\)

0.07% = \(\frac{7}{10000}=0.0007\)

Question 3. Let us express the following in percentages and arrange them in ascending order

1. \(\frac{1}{2}, \frac{1}{4}, \frac{3}{5}\)
Solution:

⇒ \(\frac{1}{2}=\frac{1}{2} \times 100 \%=50 \%\)

⇒ \(\frac{1}{4}=\frac{1}{4} \times 100 \%=25 \%\)

⇒ \(\frac{3}{5}=\frac{3}{5} \times 100 \%=60 \%\)

Ascending order: 25%; 50% ; 60%

∴ \(\frac{1}{4}, \frac{1}{2}, \frac{3}{5}\)

2. \(\frac{2}{5} ; \frac{13}{25} ; \frac{7}{10}\)
Solution:

⇒ \(\frac{2}{5}=\frac{2}{5} \times 100 \%\) = 40%

⇒ \(\frac{13}{25}=\frac{13}{25} \times 100 \%\) = 52%

⇒ \(\frac{7}{10}=\frac{7}{10} \times 100 \%\)= 70

Ascending order: 40% ; 52% ; 70%

Class 6 Math Solutions WBBSE

i.e., \(\frac{2}{5} ; \frac{13}{25} ; \frac{7}{10}\)

3. \(1 \frac{2}{5}, 1 \frac{1}{2}, 1 \frac{9}{10}\)
Solution:

\(1 \frac{2}{5}\)

= \(\frac{7}{5}=\frac{7}{5} \times 100 \%\) =140%

⇒ 1 \(\frac{1}{2}\)

= \(\frac{3}{2}=\frac{3}{2} \times 100 \%\) =150%

⇒ 1 \(\frac{9}{10}\)

= \(\frac{19}{10}=\frac{19}{10} \times 100 \%\) =190%

Ascending order: 140% ; 150% ; 190%

i.e., \(\frac{7}{5} ; \frac{3}{2} ; \frac{19}{10}\)

i.e., \(1 \frac{2}{5}, 1 \frac{1}{2}, 1 \frac{9}{10}\)

Class 6 Math Solutions WBBSE

4. 0.02,0.15,0.6
Solution:

⇒ 0.02 = \(\frac{2}{100} \times 100 \%=2 \%\)

⇒ 0.15 = \(\frac{15}{100} \times 100 \%=15 \%\)

⇒ 0.6 = \(\frac{6}{10} \times 100 \%=60 \%\)

Ascending order: 2%; 15%; 60%

i.e., 0.02; 0.15; 0.6

Question 4. 1. If 3 out of 5 candies were taken, let’s find the percentage of candies taken.
Solution:

3 out of 5 candies are taken.

∴ Percentage of candies taken = \(\frac{3}{5}\) x 20% = 60%.

2. If 6 out of 24 berries are rotten. Let’s find the percentage of berries that have rotten.
Solution:

6 out of 24 berries are rotten.

∴ Percentage of rotten berries = \(\frac{6}{24}\) x 100% =25%.

WBBSE Math Solutions Class 6

3. Today 7 students of our class are absent. There are 35 students in our class. Let us find the percentage of students present today in our class.
Solution:

7 students out of 35 students are absent.

∴ 35 -7 = 28 students are present.

∴ Percentage of present students = \(\frac{28}{35}\) x100% = 80%.

4. A bamboo is of length 55 m. 11 m of it is under muddy water. Let’s find what percentage of the bamboo is above the muddy water.
Solution:

Total length of bamboo = 55 m.

Under muddy water = 11 m.

Length of bamboo above muddy water = (55-11)m = 44m.

Percentage of the bamboo above the muddy water

= \(\frac{44}{55}\) 100% = 80 %

WBBSE Math Solutions Class 6

Question 5. There are 2100 story books in our local library. If 30% more storybooks are bought, what will be the total number of storybooks in the library now also find out the additional number of storybooks bought.
Solution:

Total number of story books in our library = 2100.

30% of 2100 story books = \(\frac{30}{100}\) x 2100 = 630.

∴ Total number of books now = 2100 + 630 = 2730.

Question 6. It’s raining heavily since morning. Hence only 20% of students were present in Alam’s school. Let’s find how many students were actually present in school if the total number of students in Alam’s school is 1230.
Solution:

Total number of students in Alam’s school = 1230.

Number of students present today = 20% of 1230

= \(\frac{20}{100}\) X 1230 = 246.

Question 7. 10. Today I shall prepare orange juice, ‘sherbet’ (drink) myself. For preparing 300 ml of ‘sherbet’ I added 18% orange juice to it. Let’s find how many milliliters of orange juice I added to the ‘sherbet’.
Solution:

Quantity of sherbet = 300 ml.

Orange juice added = 18% of 300ml

= \(\frac{18}{100}\) x 300 ml = 54 ml.

WBBSE Math Solutions Class 6

Question 8. Shobhan used organic fertilizer this year in his land This increased the production of paddy by 25% than last year. If last year’s production of paddy is 12 quintals, let’s find the production of oaddv this year.
Solution:

Last year production of paddy = 12 Quintals.

This year production increased by 25%

= \(\frac{25}{100}\) x 12 Quintals = 3 Quintals.

∴ Total production = (12 + 3) Quintals = 15 Quintals.

Question 9. In Rasulpur village, the population has increased by 12% than the previous year. Till last year, the population of the village was 775. Let’s find the present population of the village.
Answer:

Till last year, the population of the village was 775.

The population increased by 12%

= 12/100 x 775 = 993

∴ Present population of the village = 775 + 93 = 868.

Question 10. 80 students appeared for the Madhyamik Examination this year from our school. If 65% of students have passed, let us find how many students have actually passed.
Answer:

Number of students appeared for the Madhyamik Examination this year = 80.

Percentage of passed = 65%.

∴ Number of students passed = \(\frac{65}{100}\) x 80 = 52.

Class 6 WBBSE Math Solutions

Question 11. A certain alloy contains 70% popper and rest zinc. In 20 kg of such alloy let’s find how many kg of zinc is required.
Solution:

Total weight of alloy = 20 kg.

Percentage of copper present = 70%

= \(\frac{70}{100}\) X 20 kg = 14 kg.

∴ Quantity of zinc required = 20 – 14 = 6 kg.

Question 12. Due to the rise in the price of sugar we decided to reduce the consumption of sugar by 4%. Presently, we consume 625 gm of sugar every day. Let’s find how many grams the consumption of sugar is reduced and also the amount of sugar consumed each day now.
Solution:

At present we consume sugar every day = 625 gm.

Consumption of sugar reduced by 4% = \(\frac{4}{100}\) x 625 = 25gm

∴ Amount of sugar consumption today = 625gm – 252gm = 600gm

Question 13. Anilbabu pays 22% of his salary for house rent. If he pays Rs. 1870 per month for rent, let’s find his monthly salary.
Solution:

Monthly rent
22
1870

Monthly salary
100
?

When Rs. 22 is the rent, monthly salary = Rs. 100.

When Rs. 1 is the rent, monthly salary = Rs. \(\frac{100}{22}\)

When Rs. 1870 is the rent, monthly salary = Rs. \(\frac{100}{22}\) x 1870 = Rs. 8500.

Class 6 WBBSE Math Solutions

Question 14. Yasina Khatoon cultivates jute in 55% of her agricultural land. If she cultivates jute in 11 bighas of land, let’s find her total agricultural land.
Solution:

If Yasina cultivated 55 bighas of land, she had 100 bighas of agricultural land.

Cultivated land
55
11

Total agricultural land
100
?

If 55 bighas of cultivated land, total agricultural land = 100 bighas.

If 11 bighas of cultivated land, total agricultural land = 100 bighas = \(\frac{100}{55}\) x 11 bighas = 20 bighas.

Question 15. Out of total monthly expenses of our family, Rs. 4750 is spent on food and 5900 for all other expenses. If expenses on food is increased by 10% and other expenses is decreased by 16%, then let’s calculate whether the total monthly expenses will increase or decrease.
Solution:

Expenses for food = Rs. 4750

and other expenses = Rs. 5900

Expenses for food increased by 10%

= Rs. 4750 x \(\frac{10}{100}\)

= Rs. 475

Now expenses for food = Rs. 4750 + Rs. 475 = Rs. 5225

Expenses for other decreased by 16%

= Rs. 5900 x \(\frac{16}{100}\) = Rs. 944

Now other expenses = Rs. 5900 – Rs. 944 = Rs. 4956

Previously total expenses were = Rs. 4750 + Rs. 5900 = Rs. 10650

Presently total expenses are = Rs. (5225 + 4956) = Rs. 10181

Expenses Decrease in = Rs. (10650 -10181) = Rs. 469

Class 6 WBBSE Math Solutions

Question 16. The present population of the city is 26250. If the population increases at the rate of 4% every year, then let’s find what will the population of the city be next year. Also, let’s find what the population will be after two years.
Solution:

The present population of a city = 26250.

Population increases by 4%

= \( x 26250 = 1050.

∴ The population of the city in 1st year = 26250 + 1050 = 27300.

Next year population again increased by 4%

= [latex]\frac{4}{100}\) x 27300 = 1092

∴ The population of the city in 2nd year = 27300 + 1092 = 28392.

WBBSE Class 6 Maths Solution for Question 17. During harvesting the price of paddy was Rs. 1080 per quintal. In monsoon the price increased by 15%, let’s find out how much more the farmer who sold 12 quintals of paddy earlier could have earned, had he sold the same amount during monsoon.
Solution:

During harvesting, the price of paddy per quintal was Rs. 1080.

In monsoon, the price increased by 15%

= \(\) x Rs. 1080

= Rs. 162.

∴ During monsoon, he sold paddy at

= Rs. (1080 + 162) per quintal.

= Rs. 1242 per quintal

During monsoon, he will earn by selling 12 quintals

= Rs. 162×12

= Rs. 1944 more.

WBBSE Solutions For Class 6 Maths Chapter 6 Multiplication And Division Of A Fraction By Whole Number And By Fraction

Class 6 Math Solutions WBBSE Chapter 6 Multiplication And Division Of A Fraction By Whole Number And By Fraction Exercise 6

Question 1. Hands-on trial Four circular pieces of paper of equal size are taken. Half a portion of the circles are colored. Let us see what we can do with colored half positions.
Solution:

WBBSE Solutions For Class 6 Maths Chapter 6 Multiplication And Division Of A Fraction By Whole Number And By Fraction Four Circular Pieces Of Paper Of Equal Size Are Taken

4 x \(\frac{1}{2}\) → \(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\) = 2

Hence, 4 x \(\frac{1}{2}\) = \(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\) = \(\frac{1+1+1+1}{2}\) = \(\frac{4}{2}\) = 2

Or, 4x\(\frac{1}{2}\) = \(\frac{1\times1}{2}\) = 2

Read and Learn More WBBSE Solutions For Class 6 Maths

Again, let us take four circular pieces of paper of equal size. \(\frac{3}{4}\) part of it are coloured. Let us find what is the sum total of coloured portions.

WBBSE Solutions For Class 6 Maths Chapter 6 Multiplication And Division Of A Fraction By Whole Number And By Fraction Four Circular Pieces Of Papers Of Equal Size

4 x \(\frac{3}{4}\) = \(\frac{3}{4}\) + \(\frac{3}{4}\) + \(\frac{3}{4}\) + \(\frac{3}{4}\) = 3

4 x \(\frac{3}{4}\) = \(\frac{3+3+3+3}{4}=\) = 3 or 4 x \(\frac{3}{4}\) = \(\frac{4 \times 3}{4}\) = 3

Class 6 Math Solutions WBBSE

Question 2. If 8 equal pieces \(\frac{1}{2}\) of 12 equal pieces of \(\frac{3}{4}\) meter and 16 equal pieces of \(\frac{3}{8}\) meter of ribbon are needed, let us find what length of ribbon will be required in all.
Solution:

8 pieces of \(\frac{1}{2}\) meter = 8 x \(\frac{1}{2}\)m = \(\frac{8 \times 1}{2}\) = \(\frac{8}{2}\) = 4 m.

12 pieces of \(\frac{3}{4}\) metre = 12 x \(\frac{3}{4}\) m = \(\frac{16 \times 3}{4}\) = 6 m.

Total length of ribbon required = 4m + 9m + 6m = 19m.

Question 3. (1)

WBBSE Solutions For Class 6 Maths Chapter 6 Multiplication And Division Of A Fraction By Whole Number And By Fraction Four Circular Pieces 3 By 4 Of A Paper

Solution: That is 3 x \(\frac{1}{4}\) = \(\frac{3}{4}\)

(2)

WBBSE Solutions For Class 6 Maths Chapter 6 Multiplication And Division Of A Fraction By Whole Number And By Fraction 3 By 2 Pieces Of A Paper

Solution: 3 x \(\frac{2}{5}\) = \(\frac{6}{5}\) = 1 + \(\frac{1}{5}\)

(3) Let us color and deduce
Solution:

WBBSE Solutions For Class 6 Maths Chapter 6 Multiplication And Division Of A Fraction By Whole Number And By Fraction 3 By 2 Of Circular Piece

1 + \(\frac{1}{2}\) = 1\(\frac{1}{2}\) = \(\frac{3}{2}\)

Class 6 Math Solutions WBBSE

(4) Similarly, colour twice and let us bring together the colored position to conclude
Solution:

WBBSE Solutions For Class 6 Maths Chapter 6 Multiplication And Division Of A Fraction By Whole Number And By Fraction Color Twice And Let Us Bring Together The Colored Position

⇒ \(\frac{3}{2}\) + \(\frac{3}{2}\) = 3

Hence, 2 x \(\frac{3}{2}\) = 3.

WBBSE Solutions For Class 6 Maths Chapter 6 Multiplication And Division Of A Fraction By Whole Number And By Fraction 27 By 5 Coloured Paper

(5 + \(\frac{2}{5}\)) = 5\(\frac{2}{5}\) = \(\frac{27}{5}\)

(5) Let us draw three similar pictures and let us find the total color portions.
Solution:

WBBSE Solutions For Class 6 Maths Chapter 6 Multiplication And Division Of A Fraction

1. 1 + \(\frac{2}{5}\) = 1\(\frac{2}{5}\) = \(\frac{7}{5}\)

2. 1 + \(\frac{3}{5}\) = \(\frac{8}{5}\)

3. 1 + \(\frac{4}{5}\) = \(\frac{9}{5}\)

Class 6 WBBSE Math Solutions

WBBSE Solutions For Class 6 Maths Chapter 6 Multiplication And Division Of A Fraction By Whole Number And By Fraction 1 By Colored

3 x \(\frac{7}{5}\) = \(\frac{7 \times 3}{5}\) = \(\frac{21}{5}\) = 4\(\frac{1}{5}\)

7 x \(\frac{3}{5}\) = 4 + \(\frac{1}{5}\)

Let us calculate and color accordingly

(6) 3 x \(\frac{3}{4}\) = \(\frac{3 \times 3}{4}\) = \(\frac{9}{4}\) = 2\(\frac{1}{4}\)
Solution:

WBBSE Solutions For Class 6 Maths Chapter 6 Multiplication And Division Of A Fraction By Whole Number And By Fraction 2 And 1 By 4 Colored The Circular pieces

(7) 2 x \(\frac{7}{6}\) = \(\frac{2 \times 7}{6}\) = \(\frac{14}{6}\) = \(\frac{7}{3}\) = 2\(\frac{1}{3}\)
Solution:

WBBSE Solutions For Class 6 Maths Chapter 6 Multiplication And Division Of A Fraction By Whole Number And By Fraction 2 And 1 By 3 Colored Pieces

Class 6 WBBSE Math Solutions Chapter 6 Multiplication And Division Of A Fraction By Whole Number And By Fraction Exercise 6.1

Question 1. Sraboni had Rs. 100. She spent \(\frac{1}{2}\) of her money at a bookshop and \(\frac{1}{4}\) of her money at a grocery shop. Let us find how much money she paid at the book shop and at the grocery and how much money is left with her.
Solution:

At book shop, Sraboni paid \(\frac{1}{2}\) part of Rs. 100

= Rs. 100 x \(\frac{1}{2}\) = Rs. \(\frac{100 \times 1}{2}\) = Rs. \(\frac{100}{2}\) = Rs. 50

At the grocery, she paid \(\frac{1}{4}\) part of Rs. 100

= Rs. 100 x \(\frac{1}{4}\) = Rs. \(\frac{100 \times 1}{4}\) = Rs. \(\frac{100}{4}\) = Rs. 25.

Hence, Sraboni is left with Rs. 100 – (Rs. 50 + Rs. 25)

= Rs. 100 – Rs. 75 = Rs. 25 money is left with her

Question 2. In our art class, we were asked to draw a scenery. Samir took \(\frac{2}{5}\) part of 1 hr, Mita took \(\frac{5}{12}\) part of 1 hr, Ajij did it in \(\frac{1}{2}\) part of 1 hr and Sabbar finished it in \(\frac{7}{12}\) part of 1 hr. Let us calculate how much time each of them took to draw the scenery and also who took maximum time to finish the work and who took minimum time.
Solution:

∴ 1 hr = 60 min.

Samir finished his drawing in \(\frac{2}{5}\) part of 1 hr.

i.e., \(\frac{2}{5}\) of 60 min = 60 min x \(\frac{2}{5}\) = \(\frac{60 \times 2}{5}\) min = 24 min.

Mita finished her drawing in \(\frac{5}{12}\) of 1 hr, i.e., \(\frac{5}{12}\) of 60 min

= 60 min x \(\frac{5}{12}\) = \(\frac{60 \times 5}{12}\) min = 25 min.

Similarly, Ajij finished his drawing in \(\frac{1}{2}\) part of 1 hr = 30 min

Sabbar look \(\frac{7}{12}\) part of 1 hr = 60 min x \(\frac{7}{12}\) = 35 min.

Hence, to draw the same scenery Samir took minimum time and Sabbar took maximum time.

Class 6 WBBSE Math Solutions

Question 3. For a school picnic, Rehena paid \(\frac{5}{6}\) part of Rs. 30 and Javed paid \(\frac{5}{9}\) part of Rs. 45. Let us find who paid more.
Solution:

Rehana paid \(\frac{5}{6}\) part of Rs. 30 = Rs. 25 and Javed paid \(\frac{5}{9}\) part of 45 =
Rs. 25.

Hence, Rahana and Javed both paid Rs. 25.

Question 4. Let us express \(\frac{5}{8}\) part of Rs. 2 in rupees and paise.
Solution:

Rs. 2 = 200 paise

Hence, \(\frac{5}{8}\) part of Rs. 2 = \(\frac{5}{8}\) part of 200 paise

= 200 x \(\frac{5}{8}\) paisa = 125 paise = Re 1 and 25 paise.

Question 5. Let us multiply

1. 120 x \(\frac{3}{5}\)
Solution:  24 x 3 = 72

2. 2 x 215 x \(\frac{3}{5}\)
Solution: 2 x 43 x 3 = 258

3. 500 x \(\frac{17}{25}\)
Solution: 20 x 17 = 340

4. 169 x \(\frac{4}{13}\)
Solution: 13 x 4 = 52

Question 6. Lets do it

WBBSE Math Solutions Class 6

1. \(\frac{3}{4}\) of 1 year = ____ month
Solution: \(\frac{3}{4}\) of 1 year = \(\frac{3}{4}\) x 12 months = 3 x 3 = 9 months

2. \(\frac{3}{4}\) of Rs. 5 = ____ paise
Solution: \(\frac{3}{4}\) of Rs. 5 = \(\frac{3}{4}\) x 500 paise = 3 x 125 = 375 paise

3. \(\frac{3}{5}\) of 60 apples = _______ apples
Solution: \(\frac{3}{5}\) of 60 apples = 3 x 12 apples = 36 apples

4. \(\frac{3}{20}\) of 40 litre = ___ litre
Solution: \(\frac{3}{20}\) of 40 litre = 3 x 2 = 6 litre

Let Us Measure Land

Question 7. Rahim uncle planted flowering plants in half portion of his rectangular field. The length and breadth of the field are 50 metre and 40 metre respectively.
Solution:

Area of Rahim uncle’s field = (50 x 40) sq.m = 2000 sq.m.

\(\frac{1}{2}\) portion of 2000 sq.m, of field = \(\frac{1}{2}\) x 2000 sq.m = 2000 sq.m.

Hence, he planted flowering plants in 1000 sq.m, of field.

But in half a portion of his flower garden, Rahim uncle planted marigold. Therefore, marigold is planted in \(\frac{1}{2}\) part of 1000 sq.m

= 1000 x \(\frac{1}{2}\) sq.m = 500 sq.m.

Total land = 2000 sq.m.

Hence, 500 sq.m = \(\frac{500}{2000}\) part of 2000 sq.m. = \(\frac{1}{4}\) part.

i. e., he has planted marigolds in \(\frac{1}{4}\) part of his land. He planted marigolds or half of half portion of his land

= \(\frac{1}{2}\) of \(\frac{1}{2}\) portion of land = \(\frac{1}{2}\) x \(\frac{1}{2}\) = \(\frac{1}{4}\) part.

WBBSE Math Solutions Class 6 Chapter 6 Multiplication And Division Of A Fraction By Whole Number And By Fraction Exercise 6.2

Question 1. Rohit can walk 1\(\frac{1}{2}\) km in 1 hr. Let us find how far he can walk in 6\(\frac{1}{2}\) hrs. [1\(\frac{1}{2}\) km= \(\frac{3}{2}\) km; 6\(\frac{1}{2}\) hrs = \(\frac{3}{2}\) hrs]
Solution:

Rohit can walk in 1 hr 1\(\frac{1}{2}\) km = \(\frac{3}{2}\) km.

∴ Rohit can walk in 6\(\frac{1}{2}\) hr = \(\frac{13}{2}\) hr = \(\frac{3}{2}\) x \(\frac{13}{2}\) = \(\frac{39}{4}\) km = 9\(\frac{3}{4}\) km

Question 2. Rafikul’s uncle built a house on \(\frac{3}{5}\) of \(\frac{4}{7}\) part of his land and the rest of the land is left for cultivation. Let us find out on what part of land he built his house.
Solution:

Rafikul’s uncle built a house on \(\frac{3}{5}\) of \(\frac{4}{7}\) part of his land.

Part of the land he built his house on = \(\frac{3}{5}\) x \(\frac{4}{7}\) = \(\frac{12}{35}\) Part

Question 3. Let us multiply the following:

1. \(\frac{2}{3}\) x \(\frac{5}{6}\)
Solution: 5/9

2. \(\frac{7}{8}\) x \(\frac{3}{10}\)
Solution: 21/80

WBBSE Class 6 Maths Solutions

3. 19\(\frac{3}{4}\) x \(\frac{1}{7}\)
Solution:

= \(\frac{79}{4} \times \frac{1}{7}=\frac{79}{28}\)

= \(2 \frac{23}{28}\)

4. \(\frac{16}{5}\) x \(\frac{4}{7}\)
Solution: \(\frac{432}{35}\) = 12\(\frac{12}{35}\)

Question 4. The product of two proper fractions is always a ____ (proper Aim- proper) fraction.
Solution: The product of two proper fractions is always a proper fraction.

Example: \(\frac{3}{7} \times \frac{5}{5}=\frac{6}{35}\)

Question 5. The product of two improper fractions is always a/an ______ (proper/improper) fraction.
Solution: The product of two improper fractions is always an improper fraction.

Example: \(\frac{8}{3} \times \frac{11}{5}=\frac{88}{15}\)

WBBSE Class 6 Maths Solutions Chapter 6 Multiplication And Division Of A Fraction By Whole Number And By Fraction Exercise 6.3

Question 1. Few friends have come at Ayesha’s House. Ayesha brought 6 apples. Each friend ate 1— apple and no apples were left. Let’s find out how many friends of Ayesha ate apples.
Solution:

Number of apples = 6.

Each friend ate 1\(\frac{1}{2}\) apples = \(\frac{3}{2}\) apples.

∴ Number of friends = 6 ÷ \(\frac{3}{2}\)= 6 x \(\frac{2}{3}\) = 4

Question 2. There are 9 boiled eggs if each wants to have 1\(\frac{1}{2}\) eggs, let’s find how many can eat eggs.
Solution:

Total number of eggs = 9

Each want to have = 1\(\frac{1}{2}\) = \(\frac{3}{2}\) eggs

Number of persons =9÷ \(\frac{3}{2}\) = 9x\(\frac{2}{3}\)=6

WBBSE Class 6 Maths Solutions

Question 3. There are 5 packets of biscuits. If each person eats \(\frac{1}{4}\) part of a packet of biscuit, let’s find how many persons can eat the biscuits.
Solution:

Number of packs of biscuits = 5

Each person eats \(\frac{1}{4}\) part of a packet.

∴ Number of persons = 5 x \(\frac{4}{1}\) = 20

Question 4. Let’s find the values of the following

1. 4 ÷ \(\frac{1}{3}\)
Solution: 4×3 = 12

2. 3 ÷ \(\frac{1}{6}\)
Solution: 3×6 = 18

3. 8 ÷ \(\frac{1}{5}\)
Solution: 8X5 = 40

4. 5 ÷ \(\frac{1}{4}\)
Solution: 5×4 = 20

5. 6 ÷ \(\frac{1}{2}\)
Solution: 6×2 = 12

6. 15 ÷ \(\frac{5}{7}\)
Solution: 15x\(\frac{7}{5}\) = 21

7. 20 ÷ \(\frac{4}{5}\)
Solution: 20 x \(\frac{5}{4}\) = 25

8. 9 ÷ 1\(\frac{2}{25}\)
Solution: 9 \(\frac{27}{25}\)

= 9 ÷ \(\frac{27}{25}\)

= 9 x \(\frac{25}{27}\) = \(\frac{25}{3}\) = 8\(\frac{1}{3}\)

9. 7 ÷ 2\(\frac{3}{16}\)
Solution: 7 ÷ \(\frac{35}{16}\)

= 7 x \(\frac{16}{35}\) = 7x\(\frac{16}{5}\) = 3\(\frac{1}{5}\)

10. 4 ÷ 2\(\frac{10}{13}\)
Solution: 4 ÷ \(\frac{36}{13}\)

= \(4 \times \frac{13}{36}=\frac{13}{9}=1 \frac{4}{9}\)

WBBSE Class 6 Maths Solutions

11. 11 ÷ \(\frac{55}{18}\)
Solution: 11 x \(\frac{18}{55}\)

= \(\frac{18}{5}\) = 3\(\frac{3}{5}\)

12. 18 ÷ \(\frac{3}{5}\)
Solution: 18 x \(\frac{5}{3}\) = 6 x 5 = 30

Chapter 6 Multiplication And Division Of A Fraction By Whole Number And By Fraction Exercise 6.4

Question 1. Let’s find the values of the following:

1. \(\frac{7}{8} \div \frac{21}{5}\)
Solution:

= \(\frac{7}{8} \times \frac{5}{21}=\frac{5}{24}\)

2. \(\frac{3}{28} \div \frac{9}{7}\)
Solution:

= \(\frac{3}{28} \times \frac{7}{9}=\frac{1}{12}\)

3. \(5 \frac{2}{3} \div \frac{1}{6}\)
Solution:

= \(\frac{17}{3} \times \frac{6}{1}=34\)

4. ______ ÷ __________
Solution:

= \(\frac{4}{7} \div \frac{12}{21}=\frac{4}{7} \times \frac{21}{12}=1\)

WBBSE Class 6 Maths Solutions

Question 2. 1. Let’s do mentally

(1) How much is \(\frac{1}{2}\) of Rs. 10?
Solution: \(\frac{1}{2}\) of Rs. 10 = Rs. 5.

(2) How much is \(\frac{1}{5}\) part of Rs. 25?
Solution: \(\frac{1}{5}\) of Rs. 25 = Rs. \(\frac{1}{5}\) x 25 = Rs. 5.

(3) \(\frac{1}{3}\) of how much is Rs. 4?
Solution: Let \(\frac{1}{3}\) of Rs. x = 4

∴ x = Rs. 4 x 3 = Rs. 12.

(4) \(\frac{1}{4}\)part of what time is 65 min?
Solution: Let \(\frac{1}{3}\) of x min = 65

∴ x = 65 x 6 = 390 min.

(5) I have taken \(\frac{1}{3}\) part of mangoes from Mala’s basket of mangoes. If I have taken 7 mangoes, how many mangoes Mala had in her basket, let’s find.
Solution: \(\frac{1}{3}\) part of mangoes from Mala’s basket = 7.

∴ Total number of mangoes of Mala’s basket = 7÷\(\frac{1}{3}\) = 7 x 3 = 21

(6)

WBBSE Solutions For Class 6 Maths Chapter 6 Multiplication And Division Of A Fraction By Whole Number And By Fraction Number Of Mangoes Of alas Basket

Solution:

WBBSE Solutions For Class 6 Maths Chapter 6 Multiplication And Division Of A Fraction By Whole Number And By Fraction 2 Circles, 6 Triangles, 1 Quadrilaterals Is Colored

Total circles = 7; \(\frac{2}{7}\) part of 7 circles = 2.

Total triangles = 10; \(\frac{3}{5}\) part of 10 triangles = 6.

Total quadrilaterals = 7; \(\frac{1}{7}\) part of 7 quadrilaterals = 1.

(7) Today father has brought 10 I of drinking water from a nearby tubewell. Mother used \(\frac{1}{5}\) part of water for cooking and \(\frac{1}{4}\) part of remaining water was used for drinking purpose. Let us find how many liters of water is left.
Solution:

Class 6 Maths Solutions WBBSE

Quantity of drinking water = 10 litres

Mother used water for cooking = \(\frac{1}{5}\) x 10 litres = 2 litres

Remaining water = (10-2) litre = 8 litres

Used for drinking water = \(\frac{1}{4}\) x 8 liters = litres

Remaining water = (8- 2) litres = 6 litres

Question 3. Let’s find the value of:

1. \(3 \times \frac{6}{11}\)
Solution:

= \(\frac{18}{11}=1 \frac{7}{11}\)

2. \(11 \times \frac{2}{3}\)
Solution:

= \(\frac{22}{3}=7\frac{1}{3}\)

3. \(\frac{7}{3} \times 2 \frac{3}{2}\)
Solution:

= \(\frac{7}{3} \times \frac{7}{2}=\frac{49}{6}=8 \frac{1}{6}\)

4. \(\frac{3}{8} \times \frac{6}{4}\)
Solution:

= \(\frac{3 \times 6}{8 \times 4}=\frac{9}{16}\)

5. \(\frac{6}{49} \times \frac{7}{3}\)
Solution:

= \(\frac{6 \times 7}{49 \times 3}=\frac{2}{7}\)

6. \(\frac{15}{28} \times 2 \frac{1}{3}\)
Solution:

= \(\frac{15}{28} \times \frac{7}{3}=\frac{5}{4}\)

7. \(4 \frac{8}{13} \times 7 \frac{4}{5}\)
Solution:

= \(\frac{60}{13} \times \frac{39}{5}=36\)

Question 4. A bucket holds \(\frac{1}{2}\) liter of water. Let us calculate how much water 7 such buckets can hold.
Solution:

A bucket holds = \(\frac{1}{2}\) litre water

∴ 7 such buckets contain = 7 x \(\frac{1}{2}\) litres = \(\frac{7}{2}\) = 3.5 litres water.

Class 6 Maths Solutions WBBSE

Question 5. After retirement, Akhilbabu donated \(\frac{1}{4}\) part of his property to the local library, \(\frac{1}{6}\) part of the remaining property was given to his wife and rest was divided equally among his two sons. Let us calculate to find what part of his property was given to his wife and each of two sons.
Solution:

Akhilbabu donated = \(\frac{1}{4}\) part to local library.

Remaining part = 1 – \(\frac{1}{4}\) = \(\frac{4-1}{4}\) = \(\frac{3}{4}\) part.

Wife received = \(\frac{1}{6}\) x \(\frac{3}{4}\) = \(\frac{1}{8}\) part.

Remaining part= \(\frac{3}{4}\) – \(\frac{1}{8}\) = \(\frac{3 \times 2 -1}{8}\) = \(\frac{5}{8}\)

∴ Two sons will get = \(\frac{5}{8}\) part.

Each son received = \(\frac{1}{2}\) X \(\frac{5}{8}\) = \(\frac{5}{16}\) part.

Question 6. From \(\frac{1}{2}\) part of Rs. 150, how much is to be taken away so that only Rs. 30 is left?
Solution:

⇒ \(\frac{1}{2}\) part of Rs. 150 = 150 x \(\frac{1}{2}\) = Rs. 75

Remaining money = Rs. 30.

We will take = Rs. 75 — Rs. 30 = Rs. 45

Question 7. Let us find the value when 3 times of \(\frac{6}{7}\) is added to 2\(\frac{6}{7}\)
Solution:

= 3 times of \(\frac{6}{7}\) + \(\frac{20}{7}\)

= 3 X \(\frac{6}{7}\) + \(\frac{20}{7}\)= \(\frac{38}{7}\)

=5\(\frac{1}{7}\)

Question 8. In the first year the cultural program of the town had 1400 spectators. Next year the number increased by \(\frac{7}{10}\) parts let’s find the total number of next year.
Solution:

Number of spectators = 1400.

Next year spectators increased by \(\frac{7}{10}\)

i.e., \(\frac{7}{10}\) x 1400 = 980

Total number of spectators in next year = 1400 + 980 = 2380

Class 6 Maths Solutions WBBSE

Question 9. Let’s find the reciprocals of the following fractions and mark the reciprocals by ‘O’ sign which are proper fractions:
Solution:

1. Reciprocal of \(\frac{7}{5}=\frac{5}{7}\)

2. Reciprocal of \(\frac{1}{3}\) = 3

3. Reciprocal of \(\frac{5}{8}=\frac{8}{5}\)

4. Reciprocal of \(\frac{9}{7}=\frac{7}{9}\)

5. Reciprocal of \(\frac{12}{5}=\frac{5}{12}\)

6. Reciprocal of \(\frac{5}{8}=\frac{8}{5}\)

7. Reciprocal of \(\frac{1}{8}=\frac{8}{1}\)

Question 10. Which number has reciprocal as itself?
Solution: The number which has reciprocal as itself = 1.

Question 11. Sita gives me \(\frac{2}{3}\) part of the total number of stamps she has. If she gives me 18 stamps, let us find how many stamps Sita has.
Solution:

2/3 part of total number of stamps = 18.

∴ Total number of stamps of Sita = 18 \(\frac{2}{3}\) = 18 x \(\frac{3}{2}\) = 27.

Question 12. Rajia gave \(\frac{2}{5}\) part of her money to Debnath and \(\frac{3}{10}\) part of money to Sunita. If she has Rs. 180 left, let us find how much money Rajia had at the begining.
Solution:

Rajia gave \(\frac{2}{5}\) part and \(\frac{3}{10}\) part of her money to Debnath and Sunita respectively.

∴ She gave \(\frac{2}{5}+\frac{3}{10}=\frac{4+3}{10}=\frac{7}{10}\) part of money.

Remaining part of her money = \(1-\frac{7}{10}=\frac{10-7}{10}=\frac{3}{10}\)

According to problem, \(\frac{3}{10}\) part of money = Rs. 180

∴ Total amount of money she had = Rs. 180 x \(\frac{10}{3}\) = Rs. 600.

Question 13. Let’s find the values

1. \(15 \div \frac{5}{3}\)
Solution:

= \(15 \times \frac{3}{5}=9\)

2. \(14 \div \frac{7}{2}\)
Solution:

= \(14 \times \frac{2}{7}=4\)

3. \(\frac{6}{13} \div 3\)
Solution:

= \(\frac{6}{13} \times \frac{1}{3}=\frac{2}{13}\)

4. \(\frac{12}{19} \div 6\)
Solution:

= \(\frac{12}{19} \times \frac{1}{6}=\frac{2}{19}\)

5. \(5 \frac{1}{5} \div \frac{13}{2}\)
Solution:

= \(5 \frac{1}{5} \div \frac{13}{2}=\frac{26}{5} \times \frac{2}{13}=\frac{4}{5}\).

Class 6 Maths Solutions WBBSE

6. \(2 \frac{2}{5} \div 1 \frac{1}{5}\)
Solution:

= \(2 \frac{2}{5} \div 1 \frac{1}{5}=\frac{12}{5} \div \frac{6}{5}=\frac{12}{5} \times \frac{5}{6}=2\)

7. \(4 \frac{3}{7} \div 3 \frac{2}{7}\)
Solution:

= \(\frac{31}{7} \div \frac{23}{7}=\frac{31}{7} \times \frac{7}{23}=\frac{31}{23}=1 \frac{8}{23}\).

Question 14. Let’s mark ‘√’ for the correct answer

1. How many \(\frac{1}{16}\) are there in \(\frac{3}{4}\)?

  1. 64
  2. 12
  3. 4
  4. 3

Solution:

Required no. = \(\frac{3}{4}\) ÷ \(\frac{1}{16}\)

= \(\frac{3}{4}\) x \(\frac{16}{1}\)

∴ 2. 12.

2. \(\frac{3}{4}\) part of a ribbon is 56 m. Let’s calculate the original length of the ribbon.

  1. 43 m
  2. 64 m
  3. 63 m
  4. 72 m

Solution: Original length of the ribbon

= 56 m x \(\frac{8}{7}\) = 64 m

∴ 2. 64 m

3. Reciprocal of 5\(\frac{6}{7}\) will be

  1. 5\(\frac{3}{4}\)
  2. \(\frac{41}{7}\)
  3. \(\frac{7}{41}\)
  4. \(\frac{7}{56}\)

Solution: Reciprocal of 5\(\frac{6}{7}\), i.e., \(\frac{41}{7}\)= \(\frac{7}{41}\)

∴ 3. \(\frac{7}{41}\)

Question 15. From 16\(\frac{2}{3}\)m long ribbon, \(\frac{3}{8}\) part is cut off. If it is further divided into 5 equal pieces, let’s find the length of each pieces.
Solution:

Length of the ribbon = \(16 \frac{2}{3}=\frac{50}{3}\) m

⇒ \(\frac{3}{8} \text { part }=\frac{50}{3} \times \frac{3}{8}=\frac{50}{8} \mathrm{~m}=\frac{25}{4} \mathrm{~m}\)

∴ Length of each piece = \(\frac{25}{4} \div 5=\frac{25}{4} \times \frac{1}{5}=\frac{5}{4} m .=1 \frac{1}{4} m\)

Question 16. Father bought 12\(\frac{7}{10}\) m of cloth for window curtains. But there was already 5\(\frac{3}{5}\)m of cloth for curtains at home. 4\(\frac{5}{6}\) m of cloth is required to make curtains for each of 3 windows. What length of cloth will remain?
Solution:

Length of cloth

= \(12 \frac{7}{10} \mathrm{~m}+5 \frac{3}{5} \mathrm{~m}\)

= \(\frac{127}{10}+\frac{28}{5}\)

= \(\frac{127+56}{10} \mathrm{~m}\)

= \(\frac{183}{10} \mathrm{~m} .\)

Cloth required to make one curtain = \(4 \frac{5}{6} m=\frac{29}{6} m\)

∴ To make 3 curtains cloth required = \(\frac{29}{6} \times 3 \mathrm{~m}=\frac{29}{2} \mathrm{~m}\)

Remaining cloth = \(\frac{183}{10} m-\frac{29}{2} m\)

= \(\frac{183-145}{10} m=\frac{38}{10}m\)

= \(\frac{19}{5}m=3\frac{4}{5}m\)

Question 17. My grandmother prepared some pickle. She removed \(\frac{4}{7}\) part of the pickle in a glass jar for future use. Rest she divided among 6 of us. Let’s find how much each of us will get.
Solution:

Grandmother removed \(\frac{4}{7}\) part of the pickle.

Remaining part = \(1-\frac{4}{7}=\frac{7-4}{7}=\frac{3}{7} \text { part. }\)

∴ \(\frac{3}{7}\) part was divided among 6 of us.

∴ Each of us will get = \(\frac{3}{7} \div 6=\frac{3}{7} \times \frac{1}{6}=\frac{1}{14} \text { part. }\)

Question 18. Mehboob and his group have decided that in 33 days they would repair 24\(\frac{11}{15}\) km of road. They repaired \(\frac{11}{15}\) km of road each day for 25 days. If they are to finish the work in due time, at what rate they would work for the remaining days?
Solution:

They repaired in each day = \(\frac{11}{15}\) km.

∴ They repaired in 25 days = \(\frac{11}{15}\) x 25 = \(\frac{55}{3}\) km

∴ Remaining work = 24\(\frac{11}{15}\) – \(\frac{55}{3}\) = \(\frac{371}{15}\) – \(\frac{55}{3}\) = \(\frac{371-275}{15}=\frac{96}{15} \mathrm{~km}\)km.

Remaining days = 33 – 25 = 8.

∴ In 8 days they have to repair \(\frac{35}{5}\) km.

∴ In 1 day they will repair \(\frac{32}{5} \times \frac{1}{8}=\frac{4}{5} \mathrm{~km}\)

Multiplication For Class 6

Question 19. 5 is added to —\(\frac{3}{7}\) and the sum is multiplied by 4\(\frac{2}{3}\) Now, the  product is divided by 4\(\frac{4}{9}\) and the quotient is subtracted from 8\(\frac{2}{5}\). Let’s find the number after subtraction.
Solution:

⇒ \(5+\frac{3}{7}=\frac{35+3}{7}=\frac{38}{7}\)

The sum is multiplied by \(4 \frac{2}{3}\).

∴ \(\frac{38}{7} \times 4 \frac{2}{3}=\frac{38}{7} \times \frac{14}{3}=\frac{76}{3}\)

Then the product is divided by 4\(\frac{4}{9}\)

i.e., \(\frac{76}{3} \div 4 \frac{4}{9}=\frac{76}{3} \div \frac{40}{9}=\frac{76}{3} \times \frac{9}{40}=\frac{57}{10}\)

The quotient is subracted from 8\(\frac{2}{5}\).

i.e., \(8 \frac{2}{5}-\frac{57}{10}=\frac{42}{5}-\frac{57}{10}=\frac{84-57}{10}\)

= \(\frac{27}{10}=2 \frac{7}{10}\)

∴ Required No. = \(2 \frac{7}{10}\)

Multiplication For Class 6

Question 20. Let us simplify

1. \(\frac{1}{5}+\frac{2}{3}-\frac{1}{2}\)
Solution:

= \(\frac{6+20-15}{30}=\frac{26-15}{30}=\frac{11}{30}\)

2. \(\frac{1}{5}+\frac{1}{2}\)–\(\frac{2}{15}-\frac{1}{6}\)
Solution:

= \(\frac{6+15-4-5}{30}=\frac{26-9}{30}=\frac{12}{30}=\frac{2}{5}\)

3. \(\frac{7}{12}+5 \frac{2}{9}+\frac{11}{18}-2 \frac{5}{12}\)
Solution:

= \(\frac{7}{12}+\frac{47}{9}+\frac{11}{18}-\frac{29}{12} \)

= \(\frac{21+188+22-87}{36}\)

= \(\frac{231-87}{36}\)

= \(\frac{144}{36}\) = 4

4. \(3 \frac{1}{2}+\frac{7}{6} \times \frac{3}{8}-\frac{5}{24}\)
Solution:

= \(\frac{7}{2}+\frac{7}{16}-\frac{5}{24}\)

= \(\frac{168+21-10}{48}=\frac{189-10}{48}=\frac{179}{48}=3 \frac{35}{48}\)

5. \(\frac{3}{8} \div \frac{2}{3}\) of \(\frac{1}{9}\) of \(\frac{1}{16}\)
Solution:

= \(\frac{3}{8} \div \frac{2}{3} \times \frac{1}{9} \times \frac{1}{16}\)

= \(\frac{3}{8} \div \frac{1}{3 \times 9 \times 8}=\frac{3}{8} \times \frac{3 \times 9 \times 8}{1}=81\)

6. \(6 \frac{2}{5}+3 \frac{1}{3}+\frac{1}{2}-\frac{7}{10}\)
Solution:

= \(\frac{192+100+15-21}{30}=\frac{307-21}{30}=\frac{286}{30}=9 \frac{8}{15}\)

Division For Class 6

7. \(\left\{\frac{11}{16} \div\left(\frac{5}{6}+\frac{2}{3}\right)\right\}-\frac{1}{3}\)
Solution:

= \(\left(\frac{11}{16} \div \frac{9}{6}\right)-\frac{1}{3}=\left(\frac{11}{16} \times \frac{6}{9}\right)-\frac{1}{3}\)

= \(\frac{11}{24}-\frac{1}{3}=\frac{11-8}{24}=\frac{3}{24}=\frac{1}{8}\)

8. \(4 \frac{2}{3} \div \frac{2}{3}-\frac{3}{8}\)
Solution:

= \(\frac{14}{3} \times \frac{3}{2}-\frac{3}{8}\)

= \(\frac{7}{1}-\frac{3}{8}=\frac{56-3}{8}=\frac{53}{8}=6 \frac{5}{8}\)

9. \(\left(2 \frac{3}{4}+3 \frac{1}{2} \div 2 \frac{1}{7}\right) \div 13 \frac{1}{4}\)
Solution:

= \(\left(\frac{11}{4}+\frac{7}{2} \div \frac{15}{7}\right) \div \frac{53}{4}\)

= \(\left(\frac{11}{4}+\frac{49}{30}\right) \div \frac{53}{4}\)

= \(\frac{165+98}{60} \times \frac{4}{53}=\frac{263}{60} \times \frac{4}{53}=\frac{263}{795}\)

10. \(1-\left[\frac{1}{2} \div\left\{2-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\right\}\right]\)
Solution:

= \(1-\left[\frac{1}{2} \div\left\{2-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\right\}\right]\)

= \(1-\left[\frac{1}{2} \div\left\{2-\frac{1}{2}\left(\frac{3-2+1}{6}\right)\right\}\right]\)

= \(1-\left[\frac{1}{2} \div\left\{2-\frac{1}{2} \times \frac{2}{6}\right\}\right]\)

= \(1-\left[\frac{1}{2} \div\left\{2-\frac{1}{6}\right\}\right]\)

= \(1-\left[\frac{1}{2} \div\left\{\frac{12-1}{6}\right\}\right]\)

= \(1-1-\left[\frac{1}{2} \div \frac{11}{6}\right]=1-\left[\frac{1}{2} \times \frac{6}{11}\right]=1-\frac{3}{11}=\frac{11-3}{11}=\frac{8}{11}\)

Division For Class 6

11. \(2-\frac{1}{10} \times \frac{1}{3} \div \frac{4}{25} \div \frac{1}{8}\)
Solution:

= \(2-\frac{1}{10} \times \frac{1}{3} \times \frac{25}{4} \times \frac{8}{1}\)

= \(2-\frac{5}{3}=\frac{6-5}{3}=\frac{1}{3}\)

12. \(\frac{1}{2}\left[3 \frac{1}{2} \div 2 \frac{1}{3}\left\{1 \frac{1}{4} \div\left(2+3 \frac{2}{3}\right)\right\}\right]\)
Solution:

= \(\frac{3}{2}\left[\frac{7}{2} \div \frac{7}{3}\left\{\frac{5}{4} \div \frac{17}{3}\right\}\right]\)

= \(\frac{3}{2}\left[\frac{7}{2} \div \frac{7 \times 5}{17 \times 4}\right]\)

= \(\frac{3}{2}\left[\frac{7}{2} \times \frac{17 \times 4}{7 \times 5}\right]\)

= \(\frac{3}{2} \times \frac{17 \times 2}{5}=\frac{51}{5}=10 \frac{1}{5}\)

13. \(\left(1 \frac{1}{13} \times 2 \frac{3}{5}\right) \div\left(7 \frac{1}{2} \times 3 \frac{1}{10}\right) \div \frac{28}{279}\)
Solution:

= \(\left(\frac{14}{13} \times \frac{13}{5}\right) \div\left(\frac{15}{2} \times \frac{31}{10}\right) \div \frac{28}{279}\)

= \(\frac{14}{5} \div \frac{93}{4} \div \frac{28}{279} \)

= \(\frac{14}{5} \times \frac{4}{93} \times \frac{279}{28}\)

= \(\frac{6}{5}=1 \frac{1}{5}\)

Division For Class 6

Question 21. (1) In the figure given below, let \(\frac{2}{3}\) be multiplied by the fractions
at the angular portion and the product be written in the blank circles outside.

(2) Let \(\frac{2}{3}\) be multiplied by the fractions in horizontal and vertical circles and let the product be placed in the corresponding blank circles outside.

WBBSE Solutions For Class 6 Maths Chapter 6 Multiplication And Division Of A Fraction By Whole Number And By Fraction 2 By 3 Be Multiplied By The Fractions At The AngularPortion

Solution:

WBBSE Solutions For Class 6 Maths Chapter 6 Multiplication And Division Of A Fraction By Whole Number And By Fraction 2 By 3 Be Multiplied By Fractions At The Angular Portion

(s) is multiplied by the fraction of the angular points we get the following results (K, M, O, P).

1. sxk= \(\frac{2}{3} x \frac{1}{5}=\frac{2}{15}\)…. (A)

2. sxM = \(\frac{2}{3} \times 1 \frac{1}{5}=\frac{2}{3} \times \frac{6}{5}=\frac{4}{5}\)….(C)

3. sxO = \(\frac{2}{3} \times \frac{3}{5}=\frac{2}{3}\)…….(E)

4. sxQ = \(\frac{2}{3} \times \frac{9}{12}=\frac{1}{2}\)…..(G)

Division For Class 6

Again, \(\frac{2}{3}\) (s) is divided by the fractions of the horizontal & vertical mid¬points (R, N, and L, P) we get.

1. \(S \div R=\frac{2}{3} \div \frac{2}{5}=\frac{2}{3} \times \frac{5}{2}=\frac{5}{3}\) …..(H)

2. \(\mathrm{S} \div \mathrm{N}=\frac{2}{3} \div \frac{3}{2}=\frac{2}{3} \times \frac{2}{3}=\frac{4}{9}\)……(D)

3. \(\mathrm{S} \div \mathrm{L}=\frac{2}{3} \div \frac{2}{9}=\frac{2}{3} \times \frac{9}{2}\)….(B)

4. \(S \div P=\frac{2}{3} \div \frac{4}{3}=\frac{2}{3} \times \frac{3}{4}=\frac{1}{2}\)….(F)

 

WBBSE Solutions For Class 6 Maths Chapter 7 Multiplication And Division Of Decimal Fraction By Whole Number And Decimal Fraction

Class 6 Math WBBSE Solutions Chapter 7 Multiplication And Division Of Decimal Fraction By Whole Number And Decimal Fraction Exercise 7

Question 1. Let’s draw and find the product of 0.4 x 0.7 and 0.2 x 0.8 by taking a square of white paper.
Solution:

1. 0.4 x 0.7

WBBSE Solutions For Class 6 Maths Chapter 7 Multiplication And Division Of Decimal Fraction By Whole Number And Decimal Fraction Taking A Square White Paper 0.4 x 0.7

2. 0.2 x 0.8

WBBSE Solutions For Class 6 Maths Chapter 7 Multiplication And Division Of Decimal Fraction By Whole Number And Decimal Fraction Taking A Square White Paper 0.2 x 0.8

1. 0.4 x 0.7

Read and Learn More WBBSE Solutions For Class 6 Maths

WBBSE Solutions For Class 6 Maths Chapter 7 Multiplication And Division Of Decimal Fraction By Whole Number And Decimal Fraction Taking A Square White Paper 0.4 x 0.7 Graph

2. 0.2 x 0.8

WBBSE Solutions For Class 6 Maths Chapter 7 Multiplication And Division Of Decimal Fraction By Whole Number And Decimal Fraction Taking A Square White Paper 0.2 x 0.8 Graph

Question 2. The side of a square is 5.2 cm. Let’s find its perimeter and area.
Solution:

The side of a square = 5.2 cm.

1. Perimeter of the square = 4 x one side

= 4 x 5.2 cm

Perimeter of the square = 20. 8 cm

2. Area of the square = side x side

= 5.2 cm x 5.2 cm

= \(\frac{52}{10} \times \frac{52}{10} \mathrm{sq} \cdot \mathrm{cm}\)

= \(\frac{2704}{100} \mathrm{sq} . \mathrm{cm}\)

Area of the square = 27.04 sq. cm.

Class 6 Math WBBSE Solutions

Question 3. Let’s find the values of the following:

1. 6.2 x 3.1
Solution:

= \(\frac{62}{10} \times \frac{31}{10}=\frac{1922}{100}\)

6.2 x 3.1 = 19.22

2. 11.3 x 2.5
Solution:

= \(\frac{113}{10} \times \frac{25}{10}=\frac{2825}{100}\)

11.3 x 2.5 = 282.5

3. 11.25 x 7.3
Solution:

= \(\frac{1125}{100} \times \frac{73}{10}=\frac{82125}{1000}\)

11.25 x 7.3 = 82.125

4. 0.03 x 0.3
Solution:

= \(\frac{3}{100} \times \frac{3}{10}=\frac{9}{1000}\)

0.03 x 0.3 = 0.009

5. 0.04 x 0.04.
Solution:

= \(\frac{4}{100} \times \frac{4}{100}=\frac{16}{10000}\)

= 0.0016

0.04 x 0.04 = 19

Class 6 Math Solutions WBBSE Multiplication And Division Of Decimal Fraction By Whole Number And Decimal Fraction Exercise 7.1

Question 1. On a white square paper let’s draw 10×10 small squares. By using different colours let’s find the value of 0.5 x 0.9, 0.2 x 0.8.
Solution:

1. 0.5 x 0.9

WBBSE Solutions For Class 6 Maths Chapter 7 Multiplication And Division Of Decimal Fraction By Whole Number And Decimal Fraction Taking A Square White Paper 10 x10 small Squares

2. 0.2 x 0.8

WBBSE Solutions For Class 6 Maths Chapter 7 Multiplication And Division Of Decimal Fraction By Whole Number And Decimal Fraction Taking A Square White Paper Draw 10 x 10

Question 2. Mithu wants to buy 4 exercise books. If an exercise book costs Rs. 12.75, let’s find how much money Mithu will need.
Solution:

Price of each exercise book = Rs. 12.75.

∴ Price of 4 exercise books = Rs. 12.75 x 4 = Rs. 51.

∴ Mithu will need = Rs. 51.

Class 6 Math Solutions WBBSE

Question 3. Rajinabibi built a house in 0.35 part of her land. She cultivated flowers in 0.2 part of the remaining land. Let’s find in what part of her land she cultivated flowers.
Solution:

Rajinabibi built a house in 0.35 part of her land.

Remaining land = 1 – 0.35 = 0.65 part.

She cultivated flowers on 0.2 part of the remaining land

= \(0.2 \times 0.65=\frac{2}{10} \times \frac{65}{100}=\frac{13}{100}=0.13 \mathrm{part}\)

Question 4. 1 have Rs. 150. With 0.3 part of my money, I bought exercise books and with 0.4 part I bought a storybook. Let me find out what amount of money is left with me.
Solution:

I have Rs. 150.

I bought an exercise book with 0.3 part and bought a story book with 0.4 part.

∴ Total expenditure (0.3 + 0.4) = 0.7 part

= Rs. 150 x 0.7

= Rs. 150 x \(\frac{7}{10}\) = Rs. 105

∴ Remaining amount = Rs. (150 – 105) = Rs. 45.

Class 6 Math Solutions WBBSE

Question 5. Today we shall travel a distance of 94.5 km. If 0.078 liter of petrol is used per kilometer, let’s calculate the total amount of petrol required.
Solution:

To travel 1 km petrol required 0.078 litre.

∴ To travel 94.5 km petrol required = 0.078 x 94.5 litre = 7.37 litre.

Question 6. Alisha’s brother took 1.4 hours to reach Shibpur Launch pier (ghat) from his house in a cycle. If the speed of his cycle is 11.5 km per hour, then let us calculate the distance of Shibpur Launch pier from Alisha’s house.
Solution:

Speed of the cycle is 11.5 km per hour, i.e., in 1 hour he goes 11.5 km.

∴ In 1.4 hour he goes 1.4 x 11.5 km = 16.1 km.

∴ Required distance = 16.1 km.

Question 7. My mother asked me to buy 2.5 kg of pulse. The cost of 1 kg of pulse is Rs. 62.50. Let me calculate how much money I must carry to the shop.
Solution:

The cost of 1 kg of pulse is Rs. 62.50.

∴ The cost of 2.5 kg of pulse = Rs. 62.50 x 2.5.

∴ I must carry to the shop Rs. 156.25 = Rs. 156.25.

Class 6 Math Solutions WBBSE

Question 8. The perimeter of an equilateral triangle is 10.4 cm. Let’s find the length of the side of the equilateral triangle in decimal fractions.
Solution:

The perimeter of an equilateral triangle = 14.4 cm.

∴ Length of each side of this triangle = \(\frac{14.4}{3}\) cm = 4.8 cm.

Question 9. Let’s multiply

1. 0.7 x 0.9
Solution: 0.7 x 0.9 = 0.63

2. 0.6 x 0.5
Solution: 0.6 x 0.5 = 0.30

3. 0.02 x 0.2
Solution: 0.02 x 0.2 = 0.004,

4. 0.67 x 0.39
Solution:

0.67 x 0.39 = \(\frac{67}{100} \times \frac{39}{100}=\frac{2613}{10000}=0.2613\)

5. 0.52 x 0.43
Solution:

0.52 x 0.43 = \(\frac{52}{100} \times \frac{43}{100}=\frac{2236}{10000}\) = 0.2236

6. 0.07 x 0.97
Solution:

0.07 x 0.97 = \(\frac{7}{100} \times \frac{97}{100}=\frac{679}{10000}\) = 0.0679

WBBSE Math Solutions Class 6

7. 6.23 x 2.51
Solution:

6.23 x 2.51 = \(\frac{623}{100} \times \frac{251}{100}=\frac{156373}{10000}\) = 15.6373

8. 5.77 x 2.93
Solution:

5.77 x 2.93 = \(\frac{577}{100} \times \frac{293}{100}=\frac{169061}{10000}\) = 16.

9. 8.23 x 0.3
Solution:

8.23 x 0.3 = \(\frac{823}{100} \times \frac{3}{10}=\frac{2469}{1000}\) = 2.469

10. 82.03 x 0.06
Solution:

82.03 x 0.06 = \(\frac{8203}{100} \times \frac{6}{100}=\frac{49218}{10000}\) = 4.9218

11. 85.29 x 3.92
Solution:

85.29 x 3.92 = \(\frac{8529}{100} \times \frac{392}{100}=\frac{3343368}{10000}\) = 334.3368

12. 72.2 x 2.65
Solution:

72.2 x 2.65 = \(\frac{722}{10} \times \frac{265}{100}=\frac{191330}{1000}\) =191.33

WBBSE Math Solutions Class 6

13. 72.04 x 0.05
Solution:

= \(\frac{7204}{100} \times \frac{5}{100}=\frac{36020}{10000}\)

13. 72.04 x 0.05 = 3.602

14. 72.156 x 12.16
Solution:

= \(\frac{72156}{1000} \times \frac{1216}{100}\)=\(\frac{87741696}{100000}\)

72.156 x 12.16 = 877.41696

15. 0.6 x 0.3 x 0.2
Solution:

= \(\frac{6}{10} \times \frac{3}{10} \times \frac{2}{10}=\frac{36}{1000}\)

0.6 x 0.3 x 0.2 = 0.036

16. 0.2 x 0.06 x .03
Solution:

= \(\frac{2}{10} \times \frac{6}{100} \times \frac{3}{100}=\frac{36}{100000}\)

0.2 x 0.06 x .03 = 0.00036

17. 2. 14 x 0.4 x 0.9
Solution:

= \(\frac{214}{100} \times \frac{4}{10} \times \frac{9}{10}=\frac{7704}{10000}\)

2. 14 x 0.4 x 0.9 = 0.7704

WBBSE Math Solutions Class 6

18. 1.21 x0.5 x 5.2
Solution:

= \(\frac{121}{100} \times \frac{5}{10} \times \frac{52}{10}=\frac{31460}{10000}\)

1.21 x0.5 x 5.2 = 3.146

19. 3.06 x 100
Solution:

= \(\frac{306}{100} \times 100\)

3.06 x 100 = 306

20. 7.92 x 1000
Solution:

= \(\frac{792}{100} \times 1000\)

7.92 x 1000 = 7920

Question 10. Arrange the following in descending order (bigger to smaller) of their values:

1. 0.5 x 0.3, 0.5, 0.3
Solution:

0.5 x 0.3, 0.5, 0.3 = 0.15; 0.5; 0.3

WBBSE Math Solutions Class 6

In descending order = 0.5; 0.3; 0.5 x 0.3

2. 0.6 x 0.7, 0.6, 0.7
Solution:

= 0.42 ; 0.6; 0.7

In descending order = 0.7; 0.6 x 3; 0.6 x 0.7

3. 0.9 x 0.2, 0.9, 0.2
Solution:

= 0.18; 0.9; 0.2

In descending order = 0.9; 0.2 ; 0.9 x 0.2

4. 0.4 x 0.8, 0.4, 0.8
Solution:

= 0.32; 0.4; 0.8

In descending order = 0.8; 0.4; 0.4 x 0.8

5. 1.2 X 1.5,1.2,1.5
Solution:

= 1.8; 1.2; 1.5

In descending order = 1.2 x 1.5; 1.5; 1.2

6. 2.3 x 2.4, 2.3, 2.4
Solution:

= 5.52; 2.3; 2.4

In descending order = 2.3 x 2.4; 2.4; 2.3

7. 6.7 x 7.2, 6.7, 7.2
Solution:

= 48.24; 6.7; 7.2

In descending order = 6.7 x 7.2 ; 7.2 ; 6.7

8. 8.2 x 1.9, 8.2,1.9
Solution:

= 15.58; 8.2; 1.9

In descending order = 8.2 x 1.9; 8.2; 1.9

Question 11. Let’s find values in decimal numbers

1. 0.625÷ 5
Solution:

⇒ \(\frac{625}{1000} \times \frac{1}{5}=\frac{125}{1000}=0.125\)

2. \(0.627 \div 3\)
Solution:

⇒ \(\frac{627}{1000} \times \frac{1}{3}=\frac{209}{1000}\)

= 0.209

Class 6 WBBSE Math Solutions

3. \(0.343 \div 7\)
Solution:

⇒ \(\frac{343}{1000} \times \frac{1}{7}\)=\(\frac{49}{1000}\)

= 0.049

4. \(651.2 \div 4\)
Solution:

⇒ \(\frac{6512}{10} \times \frac{1}{4}=\frac{1628}{10}\).

= 162.8

5. \(7 \div 3.5\)
Solution:

⇒ \(7 \div \frac{35}{10}=\frac{7 \times 10}{35}=\frac{70}{35}\)

= 2

6. \(18 \div 0.2\)
Solution:

⇒ \(18 \div \frac{2}{10}=\frac{18 \times 10}{2}\)

= 90

7. \(28.8 \div 1.2\)
Solution:

⇒ \(\frac{288}{10} \div \frac{12}{10}\)

= \(\frac{288}{10} \times \frac{10}{12}\)

= 24

8. \(11.7 \div 1.3\)
Solution:

⇒ \(\frac{117}{10} \times \frac{10}{13}\)

= 9

Class 6 WBBSE Math Solutions

9. \(1.35 \div 1.5\)
Solution:

⇒ \(\frac{135}{100} \div \frac{15}{10}\)

= \(\frac{135}{100} \times \frac{10}{15}\)

= \(\frac{9}{10}\)

= 0.9

10. \(0.65 \div 0.5\)
Solution:

⇒ \(\frac{65}{100} \div \frac{5}{10}\)

= \(\frac{65}{100} \times \frac{10}{5}\)

= \(\frac{13}{10}\)

= 1.3

11. \(0.07 \div 0.5\)
Solution:

⇒ \(\frac{7}{100} \div \frac{5}{10}\)

= \(\frac{7}{100} \times \frac{10}{5}\)

= \(\frac{14}{100}\)

= 0.14

Class 6 WBBSE Math Solutions

Question 12. Let’s make a story and then work out:

1. 2.50 x 5 + 3.25 x 6
Solution:

= 12.50 +19.50 = 32

2. 3.75 x 8 – 2.50 x 3
Solution:

Find the difference of prices of 8 pens (each cost Rs. 3.75) and 3 pencils (each cost Rs. 2.50)

= Rs. 3.75 x 8 – Rs. 2.50 x 3

= Rs. (30 – 7.50) = Rs. 22.50

3. 22.50 x 3.50
Solution:

If the cost price of 1 kg rice is Rs. 22.50, then the cost price of 3.50 kg rice

= Rs. 22.50 x 3.50 = Rs. 78.75

4. 13.75 x 2 + 12.50 x 3
Solution:

The total cost price color boxes (the price of each Rs. 13.75) and 3 pencils (price of each Rs. 12.50)

Rs. 13.75 x 2 + Rs. 12.50 x 3

= Rs. 27.50 + Rs. 37.50 = Rs. 65.

Question 13. Let’s find the values of the following simplification:

1. 13.28 – 4.07 + 2.7 x 0.02
Solution:

13.28-4.07 + 0.054 = 13.334-4.07

= 9.264

2. {45.85 – (6.29 + 15.06)} ÷ 5
Solution:

{45.85 -21.35} ÷ 5

= 24.5-5

{45.85 – (6.29 + 15.06)} ÷ 5 = 4.9

Class 6 WBBSE Math Solutions

3. (7.8 – 7.8 x 0.2) ÷ 1.2
Solution:

= (7.8-1.56) ÷ 1.2

= 6.24 ÷ 1.2

(7.8 – 7.8 x 0.2) ÷ 1.2 = 5.2

4. 0.35 x 0.35 + 0.15 x 0.15 + 2 x 0.35 x 0.15
Solution:

0.1225 + 0.0225 + 0.105

0.35 x 0.35 + 0.15 x 0.15 + 2 x 0.35 x 0.15 = 0.25

5. {(4 – 2.07) x 2.5} -1.93
Solution:

= (1.93 x 2.5) -1.93

= 4.825 – 1.93

{(4 – 2.07) x 2.5} -1.93 = 2.5

WBBSE Solutions For Class 6 Maths Chapter 2 Concept Of Seven And Eight Digit Numbers

Class 6 Math WBBSE Solutions Chapter 2 Concept Of Seven And Eight Digit Numbers

Example: Population of Howrah District 48,41,638 Population of Kolkata District 44, 86, 679 Population of Howrah District
Solution:

WBBSE Solutions For Class 6 Maths Chapter 2 Concept Of Seven And Eight Digit Numbers Popilation Of Howrah District

48 lac 41 thousand 6 hundred thirty-eight

Population of Kolkata District

Forty-four lac eighty-six thousand six hundred seventy-nine

Population of Purulia District 29,27, 975

Population of Jalpaiguri District 38,69,675

Read and Learn More WBBSE Solutions For Class 6 Maths

Population of Purulia District

WBBSE Solutions For Class 6 Maths Chapter 2 Concept Of Seven And Eight Digit Numbers Popilation Of Purulia District

29 lac 27 thousand 9 hundred seventy-five

Population of Jalpaiguri District,

WBBSE Solutions For Class 6 Maths Chapter 2 Concept Of Seven And Eight Digit Numbers Popilation Of Jalpaiguri District

38 lac 69 thousand 9 hundred seventy-five

WBBSE Solutions For Class 6 Maths Chapter 2 Concept Of Seven And Eight Digit Numbers Place Values In Words

Let’s write a seven-digit number and expand it.

WBBSE Solutions For Class 6 Maths Chapter 2 Concept Of Seven And Eight Digit Numbers Seven Digit Number And Expanding

Using each digit only once, let us write four numbers of seven digits and then write the numbers in words.

WBBSE Solutions For Class 6 Maths Chapter 2 Concept Of Seven And Eight Digit Numbers Four Numbers Of Seven Digits Then Write In Numbers In Words

Question 1. With the help of sticks and colored balls, a number of children who took the Polio Vaccine in four districts are shown below.
Solution:

28,22,000 (approx) children in North 24-Parganas

WBBSE Solutions For Class 6 Maths Chapter 2 Concept Of Seven And Eight Digit Numbers Polio Vaccine Children In North 24 Paraganas

26,10,500 (approx.) children in South 24 – Parganas.

WBBSE Solutions For Class 6 Maths Chapter 2 Concept Of Seven And Eight Digit Numbers Polio Vaccine Children In South 24 Paraganas

24,12,300 (approx.) children in Kolkata

Class 6 Math WBBSE Solutions

WBBSE Solutions For Class 6 Maths Chapter 2 Concept Of Seven And Eight Digit Numbers Polio Vaccine Children In Kolkata 24, 12, 300

22,34,200 (approx.) children in Kolkata

WBBSE Solutions For Class 6 Maths Chapter 2 Concept Of Seven And Eight Digit Numbers Polio Vaccine Children In Kolkata 22,34,200

Total number of children from Kolkata and Howrah who were given the Polio Vaccine.

WBBSE Solutions For Class 6 Maths Chapter 2 Concept Of Seven And Eight Digit Numbers Polio Vaccine Children In Kolkata And Howrah

Total number of children from 24 – Parganas, North, South, Kolkata, and Howrah, who all got Polio Vaccine.

Class 6 Math WBBSE Solutions

In Ten Lac stick, more than 10 balls can not be placed. Hence another strict or ‘place’ is needed. The name of this stick or ‘Place’ is “Crore”(c). Instead of 10 orange balls, I take 1□ ball.

WBBSE Solutions For Class 6 Maths Chapter 2 Concept Of Seven And Eight Digit Numbers Ten Lac Stick More Than 10 Balls Cannot Be Placed

Let us put balls in respective sticks and write the number:

WBBSE Solutions For Class 6 Maths Chapter 2 Concept Of Seven And Eight Digit Numbers Put Balls In Respective Sticks

  1. One crore eleven lac six hundred five
  2. Five crore two lac eigth thousand sixty-nine

Question 4. In the morning, people of our locality are very busy. They have queued up before our school to cast a vote. I could not do so, because my age is less than 18 years. I later came to know the results of the election.
Solution:

Party A got 46, 87, 905 votes

Party B got 44, 50, 896

Party C got 42, 95, 739 votes

The total number of votes parties have got is 13, 4, 34, 540 votes

Total number of voters are 135, 07, 375 in this locality

Total number of voters 135,07, 375

Votes that are cast 134,34, 540

Votes that are not cast or canceled = 72,835

Let us find which party won the election and by how many votes.

Class 6 Math WBBSE Solutions

Out of 46, 87, 905; 44, 50, 896, and 42, 95, 739, 46, 87, 905 is the greatest number.

Hence, Party A won the elections.

Let us find how many votes Party A got more than Party B.

Party A got 46, 87, 905 Party B got 44, 50, 896 Party A won by 2, 37, 009

WBBSE Class 6 Maths Solutions Chapter 2 Concept Of Seven And Eight Digit Numbers Exercise 2

Question 1. 30439872 ÷ 516
Solution:

⇒ \(\text { 516) } 30439872(58992 \\
2580 \\
4639 \\
\underline{4128} \\
5118 \\
\frac{4644}{4747} \\
\frac{4644}{1032} \\
\frac{1032}{\mathrm{x}}\)

Answer: 58992

Question 2. ▭ x 32=75285600
Solution: Required number =75285600 ÷ 32

⇒ \(\text { 32) } 75285600 \text { ( } 2352675 \\
\frac{64}{112} \\
\frac{96}{168} \\
\frac{160}{85} \\
\frac{64}{216} \\
\frac{192}{240} \\
\frac{224}{160} \\
\frac{160}{x}\)

Answer: 2352675

Question 3. 9128088 ÷ 388=
Solution:

⇒ \(\begin{aligned}
& 388) 9128088(23526 \\
& \frac{776}{1368} \\
& \frac{1164}{2040} \\
& \frac{1940}{1008} \\
& \frac{776}{2328} \\
& \frac{2328}{x}
\end{aligned}\)

Answer: 23526

Question 4. ▭ ÷ 297=65833
Solution: Required number = 65833 x 297

⇒ \(\begin{array}{r}
& 65833 \\
& 297 \\
& 460831 \\
& 592497 x \\
& 131666 x x \\
& \hline 19552401
\end{array}\)

WBBSE Class 6 Maths Solutions

Question 5. 30,08,00000 ÷ ▭=2,00000
Solution:

Required number =300800000 ÷ 200000 =1504

⇒ \(\begin{aligned}
& 200000) 300800000(1504 \\
& \frac{200000}{1008000} \\
& \frac{1000000}{800000} \\
& \frac{800000}{x}
\end{aligned}\)

Question 6. 752 x ▭ =3556208
Solution:

Required number =3556208 ÷ 752 =4729

⇒ \(\begin{aligned}
& \text { 752) } 3556208(4729 \\
& \frac{3008}{5482} \\
& \frac{5264}{2180} \\
& \frac{1504}{6768} \\
& \frac{6768}{x}
\end{aligned}\)

Class 6 Maths Solutions WBBSE Chapter 2 Concept Of Seven And Eight Digit Numbers Exercise 2.1

Question 1. Let us write in words

1. 782005
Solution: Seven lacs eighty-two thousand five

2. 4207029
Solution: Forty-two lacs seven thousand twenty-nine.

3. 30030030
Solution: Three crores thirty thousand thirty

4. 50505005
Solution: Five crores five lacs five thousand five

WBBSE Class 6 Maths Solutions

5. 42034047
Solution: Four crores twenty lacs thirty-four thousand forty-seven

Question 2. Let’s write these numbers using numerals

1. Seventy-eight lac eight hundred eight
Solution: 7800808

2. Ninety-three lac forty-four thousand six hundred five
Solution: 9344605

3. Three crore three lac three thousand three hundred three
Solution: 30303303

4. Thirty-three crore thirty-three lac thirty-three thousand thirty-three.
Solution: 333333033

5. Seventy-seven crore seven thousand seven
Solution: 770007007

Question 3. Let’s match the two sides

WBBSE Solutions For Class 6 Maths Chapter 2 Concept Of Seven And Eight Digit Numbers Let Match The Two Sides

Solution: 1. D, 2. C, 3. A, 4. E, 5. B

Question 4. Let us write by choosing the correct answer

1. Twenty lac ten thousand eight

  1. 2001008
  2. 2010008
  3. 2100008

Solution: 2. 2010008

WBBSE Class 6 Maths Solutions

2. One crore eleven lac eight thousand forty-one 

  1. 11018041
  2. 11010841
  3. 11108041

Solution: 3. 11108041

3. Two crore three lac sixty thousand five hundred twenty-six

  1. 20360526
  2. 20365026
  3. 20360562

Solution: 1. 20360526

Question 5. Let us write the expanded from of the following numbers

1. 4627593
Solution: 4000000 + 600000 + 20000 + 7000 + 500 + 90 + 3

2. 2213101
Solution: 2000000 + 200000 + 10000 + 3000 + 100 + 1

3. 9999999
Solution: 9000000 + 900000 + 90000 + 9000 + 900 + 90+9

4. 7007007
Solution: 7000000 + 7000 + 7

5. 2406739
Solution: 2000000 + 400000 + 6000 + 700 + 30 + 9.

Question 6. In the number 37452129, let us find the difference between the place values of the digit ‘2’.
Solution:

In the number 37452129,

Place values of 2 are 2000 and 20.

∴ The difference between 2000 and 20 is (2000-20) = 1980.

Question 7. In the number 27946138, let us find the difference between the place value and the actual value of ‘9′
Solution:

In the number 27946138, the place value of 9 is 900000.

∴ The difference between 900000 and 9 is (900000 – 9) = 899991.

Question 8. Let’s form greatest and least numbers of 8 digits with the digits given below:

1. 3, 5, 7, 9, 2, 6, 5, 6
Solution:

Greatest number = 9 7 6 6 5 5 3 2

Least number = 2 3 5 5 6 6 7 9

2. 6, 4, 8, 5,1,2, 0, 3
Solution:

Greatest number — 8 6 5 4 3 2 1 0

Least number- 1 0 2 3 4 5 6 8

3. 7, 3, 2,1, 9, 5, 6, 0
Solution:

Greatest number – 9 7 6 5 3 2 1 0

Least number – 1 0 2 3 5 6 7 9

4. 8, 9, 2, 4, 7, 3, 2, 1
Solution:

Greatest number – 9 8 7 4 3 2 2 1

Least number – 1 2 2 3 4 7 8 9

Class 6 Math Solutions WBBSE

Question 9. Let us write the following numbers in ascending (increasing) order:

1. 7525762,7525662,7526762,7525652
Solution: 7525652 < 7525662 < 7525762 < 7526762

2. 8705321,8702358, 8707341,8703741
Solution: 8705321 <8703741 <8705321 <8707341

3. 518896,872300, 27562, 300252
Solution: 27562 < 300252 < 518896 < 872300.

Question 10. Let us write the following numbers in descending (decreasing) order:

1. 4503210,4503201,4503120, 4502210
Solution: 4503210 > 4503201 > 4503120 > 4502210

2. 301516,8640051,302560, 6352289
Solution: 8640051 > 6352289 > 302560 >301516

3. 5102080, 5108200, 5100280, 5182000
Solution: 5182000 > 5108200 > 5102080 > 5100280

Question 11. The sum of two numbers is 82945195. If one number is 69100278, let us find the other number.
Solution: Other number = 82945195 – 69100278 = 13844917

Question 12. The difference between two numbers is 28351036. If one number is 30529179, let us find the other number.
Solution: Other number = 30529179 + 28351036 = 58880215

Question 13. From a factory in Bakultala, last year there was an income of Rs. 7521200. This year there is an additional earning of Rs. 3250325. Let us find the total earning in two years.
Solution:

The total earning in two years

= Rs. 7521200 + Rs. 3250325

= Rs. 10771525

Class 6 Math Solutions WBBSE

Question 14. The product of the numbers is 15050490. If one number is 5, let us find the other number.
Solution: Other number = 15050490 ÷ 5 = 3010098.

Question 15. By selling certain property Samirbabu got Rs. 35629850. He gave Rs. 1006200 to his wife and divided Rs. 13050000 among his three children. Rent of the money he donated for constructing a village school. Let us find
Solution:

1. Each child got = 13050000 ÷ 3 = 4350000

2. He donated for the school

= Rs. 35629850 – (1006200 + 13050000)

= Rs. (35629850-23112000)

= Rs. 12517850

Question 16. The population of a city is two crore ninety-eight lac seventy-two thousand six hundred. Of them 12500500 are men and 8872435 are women. Let us calculate the number of children.
Solution:

Total population = 29872600

⇒ \(\begin{array}{ll}
\text { Number of men } & =12500500 \\
\text { Number of women } & \equiv 8872435 \\
\hline \text { Total number of men and women } & \equiv 21372935
\end{array}\)

∴ Number of children = 29872600 – 21372935 = 8499665

Question 17. Let us find what least number must be added to 234567 so that it is divisible by 835.
Solution:

⇒ \(\begin{aligned}
& 835) 234567(280 \\
& \frac{1670}{6756} \\
& \frac{6680}{767}
\end{aligned}\)

∴ Required number = 835 – 767 = 68

Question 18. An organization bought two paintings of an artist for Rs. 900000 and Rs. 2000000 and two more paintings of another artist for Rs. 30,21636 and Rs. 1761084. Let us find how much has been spent on paintings by the organization.
Solution:

Total price of 4 paintings

= Rs. 900000 + Rs. 2000000 + Rs. 3021636 + Rs. 1761084

= Rs. 7682720

∴ Organisation spent on paintings = Rs. 7682720

Class 6 Math Solutions WBBSE

Question 19. The area of a country is 3287263 sq. km. It has 754740 sq. km of forest area and 2503000 sq. km area is river basin. Let us find the area of the remaining land leaving forest and river basin areas.
Solution:

Total area of the country = 3287263 sq. km.

Area of forest = 754740 sq. km.

and area of river basin = 2503000 sq. km.

Total area of forest = 3257740 sq. km.

and river basin

Remaining Land = (3287263 – 3257740) sq. km = 29523 Sqkm.