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Class 7 Math Solution WBBSE Geometry Chapter 3 Properties Of Triangle Exercise 3 Solved Problems

⇒ Triangle: A plane figure bounded by three line segments is called a triangle.

 

 

⇒ In triangle ABC, three vertices are A, B and C. There are three sides AB, BC, and CA.

⇒ The side opposite to a vertex of a triangle is its base.
⇒ The sum of the measurement of three angles is 180°.

Read and Learn More WBBSE Solutions for Class 7 Maths

Classification of Triangles:

According to their sides, three triangles are

1. Equilateral triangle
2. Isosceles triangle
3. Scalene triangle

According to their angles, three triangles are

1.  Acute angled triangle
2. Right-angled triangle
3. Obtuse angled triangle

Equilateral triangle:

⇒ A triangle whose length of each side are equal is called an equilateral triangle.

 

 

⇒ In the equilateral triangle ABC,

⇒ AB = BC = CA and ∠A = ∠B = ∠C = 60°

“WBBSE class 7 maths geometry chapter 3 solutions”

Isosceles triangle: An isosceles triangle is a triangle that has two sides are equal in length.

 

 

In isosceles triangle ABC, AB = AC

Wbbse Class 7 Maths Solutions

Scalene triangle:

⇒ A scalene triangle is a triangle whose all sides are unequal in lengths.

 

 

⇒ In ΔABC, AB ≠ BC ≠ CA

“properties of triangle WBBSE class 7 maths notes”

⇒ Acute angled triangle: A triangle in which all three angles are acute (i.e. less than 90°) is called an acute angled triangle.

 

 

⇒ PQR is an acute-angled triangle in which all angles are acute.

⇒ Right-angled triangle: A triangle whose one angle is a right angle (i.e., 90°) is Q called a right-angled triangle.

 

 

⇒ The side opposite to the right angle is called the hypotenuse.

⇒ ΔABC is a right-angled triangle whose ∠ABC= 90° and AB² + BC² = AC² [Pythagorus theorem]

Obtuse angled triangle: A triangle in which one angle is obtuse (i.e. more than 90°) is called an obtuse angled triangle.

 

 

⇒ ABC is an obtuse-angled triangle.

Right-angled isosceles triangle:

⇒ A triangle in which one angle is a right angle and the adjacent sides of a right angle are equal in length is called a right-angled isosceles triangle.

In isosceles right-angled triangle ABC, ∠ABC= 90° and AB = BC.

 

“exercise 3 solved problems on properties of triangles class 7”

⇒ The altitude or the height of a triangle: The length of the perpendicular from the vertex to the opposite side of a triangle is called the altitude or the height of the triangle.

 

 

⇒ In the adjacent image, AD is the altitude of the ΔABC.

⇒ The median of the triangle:

⇒ The median of a triangle is the line segment drawn from any vertex to the midpoint of the opposite side.

 

 

⇒ In ΔABC, AD is a median.

⇒ Area of the triangle = \(/frac{1}{2}\) × base x height.

⇒ The sum of length of any two sides of any triangle is greater than the length of the third side.

Question 1. Choose the correct answer

1. The number of medians of a triangle is

1. 1
2. 2
3. 3
4. None of these

 

 

Solution: There are three medians of a triangle ABC
⇒ AD, BE, and CF.

⇒ So the correct answer is 3.3

2. If the measurement of one angle of a triangle is 105° the triangle is

1.  Acute angled triangle
2. Right-angled triangle
3. Obtuse angled triangle
4. None of these

Solution: 105° is an obtuse angle. The triangle is an obtuse-angled triangle.

So the correct answer is 3. Obtuse angled triangle

“WBBSE class 7 maths chapter 3 important questions”

3. In which triangle the length of the median and height are same?

1. Equilateral triangle
2. Isosceles triangle
3. Scalene triangle
4. None of these

Solution: In an equilateral triangle height and median are the same.

So the correct answer is 1. Equilateral triangle

Question 2. Write true or false 

 

 

1. The position of medians of a triangle always inside the triangle.

Solution: The statement is true.

2. The hypotenuse of the right angled triangle is the smallest side.

Solution:  The statement is false.

3. These are three altitudes in a triangle.

Solution: The statement is true.

Question 3. Fill in the blanks

1. The sum of the measurement of three angles of a triangle is _____

Solution:180°

2. The number of vertices is _____

Solution: 3

3. Triangle has _____ diagonal.

Solution: No

Question 4. If the length of two adjacent side of right angle of a right angled triangle are 3 cm and 4 cm. Then find the length of its hypotenuse.

Solution:

 

Given

If the length of two adjacent side of right angle of a right angled triangle are 3 cm and 4 cm.

ΔABC its, ∠ABC = 90°

AB = 3 cm, BC = 4 cm

AC² =AB²+ BC² [by applying Pythagorus theorem]

AC²= 3² + 4²

AC²= 9+16

AC²= 25

AC = √25 cm = 5 cm

“how to solve triangle problems step-by-step class 7”

Question 5. If length of base and height of a triangle are 10 cm and 8 cm respectively then find the area of the triangle.

Solution

Given

If length of base and height of a triangle are 10 cm and 8 cm respectively

Area of the triangle = \(\)\frac{1}{2}] base x height

= 40 sq.cm

Area of the triangle = 40 sq.cm

Question 6. Whether the length of three sides of a triangle are 4 cm, 5 cm and 9 cm is possible?

Solution: The length of biggest side of any triangle is less than the sum of lengths of other two sides.
As 4+5=9
So, the given length of three sides of the triangle is not possible.

“sum of angles in a triangle theorem WBBSE class 7”

Question 7. Find the perimeter of the triangle of lengths of sides are 6 cm, 7 cm and 8 cm.

Solution: Perimeter = (6+ 7+ 8) cm = 21 cm.

Properties Of Triangle

Class 7 Math Solution WBBSE Properties Of Triangle Exercise 3.1

Question 1. Let us try to draw and find the number of heights (altitudes) for a triangle.
Solution :

In Δ ABC, AD, BE & CF are three*altitudes, They meet at O on the side of the triangle.

Question 2. Let us try to draw the heights of all the triangles in the same way according to sides and angles, with the help of a compass and scale.
Solution :

1. In case of Isosceles triangle:

Here PQR is an isosceles triangle PS, QT & RM are three heights of a triangle drawn from vertexes P, Q & R respectively. They meet at point O.

2. In case of Equilateral and Acute angled triangles:

⇒ Here ABC is an equilateral & acute angled triangle. AD, BE & CF are the three heights of the triangle, they meet at 6.

3. In case of the Scalene triangle:

⇒  ABC is a scalene triangle AD, BF & CF are the three heights of the triangle.

4. In the case of the obtuse-angled triangle:

⇒ Here ABC is an obtuse-dangled triangle perpendiculars are drawn from the three vertexes.

5. In case of a right-angled triangle:

⇒ ABC is a right-angled triangle ∠ ABC = 90°

Question 3. Which of these triangles has its height as one of its sides? Let’s draw and find.
Solution :

⇒ In the case of a right-angled triangle. AB is the height and also one side of the triangle.

Question 4. Which of these triangles have their heights and medians are the same line segment, let’s draw and find.
Solution :

⇒ ABC is an equilateral triangle whose each side is 6 cm.

⇒ Here AD, BE & CF are their height, also AD, BE & CF are the three medians of the triangle, as D, E & F are the midpoints of the side BC, CA & AB, respectively.

∴ Height & medians are the same line segment

Class 7 Math Solution WBBSE Properties Of Triangle Exercise 3.2

Question 1. Given 

1. Let’s write how many medians are there in a triangle.
Solution: 3 Medians are there in a triangle.

2. Let’s find how many do the three medians of a triangle meet.
Solution: Three medians of a triangle meet at a point.

3. Let’s write how many heights are there in a triangle.
Solution: There are 3 heights of a triangle.

4. Let’s find how many points the three heights (altitudes) of a triangle meet.
Solution: Three heights (altitudes) of a triangle meet at a point.

5. Let’s identify the triangles whose medians and heights (altitudes) are the same.
Solution:  Equilateral & Isosceles triangle.

“types of triangles and their properties class 7 maths”

Question 2. Let us draw all the types of triangles classified according to sides and angles and then draw 3 medians for each of these triangles using a scale and compass. From the figure drawn, let us try to find if the medians always remain inside the triangle.
Solution:

1. In the case of equilateral triangle:

In this case, medians meet at ‘O’ inside the triangle.

2. In the case of Isosceles + Acute angled triangle:

In this case, medians meet at ‘O’ inside the triangle.

3. In the case of scalene triangle:

⇒ In these medians meet at ‘O’ in the side of the triangle.

4. In the case of right-angled triangle:

⇒ In these medians meet at ‘O’ which is the mid-point of the hypotenuse

5. In the case of the obtuse-angled triangle:

⇒ In these medians meet at ‘O’ which is outside the triangle.

Question 3. Let us measure the height of each of the triangles given below with a scale and pencil.

Solution:

⇒ In the 1st figure, right of Δ ABC = 3.5cm, In the 2nd fig. height of

⇒ A DEF = 4cm & in the 3rd figure height of Δ PQR = 2.5 cm.

“Pythagoras theorem and right-angled triangles WBBSE”

Question 4. Let us draw the triangles classified according to the angles. Let us draw the height (altitudes) using a scale and compass and find if all the heights remain inside the triangle.
Solution :

1. In the 1st figure ABC is an acute-angled triangle Height AM = 4cm

2. In the 2nd figure, DEF is a right-angled triangle Height DE = 3cm.

3. In the 3rd figure PQR is an obtised angled triangle. Height PS = 5.2cm.

4. ln> The first two figure height is inside the triangle. But in 3rd figure oblused angled triangle height is outside the triangle)

Class 7 Math Solution WBBSE Geometry Chapter 2 Drawing Angles With Compass Exercise 2 Solved Problems

Question 1. To construct an angle of 90°

Solution:

Method Of Construction

 

 

1. BC is any straight line. With B as a centre and with any radius I draw an arc with a compass so that it cuts the line segment BC at D.
2. With centre D and with the same radius, I draw an arc which cuts the previous arc at E.
3. Again with centre E and with the same radius, I draw another arc, to cut the first arc at F.
4. With E and F as centres and with the same radius two more arcs are drawn to cut each other at A. Points A and B are joined.

Thus ∠ABC = 90°

Read and Learn More WBBSE Solutions for Class 7 Maths

Question 2. To construct an angle of 45°

Solution:

Method Of Construction

 

1. At first ∠ABC is drawn where ∠ABC = 90°

2. Then bisect angle ∠ABC with compass, so that ∠ABD = ∠DBC = 45°

“WBBSE Class 7 Maths Geometry Chapter 2 Drawing Angles with Compass solutions step-by-step”

Question 3. To construct the angles of 60°, 30° and 15°

Solution:

Method Of Construction

1. BC is any straight line with B as a centre and with any radius, an arc is drawn which cuts BC at P. P as a centre and with the same radius another arc is drawn to cut the previous arc at Q. I joined B and B produced to A. Thus ∠ABC 60°

 

 

2. I bisect the ∠ABC by BD I get 30° i.e., ∠DBC = 30°

 

 

3.  Again bisecting ∠DBC by BE I get ∠EBC= 15°

 

 

“West Bengal Board Class 7 Geometry Chapter 2 Drawing Angles with Compass solved problems”

Question 4. To construct the angles of 120° and 75°

Solution:

Method Of Construction

 

 

From the adjacent

∠CQR = 120°

∠PQR = 90° and ∠DQR = 60°

So∠PQD = ∠PQR – ∠DQR

∠PQD = 90° – 60°

∠PQD = 30°

∠PQD = 30°

Bisecting angle ∠PQD by QE we get ∠EQD, where∠EQD = 15°

So ∠EQR = ∠EQD +∠DQR

∠EQR= 15° + 60°

∠EQR= 75°

∠EQR = 75°

Drawing Angles With Compass

Drawing Angles With Compass Exercise 7

Question 1.  At point an on the line segment AB, let’s construct an angle of go with the help of scale, pencil and compass, let’s construct the angle; 120° 75° and 60°= angles at the same point,
Solution :

Given

⇒ At an AB, first, we draw an angle of 60°, 90° & 120°

⇒ Here ∠ BAC = 90°

⇒ ∠ BAD = 120°

⇒ ∠ BAE = 75°

⇒ ∠ BAF = 60°

“West Bengal Board Class 7 Maths Geometry Chapter 2 important questions and answers”

2. Using scale, pencil and compass, let’s construct the angles, 45 and 22\(\frac{1}{2}\)
Solution:

1. Take a line segment AO. Now with centre O, & with any radius draw an arc which cuts OA at P. Now with centre P, draw an arc which cuts the previous arc at Q.
2. Again with centre Q with the same radius an arc is drawn which cuts the previous arc at R. Now with centre Q and R, draw two arcs which cut each other at B. Join BO & AOB = 90° is formed.
3. Now by bisecting ∠ AOB, two angles ∠ AOC & ∠ COB are formed each equal to 45°.
4. Again by bisecting ∠ AOC, two angles ∠ AOD & ∠ COD two angles are formed, each equal to 22\(\frac{1}{2}\).

Question 3. Using scale, pencil and compass let’s draw the following angles –

1. 30°
2. 60°
3. 75°
4. 105°
5. 120°
6. 135°
7. 150°
8. 15°.

Solution:

Then, At O, draw an angle ∠ EOQ =.90°; ∠ GOQ = 60°& ∠ COQ = 120° (E).

⇒ Now by bisecting ∠ GOQ (= 60°) we get ∠ HOQ = 30s (A).

⇒  Again by bisecting ∠ HOQ (= 30°) we get ∠ KOQ = 15° (H)

⇒ Now by bisecting ∠ EOG, we get ∠ QOF = 75° (C)%

⇒ Again by bisecting ∠ COE we get ∠ DOQ =105° (D)

⇒ Now by bisecting ∠ POC (= 60°), we get Z AOQ = 150° (G)

⇒ Lastly by bisecting ∠AOC (= 30°), we get ∠ BOQ = 1 35° (F)

“WBBSE Class 7 Maths Geometry Chapter 2 exercise 2 solutions with explanations”

Question 4. Using scale, pencil and compass let’s draw an angle ∠ PQR of 60°. Let the side of RQ is produced to any point S, opposite to R. Then ∠ PQS = 120° Let’s also bisect the ∠ PQS and let’s verify using a protractor if ∠ PQS is bisected.
Solution:

Given

⇒  First ∠ PQR = 60° is drawn at Q, then RQ is produced to S, opposite to R.

⇒  ∠ PQS = 180° – 60° = 120° as, straight angle ∠RQS = 180°, Now

⇒  ∠ PQS is bisected by MQ

⇒  ∠MQP = ∠ MQS = 60°

“How to solve WBBSE Class 7 Maths Geometry Chapter 2 Drawing Angles with Compass questions”

Question 5. Using scale, pencil and compass let’s draw an angle ∠ ABC of measure 30°. Then let’s produce the side CB to a point D opposite to C. Now let’s draw a bisector BE of angle ∠ ABD. Measuring by protractor it is found ∠ DBE = 75° ∠ EBC = 105°
Solution :

Given

⇒ First ∠ ABC = 30° is drawn at B on the line segment BC.

⇒ Now CB is produced to a point D opposite to C.

⇒ Now BE is drawn as the bisector of ∠ ABD

⇒ Measuring by a protractor it is found ∠ DBE = 75° and ∠ EBC = 105°.

Class 7 Math Solution WBBSE Geometry Chapter 1 Revision Of Old Lesson Exercise 1 Solved Problems

Question 1. Draw a line segment AB of length 10 cm. Bisect the line segment AB with the help of a compass and measure each part.

Solution:

⇒ Method Of Construction:

 

1. I draw a line segment AB of length 10 cm
2. With A as a centre and with a radius more than half the length of AB, two arcs are drawn on either side of the line segment AB.
3. Again with B as the centre and with the radius two arcs are drawn on either of the line segment AB such that these two arcs intersect previous arcs at P and Q respectively. P and Q are joined with a scale and a pencil to get the midpoint O’ of the line segment AB.

Thus AO = OB = 5 cm.

“WBBSE Class 7 Maths Geometry Chapter 1 Revision of Old Lesson solutions step-by-step”

Read and Learn More WBBSE Solutions for Class 7 Maths

Question 2. Draw an angle of 75° with the help of a protractor, then bisect the angle with the help of a scale, pencil and compass.

Solution:

⇒ Steps for drawing the angle:

1. I draw an angle ∠ABC where ∠ABC = 75°
2. With centre B and with any radius, I draw an arc cutting the arms AB and BC of the ∠ABC at P and Q respectively.
3. With centre P and radius equal to more than half the length PQ, an arc is drawn.
4. Again with centre Q and with the same radius, an arc is drawn to the previous arc at R. Points B and R are joined and produced. The ray BR is the bisector of ∠ABC.

Thus ∠ABR=∠CBR = 37-5°

“West Bengal Board Class 7 Geometry Chapter 1 Revision of Old Lesson solved problems”

Question 3. With the help of a scale, pencil and compass, draw a line segment perpendicular to a line from a point outside it.

Solution:

⇒ Method Of Construction:

 

“WBBSE Class 7 Maths Geometry Chapter 1 exercise 1 solutions with explanations”

1. A straight line PQ is drawn and a point A is taken outside the straight line
2. I take any point T on the side of the straight line PQ opposite to point A.
3. With centre A and a radius AT, I draw an arc which cuts the straight line PQ at R and S.
4. With centres R and S and a radius equal to more than half of the line segment RS, two arcs are drawn on the side of the line PQ opposite to A, where two arcs intersect at B.
5. points A and B are joined with a scale and pencil. The line segment AB; cuts the straight line PQ at O.

∴ AO ⊥   and ∠AOP =∠AOQ = 90°

Algebra Chapter 8 Double Bar Graph Exercise 8 Solved Problems

⇒ A bar graph is a graphical representation of the numerical data by several rectangular bars of uniform width erected horizontally or vertically with equal spacing between them.

⇒ Each rectangle or bar represents only one value of the numerical data. The height or length of a bar indicates on a suitable scale the corresponding value of the numerical data.

This method of representing or explaining the statistical data is called the Bar diagram or Bar graph.

⇒ To draw a bar graph we draw two mutually perpendicular lines in the plane paper. The horizontal line is called the X-axis and the vertical line is known as the Y-axis.

⇒ If the bars are drawn vertically on a horizontal line, then the scale of heights of the bars is shown along the Y-axis. If the bars are drawn horizontally on the vertical line, then the scale of the height of the bars is shown along the X-axis. The bars can be shaded, hatched, or colored.

Read and Learn More WBBSE Solutions for Class 7 Maths

⇒ Double Bar Graph: When two bar graphs are drawn side by side along the same reference lines for better comparison of data as has been done in the graph, this is called a Double Bar Graph.

“WBBSE Class 7 Maths Algebra Chapter 8 Double Bar Graph solutions step-by-step”

Question 1. We prepared a single bar graph of the number of members from each 55 families in our locality

Scale: 1 unit = 1 family.

 

 

From the bar graph answers the following questions:

1. Out of 55 families, how many families have 4 members?
2. Out of 55 families, how many have a maximum number of family members?
3. From the bar graph find families have 5 members and members and families have 3 members.

Solution:

1. Out of 55 families, 15 families have 4 members.
2. Out of 55 families, the maximum number of family members is 5.
3.  From the graph we find 5 families have 5 members and 20 families have 3 members.

Question 2. Read the following double-bar graph and answer the following questions.

Scale: 1 unit = 1 ton.

 

 

1. What information is given by the bar graph?
2. Which state is the largest producer of rice?
3. Which state is the largest producer of wheat?
4. Which state has the total production of rice and wheat as its maximum?
5. Which state has the total production of rice and wheat minimum?

Solution:

1. It gives information regarding rice and wheat production in various states.
2. W. B.
3. U. P.
4. U. P.
5. Maharashtra.

“West Bengal Board Class 7 Maths Chapter 8 Double Bar Graph solved problems”

Question 3. Given below is the date of the number of clay dolls and sola dolls a potter in Krishnanagar has made in 5 months. Express the data through the double-bar graph.

Month	January	February	March	April	May
Number of clay dolls	600	550	450	750	900
Number of Sola dolls	500	450	600	650	700

Solution:

Scale: 1 Unit = 100 dolls

 

 

Question 4. After first continuous assessment, 6 friends worked in a group to grasp the subjects learnt through practical applications and through different methods. A table of the percentage of marks obtained in two continuous assessments after the second assessment.

Friends	Sumit	Rumki	Jahir	Merry	Joseph	Nazeen
1st assessments	45%	60%	55%	38%	72%	62%
2nd assessments	65%	65%	68%	60%	80%	70%

preparing the double bar graph

Solution:

Scale: 1 Unit = 1%

 

 

Question 5. The number of Male and Female people of a village in the year 2015-2018 are given the following table 

Year	2015	2016	2017	2018
Male	1200	1350	1450	1600
Female	1150	1200	1300	1450

Express the above information by a bar graph

Solution:

The number of Male and Female people of a village in the year 2015-2018 are given the table

Scale: 1 unit = 100 persons

 

 

 

Double Bar Graph

Double Bar Graph Exercise 8.1

Question 1. We prepared a single bar graph of the number of members from each of the 55 families in our locality. Let us study the bar graph and try to find the answers to the following question Scale: 1 unit = 1 family

1. Out of 55 families, how many families have 4 members?
Solution: 15

2. Out of 55 families how many have the maximum number of family members 
Solution: 3

Question 2. From the bar graph, we find families have 5 members and families have 3 members.
Solution: 5, 2

Question 2. A list of mountain peaks and their corresponding heights are given in the chart. Let us prepare a bar graph on squared paper taking 1 unit = 1000 metres.

Solution:

Single Bar graph of the Height of different Mountain peaks.

Question 3. For 55 students of class VII and 60 students of class 8 a list of their choice of games has been made. Let us express the data through a double-bar graph.

Solution:

Double bar graph of the students of Class 7 & Class 8 on their choice of games.

Question 4. Given below the date of the number of clay dolls and sola dolls a potter in Krishnakumar has made in 5 months. Let us express the data through a double-bar graph

Solution:

Bar Graph of Clay Dolls-& Sola Dolls of Krishnakumar made is 1st 5 months in a year.

“West Bengal Board Class 7 Maths Algebra Chapter 8 Double Bar Graph questions and answers”

Question 5. I made a list 50 students from my class according to the choices of colours as white, red, green, blue and black and then to express these daias through bar graph. [Let me do it myself.
Solution:

Given

⇒ I made a list 50 students from my class according to the choices of colours as white, red, green, blue

⇒ Bar graph of so students on their choice of colours

Question 6. The number of boys and girls of Tarai Tarapada Higher secondary school, in the last 4 years and also this year has been listed below. Let us express these data as a double bar graph. We all know the literacy rate has increased with time but let us find it whether the literacy rate of girls are more than boys or they are still backward

Solutions:

Given

⇒ The number of boys and girls of Tarai Tarapada Higher secondary school, in the last 4 years and also this year has been listed

⇒ Double bar graph: For increase in literacy for boys & girls for the year 2009 to 201 3 (5 years)

Question 7. After our first continuous assessment, we 6 friends worked in a group to grasp the subjects learnt through practical applications and through different methods. We prepared a table of the percentage of marks obtained in two continuous assessments after the second assessment. Preparing the double bar graph, let us understand, how much the new method helped to improve and who improved most.

Solution:

Double Bar Graph of Mark obtained (in percentage) in 1st & 2nd assessment by my 6 friends.

Question 8. The yearly production of cotton saris is woven by Utpal and Aminabibi from Phulia is shown is the double bar graph. Let us try to answer the following questions from the graph 

Given

The yearly production of cotton saris is woven by Utpal and Aminabibi from Phulia is shown is the double bar graph.

“WBBSE Class 7 Maths Algebra Chapter 8 Double Bar Graph exercise 8 solutions”

1. In which year did Utpal weave a maximum number of saris, and what is the number of saris woven? Again, in which year did he wove a minimum number of saris and how many are those, let’s find.
Solution:

2.  In which year did Aminabibi weave a maximum number of saris and how many are those? Also in which year did she wove a minimum number of saris and what is that number, let’s find out.
Solution:

3.  In which years did Utpal weave more saris than Aminabibi and in which year he wove a maximum number of saris than Aminabibi, let’s find.
Solution:

Utpal- 2008 & 2009 – 2008

“How to solve WBBSE Class 7 Maths Algebra Chapter 8 Double Bar Graph questions”

4. In which years did Aminabibi weave more saris than Utpal and in which year she wove a maximum number of saris than Utpal, let’s find.
Solution:

Aminabibi – 2010 & 201 0 – 2011

WBBSE Class 7 Math Solution Algebra Chapter 7 Formation Of Equation And Solutions Exercise 7 Solved Problems

1. Variable, constant, and ‘equal to’ signs are used to express the problem in the language of Mathematics. This method is called the framing of the equation.
2. The value of the variable used in the equation is called an unknown number.
3. The specific value of an unknown number for which the two sides of the sign are equal is the root or the solution of the equation.
4. The method of finding the value of the unknown number is called solving the equation.

Example: 3x – 5 = 7 [x is variable]
⇒ 3x = 5 +7
⇒ 3x = 12

⇒ \(x=\frac{12}{3}=4\)

Here 4 is the root of the equation 3x – 5 = 7.

Again, (x-2)² = x² – 4x + 4 is true for any value of the variable.

Let, x = 3, (x-2)² = (3-2)²= (1)² = 1

⇒ x² – 4x + 4 = (3)² – 4 x 3+4=1

It is true for x = 3

⇒ x=-5, (x-2)² = (-5-2)² = (-7)² = 49

⇒ x²-4x+4= (-5)²– 4 (-5)+ 4 = 25 +20 + 4 = 49

It is also true for x = – 5 so it is identified.

Read and Learn More WBBSE Solutions for Class 7 Maths

⇒ If both sides of the equation become equal for any value of the unknown quantity then this is called an identity.

⇒ An equation containing only one unknown quantity is known as a simple equation. Solving an equation is the process of finding its root.

Method of solving linear equations in one variable.

1. Simplify both sides of an equation and collect the like terms.
2. Multiply both sides by an appropriate factor (L. C. M of fractions) to remove fractions (if any).
3. Keep all the variable terms on LHS and the constant terms on the RHS following the rules of transposition.
[Any term may be brought from one side of the equation to the other by simply changing its sign. This is called transposition.]
4. Simplify both sides.
5. Divide both sides by the coefficient of the variable. The resulting coefficient of the variable becomes 1.
6. Thus the solution or root of the equation is obtained.

“WBBSE Class 7 Maths Algebra Chapter 7 solutions step-by-step”

Question 1. Choose the correct answer 

1. If \(\frac{x}{2}-\frac{5}{6}=1 \frac{2}{3}\) then the root of the equation is

1. 2
2. 3
3. 5
4. 6

Solution:

Given

If \(\frac{x}{2}-\frac{5}{6}=1 \frac{2}{3}\)

The root of the equation is 5.
So the correct answer is 3. 5

2. If \(\frac{x}{2}-\frac{2}{5}=\frac{x}{3}+\frac{1}{4}\) then value of x is

1. \(1 \frac{9}{10}\)

2. \(2 \frac{9}{10}\)

3. \(3 \frac{9}{10}\)

4. \(4 \frac{9}{10}\)

Solution:

Given

If \(\frac{x}{2}-\frac{2}{5}=\frac{x}{3}+\frac{1}{4}\)

So the correct answer is 3. \(3 \frac{9}{10}\)

3. If the sum of \(\frac{1}{5}\)th and \(\frac{1}{7}\) th of a number is 144, then the number is

1. 288
2. 420
3. 1008
4. 720

Solution:

Given

If sum of \(\frac{1}{5}\)th and \(\frac{1}{7}\) th of a number is 144

Let the required number be x.

According to question, \(\frac{x}{5}\) + \(\frac{x}{7}\)=144

The number is 420
So the correct answer is 2. 420

Question 2. Write ‘true’ or ‘false’ 

1. For x= \(\frac{2}{3}\) the expressions (3 + 2x) and (1-x) are equal.

Solution 3 + 2x = 1 − x
⇒ 2x + x = 1-3
⇒ 3x= -2

⇒x=\(-\frac{2}{3}\)

The statement is false.

“West Bengal Board Class 7 Algebra Chapter 7 Formation of Equations solved problems”

2. The root of equation 6 (7-3x) + 12x = 0 is 7.

Solution:

The statement is true.

Wbbse Class 7 Maths Solutions

3. 2x-3= \(\frac{3}{10}\)(5x-2), then the value of x is \(\frac{24}{5}\)

Solution: 2x-3=\(\frac{3}{10}\) (5x-2)

⇒ 20x – 30=15x – 6
⇒ 20x – 5x = 30 -6
⇒ 5x = 24

⇒ x = \(\frac{24}{5}\)

The statement is true.

Question 3. Fill in the blanks

1. If \(\frac{2x}{3}\) = \(\frac{3x}{8}\) + \(\frac{7}{12}\) then the value of x = _____

Solution:

Given

If \(\frac{2x}{3}\) = \(\frac{3x}{8}\) + \(\frac{7}{12}\)

The value of x =2

2. When x = ____, then \(\frac{ax+b}{3}\) and \(\frac{cx+d}{2}\) are equal.

Solution: \(\frac{ax+b}{3}\) = \(\frac{cx+d}{2}\)

⇒ 2ax + 2b = 3cx + 3d
⇒ 2ax – 3cx =3d – 2b
⇒ x(2a – 3c) = 3d – 2b

⇒ x = \(\frac{3d-2b}{2a-3c}\)

“WBBSE Class 7 Maths Chapter 7 Algebra equation formation exercise solutions”

3. If the measurement of the three angles of a triangle is x°, 2x, and 3x°, then the triangle is a ________ triangle.

Solution:

Given

The measurement of the three angles of a triangle are x°, 2x, and 3x°

The sum of the measurement of the angles of a triangle is 180°
So x°+ 2x° + 3x° = 180°

⇒ 6x° = 180°

⇒x° = \(\frac{180°}{6}\) = 30°

⇒ 3x° = 3 x 30° = 90°
=2x° = 2 x 30° 60°

The triangle is right-angled.

Question 4. Solve the following equations

1. \(\frac{x}{6}\) + \(\frac{3x-1}{12}\) + \(\frac{x-5}{18}\) = 4

2. 0.5x + \(\frac{x}{3}\) = 0.25+7

3. \(\frac{ax}{b}\) – \(\frac{bx}{a}\) =a² -b²

4. \(\frac{1}{6}\) (x−6) + \(\frac{1}{3}\) (x – 3)= \(\frac{1}{4}\) (x-4) + \(\frac{1}{5}\)  (x-5)

5. \(\frac{3x+1}{16}\) + \(\frac{2x-1}{7}\) = \(\frac{x+3}{8}\) + \(\frac{3x-1}{14}\)

6. \(\frac{x-a}{b}\) + \(\frac{x-b}{4}\) + \(\frac{x-3a-3b}{a+b}\)=0

Solution:

1. \(\frac{x}{6}\) + \(\frac{3x-1}{12}\) + \(\frac{x-5}{18}\) = 4

⇒ \(\frac{6x+9x-3+2x-10}{36}\) =4

⇒ \(\frac{17x-13}{36}\) = 4

⇒ 17x-13= 144
⇒ 17x= 144 + 13
⇒ 17x = 157

⇒ x = \(\frac{157}{17}\) = 9 \(\frac{4}{17}\)

2. 0.5x + \(\frac{x}{3}\) = 0.25 + 7

 

3. \(\frac{ax}{b}\) – \(\frac{bx}{a}\) =a² -b²

Wbbse Class 7 Maths Solutions

 

 

4. \(\frac{1}{6}\) (x−6) + \(\frac{1}{3}\) (x – 3)= \(\frac{1}{4}\) (x-4) + \(\frac{1}{5}\)  (x-5)

 

5. \(\frac{3x+1}{16}\) + \(\frac{2x-1}{7}\) = \(\frac{x+3}{8}\) + \(\frac{3x-1}{14}\)

 

6. \(\frac{x-a}{b}\) + \(\frac{x-b}{4}\) + \(\frac{x-3a-3b}{a+b}\)=0

Wbbse Class 7 Maths Solutions

 

“How to solve WBBSE Class 7 Maths Algebra Chapter 7 equations”

 Question 5. Frame and solve the equations

1. In the fruit shop there are \(\frac{1}{3}\) part apples, \(\frac{2}{7}\) part is oranges, and the remaining 160 are pears. Find out the total number of fruits.

Solution:

Given

In the fruit shop there are \(\frac{1}{3}\) part apples, \(\frac{2}{7}\) part is oranges, and the remaining 160 are pears.

Let the total number of fruits be x [x > 0]

∴ Number of apples is \(\frac{x}{3}\)

The number of oranges is \(\frac{2x}{7}\)

Remaining 160 are pears.

∴ Total number of fruits is 420.

2. The ratio of the length and breadth of a rectangle is 5: 2 and its perimeter is 140 cm. Find the area of the rectangle.

Solution:

Given

⇒ The ratio of the length and breadth of a rectangle is 5: 2 and its perimeter is 140 cm.

⇒ Let x be a common factor of the ratio where x > 0.

∴ Length of the rectangle is 5x cm and breadth is 2x cm.

⇒ The perimeter is 2(5x + 2x) cm = 14x cm.

⇒According to given condition, 14x= 140

⇒ x= \(\frac{140}{14}\) = 10

∴ Length is (10 x 5) cm or 50 cm and breadth is (10 x 2) cm or 20 cm.

∴ Area of the rectangle is (50 x 20) sq. cm = 1000 sq. cm.

“WBBSE Maths Class 7 Chapter 7 Formation of Equations and Solutions exercise answers”

3. Arun babu borrows some money to build his house. He returned ₹ 2000 more than \(\frac{1}{3}\)rd of the money he borrowed. However, he still has to repay amount of money he borrowed. However he still has to repay ₹ 21000. Find the amount of money he borrowed

Solution:

Given

⇒Arun babu borrows some money to build his house. He returned ₹ 2000 more than \(\frac{1}{3}\)rd of the money he borrowed.

⇒ Let the amount of money Arun babu borrowed be  ₹ x

⇒ He returned ₹ ( \(\frac{x}{3}\) + 2000)

⇒However he still has to reply ₹ 21000.

⇒ According to given condition, x=( \(\frac{x}{3}\) + 2000 + 21000) + \(\frac{x}{3}\) + 2000

⇒ x- \(\frac{x}{3}\) – \(\frac{x}{3}\) = 25000

⇒ \(\frac{3x-x-x}{3}\) = 75000

∴ The amount of money he borrowed ₹ 75000.

4. The sum of the ages of A and B is 54 years. 2 years ago \(\frac{2}{3}\)rds A’s age was greater than \(\frac{3}{4}\)ths the present age of B by 12 years. What are their present ages?

Solution:

Given

⇒ The sum of the ages of A and B is 54 years. 2 years ago \(\frac{2}{3}\)rds A’s age was greater than \(\frac{3}{4}\)ths the present age of B by 12 years.

⇒ Let present age of A is x years.

∴ Present age of B is (54 – x) years.

2 years ago A’s age was (x – 2) years.

⇒ According to given condition, \(\frac{2}{3}\)(x-2) =\(\frac{3}{4}\)(54-x)+12

The present age of A is 38 years and that of B is (5438) years or 16 years.

“West Bengal Board Class 7 Maths Algebra Chapter 7 questions and answers”

5. In a two digit number the digit in the unit’s place is 2 more than the digit in ten’s place. If the sum of the digits is 12 then find the number.

Solution:

Given

In a two digit number the digit in the unit’s place is 2 more than the digit in ten’s place. If the sum of the digits is 12

Let the digit in ten’s place be x

∴ The digit in unit place is (x + 2)

∴ The number is (10x+x+2)= (11x+2)

According to question, x + x + 2 = 12

⇒ 2x= 12-2 = 10

⇒x=\(\frac{10}{2}\) =5

∴ The required number is 11 x 5 + 2 = 57

6. A train traveled over a certain distance in a certain time at the rate of 30 km/hr. If the speed had been 25 km/hr, he would have taken 36 minutes more to cover the same distance. Find the distance.

Solution:

Given

⇒ A train traveled over a certain distance in a certain time at the rate of 30 km/hr. If the speed had been 25 km/hr, he would have taken 36 minutes more to cover the same distance.

⇒ Let the required distance is x km.

⇒ The time taken to cover the distance x km at 30 km/hr is \(\frac{x}{30}\)

⇒ The time taken to cover the distance x km at 25 km/hr is \(\frac{x}{25}\)

⇒ According to condition \(\frac{x}{25}\) – \(\frac{x}{30}\) = \(\frac{36}{60}\)

The required distance is 90 km

“WBBSE Class 7 Algebra Chapter 7 Formation of Equations important questions”

7. The difference between the numerator and denominator of a fraction is 5. If 2 is added to both the numerator and denominator, the fraction would become  \(\frac{2}{3}\) Find 3′ the fraction.

Solution:

Given

⇒ The difference between the numerator and denominator of a fraction is 5. If 2 is added to both the numerator and denominator, the fraction would become  \(\frac{2}{3}\)

⇒ Let the numerator of the fraction is x.

∴ Denominator is (x + 5). So the fraction is \(\frac{x}{x+5}\)

According to question \(\frac{x+2}{x+5+2}\) = \(\frac{2}{3}\)

⇒ \(\frac{x+2}{x+7}\) = \(\frac{2}{3}\)

⇒ 3x+6= 2x + 14
⇒ 3x – 2x = 14-6
⇒x= 8

∴ The fraction is \(\frac{8}{8+5}\) = \(\frac{8}{13}\)

WBBSE Algebra Chapter 6 Factorisation Exercise 6 Solved Problems

1. Factorisation of an algebraic expression is the process of finding two or more expressions whose product is the given expression.

⇒ These two or more expressions are called factors.
⇒ Factors of 6 are 1, 2, 3 and 6 [6 = 1 x 6 = 2 x 3]

⇒ The prime factors of 6 are 2, 3

2. Every number or expression is a factor of itself and 1 is a factor of any number or expression.

3. Generally 1 and the number itself may not be written to mention the factors of a number. The factors of 3ab are 3, a, and b where 3, a, and b are prime factors.

⇒ As 6x²y= 2 x 3 x x x x x y

Read and Learn More WBBSE Solutions for Class 7 Maths

Common factor: An expression that can divide all the terms of a polynomial is called the common factor of the polynomial.

Example: 4x³– 12x + 16x
= 4x x x² – 4x x 3 + 4x x 4
= 4x (x² – 3 + 4)

Here 4x is the common factor.

Method of factorization:

1. Dividing all the terms of the expression by the largest common factor which is written outside a bracket and putting the resulting quotient inside the bracket.

2. we arrange the terms of the given algebraic expression in groups such that each group has a common factor. Then we factorize each group separately we take out factors which is common to each group.

3. If any expression is the difference of the square of two expressions, its factors will be the sum of those two expressions and their difference.
i.e., an expression of the form a²– b² can be expressed as the product of (a + b) and (a – b).

“WBBSE Class 7 Maths Algebra Chapter 6 solutions”

4. Some expressions which are not in the form a² – b², can be expressed as the difference of two squares by adding or subtracting some quantity to or from those expressions.

Wbbse Class 7 Maths Solutions

Question 1. Choose the correct answer 

1. The number of prime factors of 9x²y is

1. 2
2. 3
3. 4
4. 5

Solution: 9x²y= 3 x 3 x x x x x y
The number of prime factors of 9x²y is 5

⇒ So the correct answer is 4. 5

2. The common factor of 14ab² and 21ab is

1. 7ab
2. 7a
3. 7b
4. 7ab²

Solution: 14ab² = 2 x 7 x a x b x b
21 ab = 3 x 7 x a x b

The common factor of 14ab² and 21ab is (7 x a x b) or 7ab

⇒ So the correct answer is 1. 7ab

3. The sum of factors of (2x – 6) is

1. 2x-5
2. x – 1
3. x – 3
4. x-8

Solution: 2x – 6
= 2(x-3)

Sum of factors is 2 + x – 3 = x – 1

⇒ So the correct answer is 2. x – 1

4. If two factors of a expression are (x + 1) and (x 1), then the expression is

1. 2x
2. 2
3. x²-1
4. None of these

Solution: The expression is (x + 1)(x − 1) = x²– 1

⇒ So the correct answer is 3. x²-1

“Factorisation Class 7 WBBSE solved problems”

Question 2. Write ‘true’ or ‘false’

1. The common factor of 10x²y and 15xy² is 5xy

Solution: 10x²y=2 x 5 x x x x x y
15xy² =3 x 5 x x x y x y

The common factor is (5 x x x y) or 5xy

⇒ So the statement is true.

2. The sum of factors of (a³-a²+ a) is (a² + 1)

Solution: a³ – a² + a = a (a² – a + 1)

The sum of factors = a2 +1 = a2 + 1

⇒ So the statement is true.

3. Sum of factors of (3a² – 12a) is (4a + 4).

Solution: 3a² – 12a = 3a(a – 4)

Sum of factors= 3+a+a-4 = 2a – 1

⇒ So the statement is false.

“Exercise 6 Algebra solutions WBBSE Class 7”

Question 3. Fill in the blanks 

1. The common factor of 18x², 27x³ and 45x is _____

Solution:
18x² = 2 x 3 x 3 x x x x
27x³ = 3 x 3 x 3 x x x x x x
45x = 3 x 3 x 5 x x

⇒ The common factor is (3 x 3 x x) or 9x.

2. The sum of factors of (ax + bx- ay- by) is

Solution: ax + bx – ay – by
= x (a + b) -y (a + b)
=(a+b) (x-y)

⇒ Sum of factors = a+b+x-y.

3. The factors of (ab – 5b + a – 5) are ____

Solution: ab – 5b + a – 5
= b(a – 5) + 1(a-5)
= (a – 5)(b + 1)

⇒ The factors are (a – 5) and (b + 1).

Question 4. Find the number of prime factors of a(x² – y²).

Solution: a(x²– y²) = a(x + y)(x − y)

The prime factors are a, (x + y), and (x − y).
∴ The number of prime factors is 3.

Question 5. Find the sum of factors of (a³ – a).

Solution: a³ – a
= a(a² – 1)
= a(a + 1)(a− 1)

⇒ Sum of factors = a + a + 1+ a -1 = 3a

“WBBSE Class 7 Maths Chapter 6 Factorisation”

Question 6. Resolve into factors

1. a² – 1 + 2b – b²

2. x²– 2x – y²+ 2y

3. 81x⁴+4y⁴

4. 3x⁴ + 2x²y²– y⁴

5. a⁴ + a²b² + b⁴

6. \(a^2+\frac{1}{a^2}+1\)

7. xy(1 + z²) +z (x² +y²)

8. a² – 5a+6

9. 2a²b²+2b²c² + 2c²a²-a⁴-b⁴-c⁴

10. x² + 2x – 899

Solution:

1. a² – 1 + 2b – b²

2. x²– 2x – y²+ 2y

3. 81x⁴+4y⁴

4. 3x⁴ + 2x²y²– y⁴

5. a⁴ + a²b² + b⁴

6. \(a^2+\frac{1}{a^2}+1\)

7. xy(1 + z²) +z (x² +y²)

8. a² – 5a+6

9. 2a²b²+2b²c² + 2c²a²-a⁴-b⁴-c⁴

10. x² + 2x – 899

“Step-by-step solutions for WBBSE Class 7 Algebra”

Question 7. Factorize the following

1. (a + b)⁴+ 4

2. a²(b-c)²-b²(c – a)²

3. 64ap²– 49a (p- 2q)²

4. a² – b² -6ap + 2bp + 8p²

5. 4a⁴ + 11a² + 25

Solution:

1. (a + b)⁴+ 4

2. a²(b-c)²-b²(c – a)²

3. 64ap²– 49a (p- 2q)²

4. a² – b² -6ap + 2bp + 8p²

5. 4a⁴ + 11a² + 25

Factorization

Factorization Exercise 6.1

Let us factorize the following :

1. 25xy = 5 × 5 × x × y

2. 18 × y² = 2 × 3 × 3 × x × y × y

3. 15q²r²= 3 × 5 × q × q× r × r

4. 10 × yz = 2 × 5 × x × y × z

5. 12 × yz = 2 × 2 × 3× x × y × z

Factorization Exercise 6.2

Let us factorize the following :

1. 12x²y (x + 2)= 2 × 2 × 3 × x × x × y × (x + 2)

2. 1 8yz² (2y + 3z) = 2 × 3 × 3 × y × z × z × (2y + 3z)

3. 1 6 × yz (x + y)= 2 × 2 × 2×2 × x × y × z × (× + y)

4. 15pq² (p + 3q) = 3 × 5 × p × q × q × (p + 3q)

5. 14mn² (2m – n) = 2 × 7 × m × n × n × (2m-n)

Question 1. Let us factorize:

1. 2+14x
Solution:

2+14x  = 2(1 +7x)

2. 5x – 20y
Solution:

5x – 20y = 5 ( x- 4y)

3. 6x – 3y
Solution:

6x – 3y = 3 (2x – y)

4. 3a² – 12a
Solution:

3a² – 12a = 3a (a -4)

Question 2. Let us find the common factors of the following algebraic expressions.

1. 6a, 2a²
Solution:

6a, 2a²= 2, a, 2a:

2. 5x, 6xy
Solution:

5x, 6xy = x

3. 4xyz, 12yz
Solution:

4xyz, 12yz = 2, 4, yz, yz, 2y, 2z, 4y, 4z, 4yz.

4. 7a²b, 14abc
Solution:

7a²b, 14abc =  7, a, b, 7a, 7b, ab, 7ab

Class 7 Math Solution WBBSE

Factorization Exercise 6.3

Question 1. Let us Factorise the following 

1. xy + y + 3x + 3
Solution:

Given

xy + y + 3x + 3

xy + y + 3x + 3= y (x +1 ) + 3 (× + 1 )

= (x + 1 )(y + 3)

xy + y + 3x + 3 = (x + 1 )(y + 3)

2. pq – q + 2q – 2
Solution:

Given

pq – q + 2q – 2

pq – q + 2q – 2 = q (P – 1 ) + 2 (p – 1 )

= (p – 1 ) (q + 2)

pq – q + 2q – 2 = (p – 1 ) (q + 2)

3. 6xy + 3y + 4× + 2
Solution:

Given

6xy + 3y + 4x + 2

6xy + 3y + 4x + 2 = 3y (2x + 1) + 2 (2x + 1)

= (2x +1) (3y + 2)

6xy + 3y + 4x + 2 = (2x +1) (3y + 2)

4. 10xy + 2y + 5x + 1
Solution:

Given

10xy + 2y + 5x + 1

10xy + 2y + 5x + 1 = 2y (5x + 1 ) + 1 (5x + 1 )

= (5x + 1 ) (2y + 1 )

10xy + 2y + 5x + 1 = (5x + 1 ) (2y + 1 )

Question 2. Let us find the factors of the following algebraic expressions :

1. 7xy 
Solution:

7xy = 7 × x × y

2. 9 x²y
Solution:

9 x²y= 3 × 3 × x × x ×y

3. 16ab²c 
Solution:

16ab²c = 2 × 2 × 2 × 2 × a × b × b × c

4. -25lmn = 5 × 5 × | × m × n
Solution:

-25lmn = 5 × 5 × | × m × n

5. 12x (2 + x) 
Solution:

12x (2 + x) = 2 × 2 × 3 × a × (2 + x )

6. – 5 pq (p² + 8)
Solution:

– 5 pq (p² + 8) = -5 × p × q × (p² +8)

7. 21 xy² (3x – 2)
Solution:

21 xy² (3x – 2) = 3 × 7 × x × y × y × (3x -2)

8. 121mn (m² – n)
Solution: 

121mn (m² – n) = 11 × 11 × m × n × (m² -n)

Question 3. Let us find the common factors of the following algebraic expressions :

1. 22xy, 33xz
Solution:

22xy, 33xz

Common factors = 11 , x, 11x

2. 14ab² , 21ab
Solution: 

14ab² , 21ab

Common factors = 7, a, b, 7a, 7b, 7ab

3. – 16mnl, – 39nl²
Solution:

– 16mnl, – 39nl²

Common factors = – 1, – n, – 1, nl, – nl, n, I

4. 12a²b, 18ab², 24abc
Solution:

12a²b, 18ab², 24abc

Common factors = 2, 3, 6, a, b, 2a, 2b, 3a, 3b, 6a, 6b, ab, 2ab, 3ab, 6ab

5. 2xy, 4yz, 6xz
Solution:

2xy, 4yz, 6xz

Common factors = 2.

6. 18x², 27x³, – 45x
Solution:

18x², 27x³, – 45x

Common factors = 3, 9, x, 3x, 9x

7. 5mn, 6n²l², 7l³m²
Solution:

5mn, 6n²l², 7l³m²

Common factors = 1.

Question 4. Let us write two such algebraic e×pressions which have the following as their Common Factors 

1. x²
Solution:

⇒ x² =   3x², 4x³3y

2. 2xy
Solution:

2xy = 4x²y²q, 6xyq

3. 4a²
Solution:

4a²= 12a³y, 16a²x

4. (mn + 2)
Solution:

(mn + 2) =  5(mn + 2), 7(mn + 2)²

5. x (y + 2)
Solution:

x (y + 2) = 7x²(y + 2), 9x(p + q)(y + 2)

“Class 7 WBBSE Maths Factorisation exercise solutions”

Question 5. Let us factorize the following 

1. 5 + 10x
Solution:

5 + 10x = 5(1 +2x)

2. 2x – 6
Solution:

2x – 6 = 2(x – 3)

3. 7m – 14n 
Solution:

7m – 14n = 7(m – 2n)

4. 18xy + 21 xz 
Solution:

18xy + 21 xz = 3x(6y + 7z)

5. 4xy + 6yz 
Soution:

4xy + 6yz = 2y(2x + 3z)

6. 7xyz – 6xy
Solution:

7xyz – 6xy = xy(7z – 6)

7. 7a² + 14a
Solution:

7a² + 14a = 7a(a + 2)

8. – 15m + 20 
Solution:

– 15m + 20 = 5(- 3m + 4)

9. 6a²b + 8a²b
Solution:

6a²b + 8a²b = 2ab(3a + 4b)

10. 3a² – ab²
Solution:

3a² – ab² = a (3a – b²)

11. abc – bcd =
Solution:

abc – bcd = bc(a – d)

12. 60xy³ – 4xy – 8 
Solution: 

60xy³ – 4xy – 8 = 4(15xy³ – xy – 2)

13.  x²yz + xy²z + xyz²
Solution:

x²yz + xy²z + xyz² = xyz(x + y + z)

14.  a³ – a² + a 
Solution:

a³ – a² + a = a(a² – a + 1 )

15. x²y²z² + x²y² + x²y²q² 
Solution:

x²y²z² + x²y² + x²y²q² = x²y²(z² +1 + q²)

Question 6. Let’s Factorise the following algebraic e×pressions 

1. xy + 2x + y + 2
Solution :

Given

xy + 2x + y + 2

= x(y + 2) + 1(y + 2) = (y + 2)(× + 1)

xy + 2x + y + 2 = (y + 2)(× + 1)

2. ab – 5b + a – 5
Solution:

Given

ab – 5b + a – 5

= b(a – 5) + 1 (a – 5) = (a – 5)(b + 5)

ab – 5b + a – 5 = (a – 5)(b + 5)

3. 6xy – 9y + 4x – 6
Solution :

Given

6xy – 9y + 4× – 6

= 3y(2x – 3) + 2(2x – 3) = (2x – 3)(3y + 2)

6xy – 9y + 4× – 6 = (2x – 3)(3y + 2)

4. 15m + 9 – 35mn – 21 n
Solution :

Given

15m + 9 – 35mn – 21 n

= 3(5m + 3) – 7n(5m + 3) = (3 – 7n)(5m + 3)

15m + 9 – 35mn – 21 n = (3 – 7n) (5m + 3)

5. ax + bx – ay – by 
Solution :

Given

ax + bx – ay – by

= x(a + b) – y(a + b) = (a +b)(x – y)

ax + bx – ay – by = (a +b)(x – y)

6. c – 9 + 9ab – abc
Solution :

Given

c – 9 + 9ab – abc

= – 1 (9 – c) + ab(9 – c) = (9 – c)(ab – 1)

c – 9 + 9ab – ABC = (9 – c)(ab – 1)

Class VII Math Solution WBBSE Factorization Exercise 6.4

Question 1. Let us factorize the following 

1. x² + 14× + 49
Solution :

Given

ײ + 14x + 49

= (x)² + 2.x.7 + (7)2 = (x + 7)² = (x + 7)(x+ 7)

ײ + 14x + 49 = (x + 7)(x+ 7)

2. 4m² – 36m + 81
Solution :

Given

4m² – 36m + 81

= (2m)² – 2.2m.9 + (9)²

= (2m – 9)² = (2m – 9)(2m – 9)

4m² – 36m + 81=(2m – 9)(2m – 9)

3. 25 x² + 30x + 9
Solution :

Given

25x² + 30x + 9

= (5x)² + 2.5 x .3 + (9)² = (5x + 3)² = (5x + 3) (5x + 3)

25x² + 30x + 9 = (5x + 3) (5x + 3)

4. 121b²– 88bc + 16c²
Solution :

Given

121b² – 88bc + 16c²

= (11b)² – 2.11 b.4c + (4c)² = (11 b – 4c)²

. = (11 b – 4c)(11 b – 4c)

121b² – 88bc + 16c² = (11 b – 4c)(11 b – 4c)

5. (xy)2 – 4x²y²
Solution :

Given

(x²y)² – 4x²y²

= x²y² – 4x²y² = .x²y²(x² – 4) = x²y²{(x)² – (2)2}

= x²y²(x + 2)(x – 2) = x.x.y.y.(x + 2)(x – 2)

(x²y)² – 4x²y² = x.x.y.y.(x + 2)(x – 2)

6. a⁴ + 4a²b² + 4b⁴
Solution :

Given

a⁴ + 4a²b² + 4b⁴

= (a²)² + 2.a².2b² + (2b)²

= (a² + 2b²)

= (a² + 2b²)(a² + 2b²)

a⁴ + 4a²b² + 4b⁴ = (a² + 2b²)(a² + 2b²)

7. 4x² – 1 6
Solution:

Given

4x²-16

= 4(x²– 4) = 4{(x)² – (2)²} = 2 × 2 × (x + 2)(x – 2)

4x²-16 = 2 × 2 × (x + 2)(x – 2)

8. 121 – 36x²
Solution :

Given

121 – 36x²

= (11)² – (6x)² = (11 + 6x)(11 – 6x)

121 – 36x² = (11 + 6x)(11 – 6x)

9. x²y² – p²q²
Solution :

Given

x²y² – p²q²

= (xy)² – (pq)²= (xy + pq)(xy – pq)

x²y² – p²q² = (xy + pq)(xy – pq)

10. 80m² -125
Solution :

Given

80m² -125

= 5(16m² – 25) = 5{(4m)² – (5)2} = 5(4m + 5)(4m – 5)

80m² -125 = 5(4m + 5)(4m – 5)

11. ax² – y²
Solution:

Given

ax² – y²

= ax² – ay² = a{(x)² – (y)²} = a (x + y)(x – y)

ax² – y²  = a (x + y)(x – y)

12.  I – (m + n)²
Solution :

Given

I – (m + n)²

= (I)² – (m + n)² = {I + (m + n)} {I – (m + n)}

= (I + m + n)(l – m – n)

I – (m + n)² = (I + m + n)(l – m – n)

13.  (2a – b – c)² – (a – 2b – c)²
Solution :

Given

(2a – b – c)² – (a – 2b – c)²

= {{2a – b – c) + (a – 2b – c)} {(2a – b – c) – (a – 2b – c)}

= (2a – b – c + a – 2b – c)(2a – b – c – a + 2b + c)

= (3a – 3b – 2c)(a + b)

(2a – b – c)² – (a – 2b – c)² = (3a – 3b – 2c)(a + b)

“Solved problems of Factorisation Class 7 WBBSE”

14.  x² – 2xy – 3y²
Solution :

Given

x² – 2xy – 3y²

= x² – (3 – 1 )xy – 3y² = x² – 3xy + xy – 3y²

= x(x – 3y) + y(x – 3y) = (x – 3y)(x + y)(xv)

x² – 2xy – 3y² = (x – 3y)(x + y)(xv)

15. x²+ 9y² + 6xy – z²
Solution :

Given

x²+ 9y² + 6xy – z²

= x + 6xy + 9y² – z²

= (x)² + 2.x.3y + (3y)² – (z)²

= (x + 3y)² – (z)²

= (x + 3y + z)(x + 3y – z)

x²+ 9y² + 6xy – z²  = (x + 3y + z)(x + 3y – z)

16. a² – b² + 2bc – c²
Solution :

Given

a² – b² + 2bc – c²

= a² – (b² – 2bc + c²) = (a)² – (b – c)²

= {a + (b – c)} {a – (b – c)} = (a + b – c)(a – b + c)

a² – b² + 2bc – c² = (a + b – c)(a – b + c)

WBBSE Class 7 Math Solution

17. a²(b-c)²-b²(c-a)²
Solution :

Given

a²(b-c)²-b²(c-a)²

= {a (b – c)}² – {b(c – a)}² = (ab – ac)² – (bc – ab)²

= {(ab – ac) + (bc – ab)} {(ab – ac) – (bc – ab)}

= (ab – ac + bc – ab) (ab – ac – be + ab)

= (bc – ac) (2ab – ac – bc)

= c(b – a) (2ab – bc – ca)

a²(b-c)²-b²(c-a)² = c(b – a) (2ab – bc – ca)

18. x² – y² – 6yz – 9z²
Solution :

Given

x² – y² – 6yz – 9z²

= x² – (y ²+ 6yz + 9z²)

= (x)² – {(y)² + 2.y.3z + (3z)²}

= (x)² – (y + 3z)²

= (x + y + 3z) (x – y – 3z)

x² – y² – 6yz – 9z²  = (x + y + 3z) (x – y – 3z)

19. x² – y² + 4x – 4y
Solution :

Given

x² – y² + 4x – 4y

= {(x)² -(y)} ² + 4(x-y)

= (x + y)(x – y) + 4(x – y) = (x – y)(x + y + 4)

x² – y² + 4x – 4y = (x – y)(x + y + 4)

20.  a² -b² + c² – d² – 2(ac – bd)
Solution :

Given

a² -b² + c² – d² – 2(ac – bd)

= a² – b²+ c² – d² – 2ac + 2bd

= a² – 2ac + c² – b² + 2bd – d²

= {(a)² – 2,a.c + (c)²} – (b² – 2bd + d²)

= (a – c)² – (b – d)²

= {(a – c) + (b – d)} {(a – c) – (b – d)}

= (a – c + b – d) (a – c – b + d)

= (a + b – c – d) (a – b – c + d)

a² -b² + c² – d² – 2(ac – bd) = (a + b – c – d) (a – b – c + d)

21.  2ab – a² – b² + c²
Solution :

Given

2ab – a² – b² + c²

= c² – a² + 2ab – b²

= (c)² – (a² – 2ab + b²) = (c)² – (a – b)²

= (c + a – b)(c – a + b) = (a – b + c)(b + c – a)

2ab – a² – b² + c² = (a – b + c)(b + c – a)

22. 36x² – 16a² – 24ab – 9b²
Solution :

Given

36x² – 16a² – 24ab – 9b²

= 36x² – (16a² + 24ab + 9b²)

= 36x² – {(4a)² + 2.4a.3b + (3b)²}

= (6x²) – (4a + 3b)²

= {6x + (4a + 3b)} {6x – (4a + 3b)}

= (6x + 4a + 3b) (6x – 4a – 3b)

36x² – 16a² – 24ab – 9b² = (6x + 4a + 3b) (6x – 4a – 3b)

23. a² – 1 + 2b – b²
Solution :

Given

a² – 1 + 2b – b²

= a² – (1 – 2b + b)²

= (a)² – (1 – b) = (a +1 – b) {a (1 b)}

= (a – b + 1 )(a + b – 1 )

a² – 1 + 2b – b² = (a – b + 1 )(a + b – 1 )

“Algebra Chapter 6 Factorisation WBBSE Class 7”

24. a² – 2a – b² + 2b
Solution :

Given

a² – 2a – b² + 2b

= a² – b² + 2b – 2a

= (a + b) (a – b) – 2(a – b)

= (a b)(a + b – 2)

a² – 2a – b² + 2b = (a b)(a + b – 2)

25. (a² – b²)(c² – d²– 4abcd)
Solution :

Given

(a² – b²)(c² – d²– 4abcd

(a² – b²)(c² – d²) – 4abcd

= a²c² – b²c²– a²d² + b²d² – 4abcd

= a²c² + b²d² -.2abcd – a²d²– 2abcd – b²c²

= {(ac)² – 2abcd + (bd)²} – {a²d² + 2abcd + b²c²}

= (ac – bd)² – (ad + bc)²

= {(ac – bd) + (ad + bc)} {(ac – bd) – (ad + bc)}

= (ac – bd + ad + bc) (ac – bd – ad – bc)

(a² – b²)(c² – d²– 4abcd = (ac – bd + ad + bc) (ac – bd – ad – bc)

26. a² – b² – 4ac + 4bc
Solution :

Given

a² – b² – 4ac + 4bc

= {(a)² – (b)²} – 4c(a – b)

= (a + b) (a – b) – 4c(a – b) = (a – b)(a + b – 4c)

a² – b² – 4ac + 4bc = (a – b)(a + b – 4c)

27. a² – b² – c²+ d²)² – 4(ad – bc)²
 Solution:

Given

a² – b² – c²+ d²)² – 4(ad – bc)²

= (a² – b² – c²+ d2)² – {2(ad – bc)}²

= (a² – b² – c² + d² + 2ad – 2bc)

= {a² + 2ad + d²– b² – 2bc – c2}

= [{(a)² + 2.a.d + (d)²} – {(b)² + 2.b.c + (c)²}]

= {(a + d)² – (b + c)²}

= {(a + d) + (b + c)} {(a + d) – (b + c)}

= (a + b + c + d) (a + d – b – c)

= (a + b + c + d)(a – b – c + d)

a² – b² – c²+ d²)² – 4(ad – bc)² = (a + b + c + d)(a – b – c + d)

28. 3x² – y² + z² – 2xy – 4xz
Solution :

Given

3x² – y² + z² – 2xy – 4xz

= 4x² – x² – y² + z² – 2xy – 4xz

= 4x² – 4xz + z² – x² – 2xy – y²

= {(2x)² – 2.2x.z + (z)²} – (x² + 2xy + y²)

= (2x – z)² – (x + y)²

= {(2x – z) + (x + y)} {(2x – z) – (x + y)}

= ( 2x – z + x + y) (2x – z – x – y)

= (3x + y – z) (x – y – z)

3x² – y² + z² – 2xy – 4xz = (3x + y – z) (x – y – z)

WBBSE Class 7 Math Solution Question 2. Let us factorize the following

1. 81x⁴ + 4v⁴
Solution :

Given

81x⁴+ 4y⁴

(9x²)² + (2y²)² + 2.9x².2y² – 36x2y²

= (9x² + 2y²)2 – (6xy)²

. = (9x² + 2y² + 6xy) (9x² + 2y² – 6xy)

= (9x² + 6xy + 2y²) (9x² – 6xy + 2y²)

81x⁴+ 4y⁴ = (9x² + 6xy + 2y²) (9x² – 6xy + 2y²)

2. p⁴ – 13p²q² + 4q⁴
Solution :

Given

p⁴ – 13p²q² + 4q⁴

= (p²)² – 2.p².2q² + (2q²)² -9p²q²

= (p² – 2q²)² – (3pq)² = (p² – 2q² + 3pq) (p²– 2q² – 3pq)

= (p² + 3pq – 2q²) (q² – 3pq – 2q²)

p⁴ – 13p²q² + 4q⁴ = (p² + 3pq – 2q²) (q² – 3pq – 2q²)

3. x⁸ – 16y⁸
Solution :

Given

x⁸ – 16y⁸

= (x⁴)² – (4y⁴)²

= (x⁴ + 4y⁴) (x⁴ – 4y⁴)

= {(x²)² + 2.x².2y² + (2y²)² – 4x²y²} {(x²)² – (2y)²}

= {(x² + 2y²)² – (2xy)²} (x² + 2y²) (x² – 2y²)

= (x² + 2y² + 2xy) (x² + 2y² – 2xy) (x² + 2y²) (x² – 2y²)

= (x² + 2xy + 2y²) (x² 2xy + 2y²) (x² + 2y²) (x² – 2y²)

x⁸ – 16y⁸= (x² + 2xy + 2y²) (x² 2xy + 2y²) (x² + 2y²) (x² – 2y²)

4. x⁴ + x²y² + y⁴
Solution :

Given

x⁴ + x²y² + y⁴

= (x²)² + 2x²y² + (y²)² – x²y²

= (x²+ y²)² – (xy)²

= (x² + y² + xy) (x² + y² – xy)

= (x² + xy + y²) (x² – xy + y²)

x⁴ + x²y² + y⁴ = (x² + xy + y²) (x² – xy + y²)

5. 3x⁴ + 2x²y² – y⁴
Solution:

Given

3x⁴ + 2x²y²– y⁴

= 3x⁴ + 3x²y² – x²y² – y⁴

= 3x²(x² + y²) – y²(x² + y²)

= (x² + y²) (3x² – y²)

3x⁴ + 2x²y²– y⁴ = (x² + y²) (3x² – y²)

6. x⁴ + x² + 1
Solution :

Given

x⁴ + x² + 1

= (x²)² + 2.x².1 +(1)²-x²

= (x² + 1 )² – (x)² = (x² + 1 +x) (x² + 1 – x)

= (x² + x + 1 ) (x² – x + 1 )

x⁴ + x² + 1 = (x² + x + 1 ) (x² – x + 1 )

7. x⁴ + 6x²y² + 8y⁴
Solution:

Given

x⁴ + 6x²y² + 8y⁴

= x⁴ + 4x²y² + 2x²y² + 8y²

= x²(x² + 4y²) + 2y²(x² + 4y²)

= (x² + 4y²)(x² + 2y²)

x⁴ + 6x²y² + 8y⁴ = (x² + 4y²)(x² + 2y²)

“WBBSE Class 7 Maths Exercise 6 Factorisation chapter”

8. 3x² – y² + z² – 2xy – 4xz
Solution :

Given

3x² – y² + z² – 2xy – 4xz

= 4x² – x² – y² + z² – 2xy – 4xz

= 4x² – x² – y² + z² – 2xy – 4xz

= 4x² – 4xz + z² – x² – 2xy – y²

= (4x² – 4xz + z²) – (x² + 2xy + y²)

= {(2x)² – 2. 2x.z + z)²} – {(x)² + 2.x.y + (y)²}

= (2x – z)² – (x + y)²

= {(2x – z) + (x + y)} {(2x – z) – (x + y)

= (2x – z + x + y) (2x – z – x – y)

= (3x + y – z) (x – y – z)

3x² – y² + z² – 2xy – 4xz = (3x + y – z) (x – y – z)

9. 3x⁴ – 4x²y² + y⁴
Solution :

Given

= 3x⁴ – 3x²y²– x²y² + y⁴

= 3x²(x² – y²) – y²(x² – y²)

= (x² – y²) (3x² – y²)

= {(x)² -(y)²} (3x² – y²)

= (x + y) (x – y) (3x² – y²)

3x⁴ – 3x²y²– x²y² + y⁴ = (x + y) (x – y) (3x² – y²)

10. p⁴ – 2p²q² – 15q²
Solution :

Given

p⁴ – 2p²q² – 15q²

= p⁴ – (5 – 3)p²q² – 15q⁵

= p⁴ – 5p² q² + 3p²q² – 15q²

= P²(P² 5q²) + 3q²(p² + 3q²)

= (p² – 5q²) (p² + 3q²)

p⁴ – 2p²q² – 15q² = (p² – 5q²) (p² + 3q²)

“How to solve Factorisation problems Class 7 WBBSE”

11. x⁸ + x⁴y⁴ + y⁸
Solution :

Given

x⁸ + x⁴y⁴ + y⁸

= x⁸ + 2x⁴y⁴ + y⁸ – x⁴y⁴

= {(x⁴)² + 2x⁴y⁴ + y⁴ + (y⁴)⁴} – (x⁴y⁴)⁴

= (x⁴ + y⁴)⁴ – (x⁴y⁴)⁴

= (x⁴ + y⁴ + x²y²) (x² + y⁴ – x²y²)

= {x⁴ + 2x²y² + y⁴ – x²y²} (x⁴ + y⁴ – x²y²)

= {(x²)² + 2x²y² + (y²)² – (xy)²} (x⁴ + y⁴ – x²y²)

= {(x² + y²) – (xy)²} (x⁴ + y⁴ – x²y²)

= (x² + y² + xy) (x² + y²– xy) (x⁴ + y⁴ – x²y²)

x⁸ + x⁴y⁴ + y⁸  = (x² + y² + xy) (x² + y²– xy) (x⁴ + y⁴ – x²y²)

Class 7 Math Solution WBBSE Algebra Chapter 5 Algebraic Formula Exercise 5 Solved Problems

Necessary Formula (Identities)

1. (a + b)² = a² + 2ab+ b²
⇒ (a + b)² = (a – b) ² + 4ab

2. (a – b) ² = a² – 2ab+b²
⇒ (a – b) ² = (a + b) ² – 4ab

3. (a+b+c) ² = a² + b² + c² + 2ab+ 2bc + 2ca

4. a²+ b² = (a + b) ² – 2ab
⇒= (a – b) ²+2ab

5. a²– b² = (a + b)(a − b)

6. 2(a² + b²) = (a + b) ² + (a – b) ²

7. 4ab = (a + b) ²– (a – b) ²

8. ab=(a+b) ² – (a=b) ²

Proof:

Read and Learn More WBBSE Solutions for Class 7 Maths

1. (a + b) ²= (a + b)(a + b).
= a(a + b) + b(a + b)
= a²+ ab + ab + b²
= a²+2ab+ b²
= (a² – 2ab+b²) + 4ab
= (a – b) ² + 4ab

2. (a – b) ² = (a – b)(a – b)
= a(a – b) – b(a – b)
= a² – ab – ab + b²
= a² – 2ab+b²
= (a²+2ab+ b²) – 4ab
= (a + b) ² – 4ab

3. (a + b + c) ² = {(a + b) + c)} ²
= (a + b) ² + 2(a + b)c + c²
= a ² + 2ab+ b ² + 2ac + 2bc + c2 a ²+ b  ² + c  ²+2ab+ 2bc + 2ca
= a² + b² = (a ²+2ab+b ²) – 2ab
= (a + b) ² – 2ab

“WBBSE Class 7 Maths Algebra Chapter 5 solutions”

4. a²+ b² = (a ²-2ab+b ²)+2ab
= (a – b)2 + 2ab

5. a² – b² = a² + ab – ab-b²
= a(a + b) – b(a + b)
= (a + b)(a – b)

6. 2(a ²+ b ²)= a² + b²+ a²+ b²
= (a ²+2ab+b ²) + (a ² – 2ab+b²)
= (a + b) ² + (a – b)²

7. 4ab = 2ab+2ab
= (a²+2ab+b²) – a²+2ab-b²
= (a²+2ab+b²) + (a²– 2ab+b²)
= (a + b) ² – (a – b)²

Class 7 Math Solution WBBSE

8. \(a b=\frac{4 a b}{4}=\frac{(a+b)^2-(a-b)^2}{4}\)

= \(\frac{(a+b)^2}{4}-\frac{(a-b)^2}{4}\)

= \(\left(\frac{a+b}{2}\right)^2-\left(\frac{a-b}{2}\right)^2\)

Question 1. Choose the correct answer

1. If (x+8)² = x² + 16x + K, then the value of K is

1. 16
2. 64
3. 8
4. None of these

Solution:

Given

(x+8)² = x² + 16x + K

⇒ x² + 2. x. 8+ (8)² = x² + 16x + K

⇒ 64 K

So the correct answer is 2. 64

2. For which value or values of K, will the expression p² + pk + \(\frac{1}{25}\) be a perfect square.

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

2. \(\frac{1}{25}\)

3. \(\pm \frac{2}{5}\)

4. \(\pm \frac{1}{5}\)

Solution:

Given

p² + pk + \(\frac{1}{25}\)

So the correct answer is 3. \(\pm \frac{2}{5}\)

3.  If a = \(\frac{1}{8}\) then the value of (64a²+ 16a+ 1) is

1. 8
2. 0
3.16
4. 4

Solution:

Given

a = \(\frac{1}{8}\)

64a² + 16a+ 1
= (8a)² + 2.8a.1 + (1)²
= (8a+ 1)²

= \(\left(8 \times \frac{1}{8}+1\right)^2\)

[Putting the value of a]
= (1 + 1)² = (2)²
= 4

⇒ So the correct answer is 4. 4

4. If 3x + \(\frac{1}{5 x}\) =6, then the value of \(25 x^2+\frac{1}{9 x^2}\)

1. 96

2. \(96 \frac{2}{3}\)

3. \(103 \frac{2}{3}\)

4. None of these

Solution:

Given

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

 

⇒ So the correct answer is 2. \(96 \frac{2}{3}\)

“Algebraic Formula Class 7 WBBSE solved problems”

5. If a + b = 7 and a b = 3, then the value of ab is

1. 21
2. 10
3. 58
4. 40

Solution:

Given

a + b = 7, a b = 3

ab = \(\left(\frac{a+b}{2}\right)^2-\left(\frac{a-b}{2}\right)^2\)

= \(\left(\frac{7}{2}\right)^2-\left(\frac{3}{2}\right)^2\)

= \(\begin{aligned}
& =\frac{49}{4}-\frac{9}{4} \\
& =\frac{40}{4}=10
\end{aligned}\)

⇒ So the correct answer is 2. 10

Question 2. Write ‘true’ or ‘false’

1. If (xy)²= (96y+ y)², then the value of x is 3.

Solution :

Given

(xy)² = 96y+ y²

⇒ (x-y)² = (3)²-2.3.y + y²

⇒ (x-y)²=(3-y)²

⇒ x – y = 3-y

⇒x=3-y+y

⇒ x = 3

⇒ So the statement is true.

2. The value of 1015 x 985 is 999085

Solution:

1015 x 985

= (1000+15) (1000-15)
= (1000)²– (15)²
= 1000000 – 225
= 999775

⇒ So the statement is false.

“Exercise 5 Algebra solutions WBBSE Class 7”

3. If (a-5)² + (b + 3)²= 0, then the value (a + b) is ²

Solution:

Given

If (a-5)² + (b + 3)²= 0

If the sum of two or more two square is zero then the value of each square will be zero.
(a-5)² = 0
⇒ a 5=0
⇒ a = 5

(b + 3)² = 0
⇒b+3=0
⇒ b = -3

a+b=5-3=2.

⇒ So the statement is true.

Question 3. Fill in the blanks

1. The value of (0.75 x 0.75+1.5 x 0.25 +0.25 x 0.25 is) _______

Solution: 0.75 x 0.75 + 1.5 x 0.25 + 0.25 × 0.25
= (0-75)² + 2 x 0.75 x 0.25 + (0.25)²
= (0·75 + 0.25)²
= (1.00)²
= 1

2. If a + b = 5 and ab= 6, then the value of (a – b) is _____

Solution:

Given

If a + b = 5 and ab= 6

(a – b)² = (a + b)² – 4ab
= (5)² – 4 x 6
= 25-24
= 1

⇒ a-b=±√l=±1

3. If a + b + c = 7 and ab+be+ca = -5, then the value of (a²+ b²+ c²) is _____

Solution:

Given

If a + b + c = 7 and ab+be+ca = -5

a² + b² + c² + 2 (ab + bc + ca) = (a + b + c)²
⇒ a² + b² + c² + 2 x (-5) = (7)²
⇒ a² + b² + c² = 49+ 10 = 59

Question 4. Find the square of the algebraic expression given below:

1. 4x+5y
2. 2a+3b-4c
3. a +2b3c4d
4. 999
5. 1005

Solution:

Given
1. (4x+5y)²
= (4x)² + 2.4x.5y + (5y)²
= 16x² + 40xy + 25y²
(4x+5y)² = 16x² + 40xy + 25y²

2. (2a+3b – 4c)²

= {(2a + 3b) – 4c)²

= (2a+3b)² – 2 x (2a + 3b) x 4c + (4c)²

= (2a)² + 2 x 2a x 3b + (3b)² – 8c(2a + 3b) + 16c²

= 4a² + 12ab+9b² -16ac – 24bc + 16c²

= 4a² + 9b² + 16c² + 12ab – 16ac – 24bc.

⇒ (2a + 3b) – 4c)² = 4a² + 9b² + 16c² + 12ab – 16ac – 24bc.

3. (a + 2b – 3c + 4d)²

= {(a + 2b) – (3c – 4d)}²

= (a + 2b)²– 2(a + 2b)(3c4d) + (3c-4d)²

= a²+2.a.2b + (2b)² – 2(3ac-4ad + 6bc-8bd) + (3c)² – 2.3c.4d + (4d)²

= a²+4ab + 4b² – 6ac + 8ad – 12bc + 16bd + 9c² – 24cd + 16d²

= a² + 4b² + 9c² + 16d² + 4ab – 6ac + 8ad -16bd – 24cd.

⇒ {(a + 2b) – (3c – 4d)}² = a² + 4b² + 9c² + 16d² + 4ab – 6ac + 8ad -16bd – 24cd.

4. (999)²

= (1000 – 1)²

= (1000)²-2 × 1000 × 1 + (1)²

= 1000000-2000 + 1

= 998001

⇒ (999)²= 998001

5. (1005)²
= (1000 + 5)²

= (1000)² + 2 x 1000 × 5 + (5)²

= 1000000 + 10000 + 25.

= 1010025

⇒ (1005)² = 1010025

Wbbse Class 7 Maths Solutions

Question 5. Express x as a difference of two square.

Solution: x = x.1

= \(\left(\frac{x+1}{2}\right)^2-\left(\frac{x-1}{2}\right)^2\)

Question 6. Express (8a²+ 50b²) as a sum of two square.

Solution:

Given

8a²+ 50b²

= 2 (4a² + 25b²)

= 2{(2a)² + (5b)²}

= (2a + 5b)² + (2a – 5b)²

Question 7. Find the square (3x+4y) with the help of (a – b)²= a² – 2ab+ b²

Solution:

(3x+4y)²

= {3x-(-4y)}²

= (3x)²– 2 x 3x(-4y) + (-4y)²

= 9x² + 24xy + 16y²

Question 8. For what values of p; will the expression (9x² + px + 16) be a perfect square.

Solution :

Given

9x² + px + 16

=(3x)²+2.3x.\(\frac{p}{6}\)+ \(\left(\frac{p}{6}\right)^2\) – \(\left(\frac{p}{6}\right)^2\) +16

= \(\left(3 x+\frac{p}{6}\right)^2-\frac{p^2}{36}+16\)

The given expression will be perfect square if

\(-\frac{p^2}{36}\)+16=0

\(-\frac{p^2}{36}\) = -16

⇒ p² = 16 x 36

⇒ p=±\(\sqrt{16 \times 36}\)

⇒p=14×6

⇒p±24.

“WBBSE Class 7 Maths Chapter 5 Algebraic Formula”

Question 9. Express the following in the product form

1. 49x⁴ – 36y⁴
2. (m + p + q)² – (m-p-q)²

Solution:

Given

1. 49x⁴ – 36y⁴

= (7x²)²– (6y²)²

= (7x²+6y²)(7x² – 6y²)

49x⁴ – 36y⁴ = (7x²+6y²)(7x² – 6y²)

Given

2. (m + p + q)²– (m-p-q)²

= (2m)(2p+ 2q)
= 2m x 2(p + q)
= 4m (p + q)

⇒ (m + p + q)²– (m-p-q)²  = 4m (p + q)

Question 10. For what value of t, will the expression \(\left(x^2-t x+\frac{1}{4}\right)\) be a perfect square.

Solution: \(x^2-t x+\frac{1}{4}\)

= \(x^2-2 \cdot x \cdot \frac{t}{2}+\left(\frac{t}{2}\right)^2-\left(\frac{t}{2}\right)^2+\frac{1}{4}\)

= \(\left(x-\frac{t}{2}\right)^2-\frac{t^2}{4}+\frac{1}{4}\)

The given expression is a perfect square.

So, \(-\frac{t^2}{4}+\frac{1}{4}=0\)

⇒ \(-\frac{t^2}{4}=-\frac{1}{4}=0\)

⇒ t² = 1

= t = ± 1

Question 11. Express the following as a perfect square and hence find the values

1. 49a² – 42ab+9b² [when a = 1, b = 2]

2. \(\frac{144}{p^2}-\frac{120}{p}+25\) [when p = -3]

Solution: 1. 49a² – 42ab+9b²
= (7a)² – 2 x 7a x 3b + (3b)²

= (7a-3b)²

= (7 x 1-3 x 2)²

= (7-6)²

= (1)² = 1

2. \(\frac{144}{p^2}-\frac{120}{p}+25\)

= \(\left(\frac{12}{p}\right)^2-2 \times \frac{12}{p} \times 5+(5)^2\)

= \(\left(\frac{12}{p}-5\right)^2\)

= \(\left(\frac{12}{-3}-5\right)^2\) [Putting p = -3]

= (-4-9)²

= (-13)² = 16

Question 13. If \(m-\frac{1}{m-5}=12\) then find the value of \((m-5)^2+\frac{1}{(m-5)^2}\)

Solution:

Given

\(m-\frac{1}{m-5}=12\)

The value of \((m-5)^2+\frac{1}{(m-5)^2}\) = 51

Wbbse Class 7 Maths Solutions

Question 14. If 3x – \(\frac{1}{x}\)=6, then find the value of \(\left(x^2+\frac{1}{9 x^2}\right)\)

Solution:

Given

3x – \(\frac{1}{x}\)=6

 

Question 15. If a² + b²= 52 and a – b = 2, then find the value of ab.

Solution:

Given

a² + b² = 52
⇒ (a – b)²+2ab = 52

⇒ (2)²+2ab = 52

⇒ 2ab = 52 – 4 = 48

⇒ ab= \(\frac{48}{2}\) =24

The value of ab = 24

“Step-by-step solutions for WBBSE Class 7 Algebra”

Question 16. If a + b = √3 and a b= √2, then find the value of

1.8ab (a² + b²)

2. \(\frac{3\left(a^2+b^2\right)}{a b}\)

Solution:

1. 8ab (a² + b²)

= 4ab x 2(a² + b²)
= {(a + b)² – (a – b)²} {(a + b)² + (a – b)²}

= {(V3)²– (√2)²}{(√3)² + (√2)²} [ Putting a + b = √3 and a b= √2 ]

= (3-2)(3 + 2)
= 1 x 5
= 5

8ab (a² + b²) = 5

2. \(\frac{3\left(a^2+b^2\right)}{a b}\)

= \(6 \times \frac{2\left(a^2+b^2\right)}{4 a b}\)

= \(6 \times \frac{(a+b)^2+(a-b)^2}{(a+b)^2-(a-b)^2}\)

= \(6 \times \frac{3+2}{3-2}=6 \times \frac{5}{1}\) = 30

\(\frac{3\left(a^2+b^2\right)}{a b}\) = 30

Question 17. Express as the difference of two squares

1. 72
2. (a + 3b)(2a – 5b)

Solution:
1. 72 = 9 x 8

\(\left(\frac{9+8}{2}\right)^2-\left(\frac{9-8}{2}\right)^2=\left(\frac{17}{2}\right)^2-\left(\frac{1}{2}\right)^2\)

 

2. (a + 3b)(2a 5b)=

Wbbse Class 7 Maths Solutions

Question 18. If x² – 3x-1= 0, then find the value of \(\left(x^4+\frac{1}{x^4}\right)\)

Solution:

Given

x² – 3x-1= 0

The value of \(\left(x^4+\frac{1}{x^4}\right)\) is 119.

“Class 7 WBBSE Maths Algebraic Formula exercise solutions”

Question 19. If x = \(a+\frac{1}{a}\) and y = \(a-\frac{1}{a}\) then find the value of (x² + y² – 2x²y²).

Solution:

Given

If x = \(a+\frac{1}{a}\) and y = \(a-\frac{1}{a}\)

The value of (x² + y² – 2x²y²) is 16.

 

Question 20. If \(\frac{a}{b}+\frac{c}{d}=\frac{b}{a}+\frac{d}{c}\) then prove that \(\frac{a^2}{b^2}-\frac{c^2}{d^2}=\frac{d^2}{c^2}-\frac{b^2}{a^2}\)

Solution:

Given

\(\frac{a}{b}+\frac{c}{d}=\frac{b}{a}+\frac{d}{c}\)

Wbbse Class 7 Maths Solutions

Question 21. Show that (a+b) (a² + b²)(a⁴ +b⁴)(a³ +b⁸ + b⁸) = \(\frac{a^{16}-b^{16}}{a-b}\)

Solution:

“Solved problems of Algebraic Formula Class 7 WBBSE”

Question 22. If a² + b² + c² = 2(a – b − 1), then find the value of (a + b + c).

Solution:

Given

a²+ b² + c² = 2(a – b-1)

⇒ a² + b² + c² = 2a-2b-2

⇒ a² – 2a + b² + 2b + c² + 2 = 0

⇒ (a² – 2a + 1) + (b² + 2b + 1) + c² = 0

⇒(a – 1)² + (b +1)² + c² = 0

If sum of square of two or more than two square is zero then each square will be zero.

(a -1)² = 0
⇒ a-1=0
⇒ a=1

(b + 1)2 = 0
⇒b+1=0
⇒ b = -1

c² = 0
⇒ c=0

a+b+c=1=1-1+0=0

The value of (a + b + c) is 0.

“Algebra Chapter 5 Algebraic Formula WBBSE Class 7”

Question 23. Prove that (a + b)⁴ – (a – b)⁴= 8ab (a² + b²).

Solution:
Proof:

(a + b)⁴ – (a – b)⁴

= \(\left\{(a+b)^2\right\}^2-\left\{(a-b)^2\right\}^2\)

= \(\left\{(a+b)^2+(a-b)^2\right\}\)\(\left\{(a+b)^2-(a-b)^2\right\}\)

=2(a²+b²) x 4ab

= 8ab(a²+ b²) (Proved)

(a + b)⁴ – (a – b)⁴= 8ab (a² + b²)

Question 24. Multiply (a+b+c) (a – b + c)(b + c – a)(a + b – c).

Solution:

Given

(a+b+c)(ab+c)(b + c a)(a + b – c)

= {(a + c) + b}{(a + c) – b}{b + (ca)} {b (c – a)}

= {(a + c)² – b²}{b²– (c – a)²}

= (a²+2ac + c²– b²) (b² – c² + 2ac – a²)

= {2ac + (a² – b² + c²)}{2ac – (a² – b² + c²)}

= (2ac)²– (a² – b² + c²)²

= 4a²c² – {(a² – b²)² + 2(a² – b²)c² + (c²)²}

= 4a²c² – {a⁴ – 2a²b² + b4 + 2a²c² – 2a²c² + c⁴) = 4a²c²-a⁴ + 2a²b² – b⁴ – 2a²c² + 2b²c² – c⁴

= 2a²b² + 2b²c² + 2a²c² – a⁴– b⁴ – c⁴

(a+b+c)(ab+c)(b + c a)(a + b – c) = 2a²b² + 2b²c² + 2a²c² – a⁴– b⁴ – c⁴

WBBSE Chapter 4 Algebraic Operations Exercise 4 Solved Problems

1. The terms 5, 6, \(\frac{3}{4}\) etc are called constant terms and are donated by a, b, c, etc. The terms which are always changing are known as variables and are donated by x, y, z etc.

2. The term (7x – 2) and 9x are called algebraic expressing.

3. Relations formed by addition, subtraction, multiplication, and division of a few variables and a few constants are called algebraic expressions.

4. In expression (5p + 4), p is variable and 5 and 4 are constant.

5. In the algebraic expression 7y, the variable y is multiplied by the constant 7. The factors of 7y are 1, 7, y, and 7y.

6. 7y has only one term and is called a Monomial.
(5p+ 4) has two terms and it is called Binomials.

Read and Learn More WBBSE Solutions for Class 7 Maths

Like terms and Unlike terms:

⇒ If the terms of algebraic expression are alike. They are called like terms and if they are not alike, they are called, unlike terms.

⇒ 3a²b and 4a²b  are like terms. But 3a²b and 4a²b are unlike terms.

⇒ Coefficients: A coefficient is a numerical factor present with any term.
⇒ In 5x², the coefficient of x² is 5 and the coefficient of 5x is x.

“WBBSE Class 7 Maths Algebra Chapter 4 solutions”

Question 1. Choose the correct answer 

1. The sum of (-5x + 3y) and (18x – 15y) is

1. 23x 12y
2. 13x 12y
3. 13x 18y
4. None of these

Solution: -5x + 3y+ 18x-15y
=-5x+18x + 3y 15y
= 13x 12y

So the correct answer is

2. The value of (- 3m² + 2m+ 2) – (m² – 2) is

1. -2m² + 2m
2. -4m² + 2m + 4
3. -2m²+ 2m – 4
4. -2m² + 2m + 4

Solution: (- 3m² + 2m+ 2) – (m² – 2)

= -3m² + 2m + 2-m² + 2
= -3m² -m² + 2m + 2 + 2
=-4m² + 2m + 4

So the correct answer is 2. -4m² + 2m + 4

3. The value of (-3a²) x (4a²b) x (-2) is

1.  24a²b
2. 24a²b³
3. -24a⁴b
4. 24a⁴b

Solution: (-3a²) x (4a²b) x (-2)
=(-3) x (4) x (-2) x a²⁺² x b
= 24a⁴b

So the correct answer is 4. 24a⁴b

Question 2. Write ‘true’ or ‘false’

1. The number of factors of x²y is 6

Solution:

 

The factors of x²y is 1, x, x², y, xy, and i.e., the number of factors is 6.

So the statement is true.

2. The coefficient of x² in (5-6x²y² + 6xy) is 6y

Solution: In the expression (5-6x²y² + 6xy) the term with x is (-6x² y²) whose coefficient of x² is -6y²

So the statement is false.

“Algebraic Operations Class 7 WBBSE solved problems”

3. The value of (-48x⁹ + 12x⁶) + 3x³ is (-16x⁶+ 4x³)

Solution: \(\frac{-48 \dot{x}^9+12 x^6}{3 x^3}\)

= \(-\frac{48 x^9}{3 x^3}+\frac{12 x^6}{3 x^3}\)

=-16x^9-3 +4x^6-3 =-16x⁶+4x³

So the statement is true.

Question 3. Fill in the blanks

1. 4x, (5x-2), (7x + 3) together are called _____

Solution: Algebrain expression

2. The product of a²b and (3a-4b) is _____

Solution: a²b (3a – 4b) = 3a³b – 4a²b²

3. If the price of x dozen of pen is ₹ (xy²-7x) then the price of 4 such pen is ₹ _____

Solution: x dozen = 12x

⇒ The price of the number of 12x pen is ₹ (xy² – 7x)

⇒ The price of 1 such pen is ₹ \(\frac{x y^2-7 x}{12 x}\)

⇒ The price of 4 such pen is ₹  \(\frac{4\left(x y^2-7 x\right)}{12 x}\)

=

= ₹ \(\left(\frac{y^2-7}{3}\right)\)

Some important rules: a is non zero integers and also m and n are integers

1. a^m. aⁿ = a^{m + n}

2. a^m ÷ aⁿ = a^{m – n}

3. (a^m)ⁿ = a^mn

4. \(a^m=\frac{1}{a^{-m}}\)

5. a⁰ = 1

6. (ab)^m = a^m.bⁿ

7. \(\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}\)

Proof: 

1. \(a^0=a^{m-m}=\frac{a^m}{a^m}=1\)

2. \(a^m=a^{0-(-m)}=\frac{a^0}{a^{-m}}=\frac{1}{a^{-m}}\)

“Exercise 4 Algebra solutions WBBSE Class 7”

Question 4. Represent the following algebraic expressions into ‘factor free’ type of figure mentioning the prime factors of each term. Also mention the types these expressions with respect to their number of terms.

1. xy + yz + zx
2. x² + x + 1

Solution:

1.

 

2.

 

Question 5. Find the numerical coefficients of the terms, other than the constant term.

1. x² + 6x + 5
2. x – 7xy + 9y

Solution: 1. x² + 6x + 5
The numerical co-efficient of x² is 1.
The numerical co-efficient of x is 6.

2. x – 7xy + 9y

The numerical co-efficient of x, xy, and y are 1, -7, and 9 respectively.

Question 6. Add

1. 7a²b+3ab²-9, 4a²b – 5ab² + 10, a²b + ab² – 15
2. xy-3yz.+4zx, 3xy+5yz-6zx, -2xy + 3yz-zx

Solution:

1. 7a²b+3ab²-9, 4a²b – 5ab² + 10, a²b + ab² – 15

= 7a²b+ 4a²b + a²b+3ab² – 5ab² +ab² – 9+ 10 – 15
= 12a²b – ab² – 14

Another method:

7a²b+3ab²-9
4a²b-5ab² + 10
+ a²b + ab² – 15

= 12a²b- ab² – 14

2. (xy 3yz+4zx) + (3xy + 5yz-6zx) + (-2xy + 3yz – zx)

=
= xy + 3xy – 2xy – 3y + 5yz + 3yz + 4zx-6zxzx
= 2xy + 5yz-3zx.

Question 7. Subtract

1. (3x² – 4xy- 5y²) from (-7x²-3xy+4y²)
2. (m³ – 3m² + 4m – 5) from (m³+ 3m² – 6m + 2)

Solution:

1. (-7x² -3xy + 4y²) – (3x²-4xy – 5y²)

= -7x²-3xy+4y²-3x² + 4xy + 5y²
= -7x²-3x² – 3xy + 4xy+4y²+5y²
= -10x² + xy + 9y²

Alternative method

2. (m³+ 3m²– 6m+ 2) – (m³-3m² + 4m -5)
= m³+3m²-6m+2 -m³ + 3m² – 4m + 5

= 6m2 10m +7

Question 8. How much must be added to (6y² – 6y+ 1) to get (-15y²+ 7y – 7)

Solution:

The required expression is (-15y² + 7y – 7) (6y² – 6y+1)
=-15y²+ 7y-7-6y²+6y+ 1
=-15y²+ 13y-6

Question 9. What must be subtracted from (9x²-3x+7) to get (-3x²+6x-13)

Solution:

The required expression is (9x²-3x+7)-(-3x²+6x-13)
= 9x² – 3x + 7 + 3x²-6x + 13
= 12x² – 9x+20

“WBBSE Class 7 Maths Chapter 4 Algebraic Operations”

Question 10. Subtract the sum of (-5y² + 3y -7) and (9y²-7y+ 12) from the sum of (y²+2y-6) and (12y – y² + 5)

Solution:

The required expression is
{(y²+2y-6)+(12y-y²+ 5)} – {(-5y2+ 3y 7) + (9y2-7y+ 12)}

=

= (14y -1)- (2y²-4y+5)
=14y -1 – 2y² + 4y – 5
=-2y²+18y-6

Question 11. Multiply

1. \(\left(-\frac{3}{5} a^2 b\right) \times\left(\frac{15}{8} a b^2 c\right) \times\left(\frac{4}{9} b c^2\right)\)

2. \(\left(\frac{5}{3} a^2 b c\right) \times\left(-\frac{4}{5} a b^2 c\right) \times\left(-\frac{6}{15} a b c^2\right)\)

3. (3a-4b)(5a + 6b)

4. (a² – b² + c²)(a² + b² – c²)

Solution:

1. \(\left(-\frac{3}{5} a^2 b\right) \times\left(\frac{15}{8} a b^2 c\right) \times\left(\frac{4}{9} b c^2\right)\)

2. \(\left(\frac{5}{3} a^2 b c\right) \times\left(-\frac{4}{5} a b^2 c\right) \times\left(-\frac{6}{15} a b c^2\right)\)

3. (3a-4b)(5a + 6b)

= 3a (5a + 6b) – 4b (5a + 6b)
= 15a² + 18ab – 20ab – 24b²
=15a²-2ab-24b²

4. (a²-b²+c²)(a²+b²-c²)

“Step-by-step solutions for WBBSE Class 7 Algebra”

Question 13. Divide the first expression by the second.

1. 20a²b-25ab³c²-30a²bc³, 5ab
2. 6abc-18a²bc² + 24a³b²c³, -6abc
3. a²b⁴ + a4b³ – a³b⁵4, -a⁴b

Solution:

1.

2.

 

3.

 

Question 14. If A = 2x + 3y – 4z, B= 2y+ 3z4x and C2z+3x4y. Find the sum of (A+B+C) and (A B+ C)

Solution:

Given

If A = 2x + 3y – 4z, B= 2y+ 3z4x and C2z+3x4y.

(A+B+ C) + (A- B+ C)
= 2A + 2C
= 2(2x + 3y 4z) + 2(2z+3x-4y)
=4x+6y8z+4z+6x-8y
= 10x 2y4z.

Question 15. If P = 3x² + 2xy + y², Q=-3x² – 2xy + y² and R = x² + y² and x = y – 2. Find the value of (P+Q + R).

Solution:

Given

If P = 3x² + 2xy + y², Q=-3x² – 2xy + y² and R = x² + y² and x = y – 2.

= x² + 3y²
=(-2)² + 3 (-2)² [Putting the value of x and y]
= 4+3×4
= 16.

Question 16. Simplify the following

1. 3x²+ 2x(x + 3) -4x (3x-5)
2. (a² + b²)(a² – b²) + (b² + c²)(b² – c²) + (c² + a²)(c² – a²)
3. (x²+5x)(x -1) – (x² + 2x)(x 1) – 5x (x + 1)
4. (x+3)(x-3)+(x+4)(x-4) – 2x² + 25
5. (x² + 2x – 1)(x − 1) + (x² – 2x + 1)(x + 1) − 2x(x − 1)

Solution:

1. 3x²+ 2x(x + 3) -4x (3x-5)

= 3x²+2x²+6x-12x² + 20x
= −7x² + 26x

2. (a² + b²)(a² – b²) + (b² + c²)(b² – c²) + (c² + a²)(c² – a²)

= a²(a²-b²) + b²(a² – b²) + b²(b² – c²) + c²(b² – c²) + c² (c² – a²) + a² (c² – a²)

= 0

3. (x² + 5x)(x – 1) – (x² + 2x)(x – 1) – 5x (x + 1)

= x²(x -1) + 5x(x – 1) – x² (x – 1) – 2x(x – 1)- 5x(x + 1)

= -2x² – 8x

4. (x+3)(x-3)+(x+4)(x-4)- 2x² + 25.

= x(x-3)+3(x-3) + x(x-4)+4(x-4) – 2x² + 25

= 0

5. (x² + 2x-1)(x 1) + (x² – 2x + 1)(x + 1) 2x(x-1)

= x²(x-1) + 2x(x − 1) − 1(x − 1) + x² (x + 1) − 2x(x + 1) + 1(x + 1) − 2x(x-1)

= 2x³ – 2x² – 2x + 2

 

WB Class 7 Math Solution Algebraic Operations

Algebraic Operations Exercise 6.1

Let’s write the algebraic expression and find the number of terms.

4x, 3x + 1, 2x + 1, 6p – 1, 3y + 6

Class 7 Math Solution WBBSE Algebraic Operations Exercise 6.2

Question 1. Let’s draw ‘factor tree’ type figures and from there let’s find the number of terms and factors of the following algebraic expression :

1. 2x + 1
Solution:

2. 3y+6
Solution:

Question 2.  Let’s study the algebraic expressions given below and fill in the gaps accordingly.

Class 7 Math Solution WBBSE Algebraic Operations Exercise 6.3

Question 1. Let’s form algebraic expressions from the statements given below-

1. y is added to x
Solution: x + y

2. x is subtracted from z
Solution: z – x

3. q is added to twice of p
Solution: 2p + q

4. Multiply x with a square of x.
Solution: x².x = x³

5. \(\frac{1}{4}\) th of the sum of x and y
Solution: \(\frac{x+y}{4}\)

“Class 7 WBBSE Maths Algebraic Operations exercise solutions”

6. 7 is added to 4 times the product of a & b
Solution:  4ab + 7

7. Half of y is added to twice of x
Solution: 2x +\(\frac{1}{2}\) y

8. The product of x and y is subtracted from sum of x and y.
Solution: (x + y) – xy.

Question 2. Let’s Let’s observe the patterns match and fill in the chart

1. 

General Expression 5x + 2 = Number of Match Sticks.

2. Let’s form the general expression with a variable

General Expression, Number of sticks = 4x + 1 where x = No. of trapezium

Question 3. Let us represent the following algebraic expressions into ‘factor tree’ type of figure mentioning the prime factor of each term. Also mention thetypes of these expressions with respect to their number to terms.

1. 5x (Monomial)
Solution:

2. 7 + 2x + x² (Trinomial)
Solution:

3. x² + x + 1 (Trinomial)
Solution:

4. 2x²y + 7 (Binomial)
Solution:

5. 2y³ + y (Binomial)
Solution:

6. x²y+xy²+xyz
Solution:

7. x²y + 2x²y² (Binomial)
Solution:

8. 5x + 2y (Binomial)
Solution:

Question 4. Let’s find the numerical coefficients of the terms, other than constant
term:

1. 2x + 3y
Solution: 2, 3.

2. x² + 2x + 5
Solution: 1, 2.

3. x + 5xy – 7y
Solution: 1, 5, – 7.

4. – 5 – z
Solution: – 1

5. x³ + x – y 
Solution: – 1,1, 1,

6. \(\frac{x}{2}+4\)
Solution: \(\frac{1}{2}\)

Question 5. In the following algebraic expression let’s find the coefficient of x in the terms or terms which have ‘x’ as their factor.

1. y³x + y²
Solution: y³x, y³

2. 5z² – 8zx
Solution: – 8zx, – 8z

3. – x – y + 2
Solution:– x, – 1

4. 4 + y +.yx
Solution: yx, y

5. 2 + x + xy²
Solution: x, 1 , xy2, y2

6.  5xy⁴-1 4
Solution: 15xy⁴, I5y⁴

Question 6. Let’s group the like terms from the algebraic expressions given below. 2x, y, 12xy, 13y², – 5x, 18y, – 4xy, – 2y², 21x²y, 3x, 3xy, – xy, – y, – 6x², -1 5x².
Solution:

2x, y, 12xy, 13y², – 5x, 18y, – 4xy, – 2y², 21x²y, 3x, 3xy, – xy, – y, – 6x², -1 5x²=

(2x, – 5x, 3x); (y, 18y, – y), (12xy, – 4xy, 3xy, – xy), (13y², – 2y²), (21x²y), (- 6x², – 15x²)

Question 7. Let’s identify the like and unlike pairs of terms with reasons from the pairs of terms given below.

1. 2x, 3y
2. 7x, 8x
3. 29x, 6Tx
4. 4xy, 6yz
5. 15 yx, 8xy
6. 5xy, 6x²y²

Solution:

Like term: 2,3,5

Unlike term: 1,4,5

Question 8. From the algebraic expressions given below, let’s write the terms which contain x². And also find the coefficient of x².

1. 5- xy²
Solution: No

2. – 6x² – 8y
Solution: -6

3. 3x² – 1 5xy² – 8y²
Solution: 3

4. 2 + 3k²y+4x
Solution: + 3y

5. 5 – 6x²y² + 6xy
Solution: – 6y²

Class Vii Math Solution WBBSE Algebraic Operations Exercise 6.4

Question 1. Let’s add the following

1. (- 5x + 3y) and (1 8x – 1 5y)
Solution :

Given

(- 5x + 3y) and (1 8x – 1 5y)

⇒ (- 5x + 3y) + (1 8x – 1 5y)

= – 5x + 1 8x + 3y – 1 5y

= 1 3x – 1 2y

2. (7a – 8b + 2c) and (2a + 3b – d)
Solution :

Given

(7a – 8b + 2c) and (2a + 3b – d)

⇒ (7a – 8b + 2c) + (2a + 3b – d)

= 7a + 2a + (- 8b + 3b) + 2c – d

= 9a – 5b + 2c – d

Question 2. Let’s subtract:

1.  (-mn – m + n) from (4mn + m + n)
Solution :

Given

(-mn – m + n) and (4mn + m + n)

⇒ (-mn – m + n) – (4mn + m + n)= (4mn + m + n) – (-mn – m + n)

= 4mn + m + n + mn + m- n

= 4mn + mn + m + m + n- n

= 5mn + 2m

2. (2q² + 3P² – qp + pq²) from (p² + q² – pq + p²q)
Solution:

Given

(2q² + 3P² – qp + pq²) and (p² + q² – pq + p²q)

= (p² + q² – pq + p²q) – (2q² + 3p² – pq + pq²)

= p² + q² – pq + p²q – 2q² – 3p² + pq – pq²

= p² – 3r² + q² – 2q² – pq + pq + p²q – pq²

= – 2p² – q² + p²q – pq²

Class Vii Math Solution WBBSE Algebraic Operations Exercise 6.5

Question 1. Lefs add the algebraic expressions (2x² + x + 2) and (x² + 2x + 2) with the use of colored cards.
Solution:

Let’s add :

Given

(2x² + x + 2) and (x² + 2x + 2)

⇒ (2x² + x + 2) + (x² + 2x + 2)

= 2x² + x² x + 2x + 2 + 2

= 3x² + 3x + 4

“Solved problems of Algebraic Operations Class 7 WBBSE”

Question 2. Let’s subtract the algebraic expressions (3x² + 3x – 2) from (5x² – 2x – 3)
Solution:

Given

(3x² + 3x – 2)and (5x² – 2x – 3)

Let’s subtract the algebraic expressions

(3x² + 3x – 2) from (5x² – 2x – 3)

= (5x² – 2x – 3) – (3x² + 3x – 2)

= 5x² – 2x – 3 – 3x² – 3x + 2

= 5x² – 3x² – 2x – 3x – 3 + 2

= 2x² – 5x 1

Algebraic Operations Exercise 6.6

Question 1. Let’s add the following :

1. (- 5x + 3y) and (18x – 1 5y)
Solution:

Given

(- 5x + 3y) and (18x – 1 5y)

– 5x + 3y + 18x – 15y = – 5x + 18x + 3y – 15y

= 13x – 1 2y

2. (7a – 8b + 2c) and (2a + 3b – d)
Solution:

Given

(7a – 8b + 2c) and (2a + 3b – d)

7a – 8b + 2c + 2a + 3b – d = 7a + 2a – 8b + 3b + 2c – d

= 9a – 5b + 2c> d

Question 2. Let’s Let’s subtract.:

1. (- mn – m + n) from (4mn + m + n)
Solution:

Given

(- mn – m + n) and (4mn + m + n)

(4mn + m + n). – (- mn – m + n)

= 4mn + m + n + mn + m- n

= 4mn + mn + m + m + n- n

= 5mn + 2m

2. (2q² + 3p² – qp + pq²) from (p² + q² – pq + p2q)
Solution:

Given

(2q² + 3p² – qp + pq²) and (p² + q² – pq + p2q)

(p² + q² – pq + p²q) – (2q² + 3p²– qp + pq²)

p² + q² – pq + p²q – 2q² – 3p² + pq – pq²

= p² – 3p² + q² – 2q² – pq + pq + p²q – pq²

= – 2p² – q² + p²q – pq²

= p²q – 2p² – q² – pq²

Algebraic Operations Exercise 6.7

Question 1. Let’s calculate mentally 

1. 5x + 3x
Solution:

5x + 3x = 8x.

2. – 4y + 7y
Solution:

– 4y + 7y = 3y

3. 9y – 3y
Solution:

9y – 3y = 6y.

4. – 10x – 2x
Solution: –

10x-2x = – 12x

5. 3a + 4a – 2a
Solution:

3a + 4a – 2a = 3a – 2a

= 5a

6. – 7x-2x +5x
Solution:

– 7x-2x +5x = – 9x + 5x

= – 4x

7. 6p – 2p + 3p
Solution:

6p – 2p + 3p = 9p – 2p

= 7p

8. 4x² – 2x² – 3x² + x²
Solution:  

4x² – 2x² – 3x² + x²

= 5x² – 5x²

= 0

9. 5a²b – 2a²b – 3a²b + 8a²b
Solution:

5a²b – 2a²b – 3a²b + 8a²b = 13a²b – 5a²b

= 8a²b

10. 3x² – 6x² – 2x² – x² + 6x²
Solution: 

3x² – 6x² – 2x² – x² + 6x² = 9x² – 9x²

= 0

= 2.

Question 2. 

1. My age is ‘x’ years. Pallabi is 2 years older than me. Let’s find the sum of our ages.
Solution: 

Given

My age is ‘x’ years. Pallabi is 2 years older than me.

My age  = x years

Pallabi’s age = (x + 2) years

∴ Sum of our ages = (2x + 2) years

2. Today made ’x’ number of flower garlands. Mir made 6 more than twice the number of garlandsI made. In total, how many garlands we two have made.
Solution :

Given

Today made ’x’ number of flower garlands. Mir made 6 more than twice the number of garlandsI made.

I made = x  flower garlands

Mir made = 2x + 6 flower garlands

∴ Total number of flower

Garlands we made = 3x + 6

3. Today Ratul bought guava for Rs. x, apple for Rs. (x + 15) and cucumber for Rs. (2x + 3). Let’s find, how much money did Ratul spent today on fruits.
Solution:

Given

Today Ratul bought guava for Rs. x, apple for Rs. (x + 15) and cucumber for Rs. (2x + 3).

Ratul bought guava for = Rs. x

Apple for = Rs. (x + 15)

Cucumber for = Rs. (2x + 3)

∴ Ratul spent today on fruits = (x + x + 15 +2x + 3)

= Rs. (4x + 18)

4. Last year Firoza was present in school for x days. Firoza’s friend Mohini was present for (3x + 13) days, Let’s find, how much more days Mohini was present in school than Firoza last year.
Solution :

Given

Last year Firoza was present in school for x days. Firoza’s friend Mohini was present for (3x + 13) days

Mohini was present for = 3x + 13 days

Firoza was present for = x days

Mohini was present in school than Firoz = 3x + 13 = x

(3x + 13 –  x)  = 3x –  x+ 13

= (2x + 13) days

5. Today, Dipuda sold (2x + 19) newspapers. But yesterday he sold (5x – 8) newspapers. Let’s find, how many more newspapers did Dipuda sold today than yesterday.
Solution :

Given

Today, Dipuda sold (2x + 19) newspapers. But yesterday he sold (5x – 8) newspapers.

Yesterday he sold = (5x-8) newspaper

Today Dipuda sold = (2x+19) newspaper

∴ Dipuda sold more newspaper on yesterday  = (5x-8) = (2x+19)

(5x-8-2x-19) = 5x- 8- 2x -19

= 3x – 27

6. Pareshbabu earns Rs. 8x per month. But he spends Rs. (3x – 15) per month. Let’s find the amount of money he saves per month.
Solution :

Given

Pareshbabu earns Rs. 8x per month. But he spends Rs. (3x – 15) per month.

Monthly Incom of Pareshbabu = Rs. 8x

Monthly expenditure of Pareshbabu = Rs. 3x-15.

Monthly saving of Pareshbabu = 8x= 3x-15.

(8x – 3x +15 )=  8x – 3x+15 = 0

= 5x +15

Question 3. Let’s add:

1. 3a + b; 2a + 4b; 5a – b
Solution :

Given

3a + b; 2a + 4b; 5a – b

(3a +b) + (2a + 4b) + (5a – b)

= (3a + 2a + 5a) + (b + 4b-b)

= 1 0a + 4b

2. 5a – 4; 2a + 3; 2a – 4
Solution :

Given

5a – 4; 2a + 3; 2a – 4

(5a – 4) + (2a + 3) + (2a – 4)

= (5a + 2a+ 2a) + (- 4 + 3 – 4)

= 9a – 5

3. 6a + 7a +3; 9a² – 2a + 7; 4a² – 2a + 9
Solution :

Given

6a + 7a +3; 9a² – 2a + 7; 4a² – 2a + 9

(6a² + 7a + 3) + (9a² – 2a + 7) + (4a² – 2a + 9)

= (6a² + 9a² + 4a²) + (7a – 2a – 2a) + (3 + 7 + 9)

= 19a² + 3a + 19

4. 2a²b + 5b²a + 7; 3a²b – 2b²a + 6; 8a²b – b²a + 9
Solution:

Given

2a²b + 5b²a + 7; 3a²b – 2b²a + 6; 8a²b – b²a + 9

(2a²b + 5b²a + 7) + (3a²b – 2b²a + 6) + (8a²b – b²a + 9)

= (2a²b + 3a²b + 8a²b) + (5b²a – 2b²a – b²a) + (7 + 6 + 9)

= 13a²b + 2b²a + 22

5. 4xy + 5x + 7y; – 4xy – y – 3x; 3xy – 3y + 2x
Solution :

Given

4xy + 5x + 7y; – 4xy – y – 3x; 3xy – 3y + 2x

(4xy + 5x 7y) + (- 4xy y – 3x) + (3xy – 3y + 2x)

= (4xy – 4xy + 3xy) + (5x – 3x + 2x) + (7y – y – 3y)

= 3xy + 4x + 3y

Question 4. Let’s subtract:

1. (2x + 3y) from (8x + 6y)
Solution :

Given

(2x + 3y) and (8x + 6y)

(8x +6y) – (2x + 3y)

= 8x – 2x + 6y – 3y

= 6x + 3y

2. (m² – 2) from (- 3m² + 2m + 2)
Solution :

Given

(m² – 2) and (- 3m² + 2m + 2)

(- 3m² + 2m + 2) – (m² – 2)

= – 3m² – m² + 2m + 2 + 2

= – 4m² + 2m + 4

3. (8x + 4y + 7) from (2x + 3y)
Solution :

Given

(8x + 4y + 7) and (2x + 3y)

(2x + 3y) – (8x + 4y + 7)

= 2x – 8x + 3y – 4y – 7

= – 6x – y – 7

4.  (5a² + 2a – 1 ) from (- 9a² + 3a + 2)
Solution:

Given

(5a² + 2a – 1 ) and (- 9a² + 3a + 2)

(- 9a² + 3a + 2) – (5a² + 2a – 1 )

= -9a²-5a²+ 3a-2a + 2 + 1 .

= – 14a² + a + 3

5. (- 2x² + 3y² ) from x
Solution :

Given

(- 2x² + 3y² ) and x

x – (- 2x² + 3y²) = x + 2x² – 3y²

6. (2x² + xy + 3y²) from (3x² + 5xy)
Solution :

Given

(2x² + xy + 3y²) and (3x² + 5xy)

(3x² + 5xy) – (2x² + xy + 3y²)

= 3x² + 5xy – 2x² – xy – 3y²

= 3x² – 2x² + 5xy – xy – 3y²

= x² + 4xy – 3y²

“Algebra Chapter 4 Algebraic Operations WBBSE Class 7”

Question 5. Let us Simplify the following :

1. 17x²y – 3xy² + 14x² y + 2xy²
Solution:

17x²y – 3xy² + 14x²y + 2xy²= 17x²y + 1 4 x²y – 3xy² + 2x²y

= 31x²y-xy²

2. – 5b + 18a + 6b – 2a
Solution :

– 5b + 1 8a + 6b – 2a = 1 8a – 2a + 6b – 5b

= 16a+b

3. 4m² + 3n² – (6m² + 7n²)
Solution :

4m² + 3n² – (6m² + 7n²)- 4m² – 6m² + 3n² – 7n²

= – 2m² – 4n²

4. a-b-(b-a)
Solution :

a – b – (b – a) = a – b – b + a

= a + a – b – b

= 2a – 2b

5.  (6p – 4q + 2r) + (2p + 3q – 4r)
Solution :

(6p – 4q + 2r) + (2p + 3q – 4r) = 6p + 2p – 4q + 3q + 2r – 4r

= 8p-q -2r

6.  – x + y + z – (2x – 3y + z)
Solution :

– x + y + z – (2x – 3y + z) = – x + y + z – 2x + 3y – z

= – x -2x + y + 3y + z – x

= – 3x + 4y (Ans )

7.  (x² + 2x – 5) + (3x² – 8x + 5)
Solution :

(x² + 2.x – 5) + (3×2 – 8x + 5) = x² + 2x – 5 + 3x² – 8x + 5

= x² + 3x² + 2x – 8x – 5 + 5

= 4x² – 6x

8. (7x² – 3x + 3) – (2x² – 13x – 7)
Solution :

(7x² – 3x + 3) – (2x² – 13x – 7) = 7x² – 3x + 3 – 2x² + 13x + 7

= 7x² – 2x² – 3x + 13x + 3 + 7

= 5x² + 1 0x + 1 0

9.  6a – 2a – ab – (3a + b – ab) + 2ab – b + a
Solution:

6a – 2a – ab – (3a + b – ab) + 2ab – b + a

= 6a – 2b – ab – 3a – b + ab + 2ab – b + a

= 6a – 3a + a – 2b – b – b – ab + ab + 2ab

= 4a – 4b + 2ab

Question 6. Ramu had Rs. (13x² + x – 3). He $pent Rs. (4x² – 3x – 12). Let’s find, how much money Ramu has got.
Solution :

Given

Ramu had Rs. (13x² + x – 3). He $pent Rs. (4x² – 3x – 12).

Ramu had =  Rs. 13x² + x – 3

He spent = Rs. 4x² – 3x -12

Remaining amount = Rs. (13x² + x – 3 – 4x² + 3x +12)

=   Rs. 9x² + 4x + 9

∴ Ramu has got Rs. 9x² + 4x + 9

Question 7. The lenght of three sides of a triangle are (x + 4) cm, (2x + 1) cm, and (4x – 8) cm, Let’s find the perimeter of the triangle.
Solution:

Given

The lenght of three sides of a triangle are (x + 4) cm, (2x + 1) cm, and (4x – 8) cm,

The perimeter of the triangle = Sum of the three side of the triangle.

= x + 4 + 2x + 1 +4x-8

= x + 2x + 4x + 4 + 1 – 8

= x + 2x + 4x + 5 – 8

= 7x – 3

Question 8. How much be added to – 8x² + 8x + 1 to get – 14x² + 11x – 3
Solution:

Required No. = (- 14x² + 1 1x – 3) – (- 8x² + 8x + 1)

= – 1 4x² + 11 x – 3 + 8x² – 8x – 1

= – 1 4x² + 8x² + 11 x – 8x – 3 – 1

= – 6x² + 3x – 4

“WBBSE Class 7 Maths Exercise 4 Algebraic Operations chapter”

Question 9. Let’s find, what must be subtracted from – 11 x – 7y – 9z to get – 7x +3y-5z.
Solution:

Required No. = (- 1 1x – 7y – 9z) -(- 7x + 3y – 5z)

= – 1 1 x – 7y – 9z + 7x – 3y + 5z

= – 1 1 x + 7x – 7y – 3y – 9z + 5z

= – 4x – 10y – 4z

Question 10. How much is the sum of (3x² + 4x) and (5x² – x) more than (3x – 5x²), let’s calculate.
Solution :

Required No. = (3x² + 4x) + (5x² – x) – (3x – 5x²)

= 3×2 + 4x + 5x² – x – 3x + 5x²

= 3x² + 5x² + 5x² + 4x – x – 3x

= 13x² + 4x – 4x

= 13x²

Question 11. Let’s subtract the sum (x² – 9x) and (- 2x² + 3x + 5) from the sum of (5 + 9x) and (6 – 7x + 4x²).
Solution : 

Required No. = {(5 + 9x) + (6 – 7x + 4x²)} – {(x² – 9x) + (- 2x² + 3x + 5)}

= (5 + 9x + 6 – 7x + 4x²) – (x² – 9x – 2x² + 3x + 5)

= (4x² + 9x – 7x + 5 + 6) – (x² – 9x – 2x² + 3x + 5)

= (4x² + 2x + 1 1 ) – (- x² – 6x + 5)

= 4x² + 2x + 1 1 + x² + 6x – 5

= 4x² + x² + 2x + 6x + 1 1 – 5

= 5x² + 8x + 6

WBBSE Class 7 Math Solution Algebraic Operations Exercise 6.8

Question 1. If x = 5 let’s find the values of the following algebraic expressions. When x = 5

1. 6x + 11
Solution:

6x+ 11 = 6 × 5 + 11 = 30 + 11 = 41.

2. \(\frac{x}{5}\) +2 
Solution:

⇒ \(\frac{x}{5}\)+2

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

= 1+2 = 3

3. x² + 2x – 1
Solution :

x² + 2x – 1 = (5)²+ 2× 5-1

= 25 + 10-1

= 35- 1

= 34

4. x³> 8
Solution :

x³ + 8 = (5)³ + 8

= 125 + 8

= 133

5. 10 -x
Solution:

10 – x = 10 – 5

= 5

Question 2. If y= – 3, let’s find the values of the following algebraic expressions. \(\frac{y+5}{4}\)
Solution :

5-y = 5-(-3)

= 5 + 3

= 8

3. y + 8
Solution :

y + 8 = -3 + 8

= 5

4.  y² + 2y + 3
Solution :

y² + 2y + 3 = (- 3)² + 2 (- 3) + 3 =

9 – 6 + 3

= 12 – 6

= 6

5. y³ – 1
Solution :

= (-3)3 – 1

= – 27 – 1

= -28

Question 3. Let’s find the values of the following, when x = 2, y = – 1

1. 2x + 7y
Solution :

2x + 7y = 2 (2) + 7 (- 1)

= 4-7

= – 3

2x + 7y = – 3

2. x² + y²
Solution:

x² + y²= (2)² + (- 1)²

= 4 + 1

= 5

x² + y²  = 5

3. x² + 7xy + y²
Solution :

x² + 7xy + y²= (2)² + 7. 2 (- 1) + (- 1)²

= 4 – 14 +1

= – 14 + 5

= – 9

4. x³ – 8y³
Solution:

x³ – 8y³ = (2)³ – 8(- 1)³ = 8 – 8 (- 1)

= 8 + 8

= 16

5. \(\frac{x}{9}+\frac{y}{4}\)
Solution:

⇒ \(\frac{x}{9}+\frac{y}{4}\)

= \(\frac{2}{9}+\frac{-1}{4}\)

= \(\frac{8-9}{36}\)

= \(\frac{1}{36}\)

WBBSE Class 7 Math Solution Algebraic Operations Exercise 6.9

Question 1. Let’s find the product of the following.

1. 7,2x
Solution:

7,2x = 7 × 2x = 14x (Ans.)

2. – 3x, 4x
Solution:

– 3x, 4x = (- 3x) × (4x)

= – 12 x²

3. – 2x, – 3x²
Solution:

– 2x, – 3x²= (- 2x) × (- 3x²)

= 6x³

4. 7x, 0
Solution :

7x, 0 = 7x  × 0 = 0

5. 3ab, 4ac
Solution :

3ab, 4ac= (3ab) × (4ac)

= 12a²bc (Ans.)

6. 8x², 2y²
Solution:

8x², 2y² = (8x²) × (2y²)

= 16x²y²

7. 2a²b, 3ab²
Solution :

2a²b, 3ab² = 2a²b × 3ab²

= 6a³b³

8.  (- 4xy), (- 4xy)
Solution:

(- 4xy), (- 4xy)  = (- 4xy) × (- 4xy) = 16x²y²

Question 2. Let’s muliply first monomial with second monomial and write the product in corresponding blanks spaces.
Solution :

Algebraic Operations Exercise 6.10

Question 1. In each of the following cases, let’s multiply and find their product.

1.  ab, (a² – b²)
Solution:

ab, (a² – b²) = ab x (a² – b²)

= ab³ – ab³

2. 4a, (a + b – c)
Solution :

4a, (a + b – c)= 4a × (a + b – c)

= 4a² + 4ab – 4ac

3. 6a²b², (2a + b)
Solution :

6a²b², (2a + b) = 6a²b² × (2a + b)

= 12a³b² + 6a²b³ (Ans.)

4. xyz, (x²y – y²x + z²y)
Solution :

xyz, (x²y – y²x + z²y) = xyz x (x²y – y²x + z²y)

= x³y²z – x²y³z + xy²z³

5. 0, (ab + bc – ca)
Solution :

0, (ab + bc – ca) = 0 x (ab + be + ca)

= 0

Question 2. Let’s simplify

1.  7x (2x + 3) – 5x (3x – 4)
Solution :

7x (2x + 3) – 5x (3x – 4) = 7x (2x + 3) – 5x (3x – 4)

= (14x² + 21 x) – (1 5x² – 20x)

= 1 4x² + 21 x – 1 5x² + 20xy

= 14x² – 15x² + 21 x + 20x

= – x² + 41x

2. x(x – y) + y (y – z) + z (z – x)
Solution :

x(x – y) + y (y – z) + z (z – x) = x² – xy +y² – yz + z² – zx

= x² + y² + z² – xy – yz – zx

3. 2x – 6x (5 – 8x – 3y)
Solution :

2x – 6x (5 – 8x – 3y) = 2x – 30x + 48x² + 18xy

= 48x² + 18xy – 28x

4. 7a – 2 (5a + 6b – 7)
Solution:

7a – 2 (5a + 6b – 7) = 7a – 10a – 12b + 14

= 14 -7a -12b.

WBBSE Class 7 Math Solution Algebraic Operations Exercise 6.11

Question 1. Let’s multiply :

1.  (10 – 3x) (7 + x)
Solution :

(10 – 3x) (7 + x) = 10 (7 + x) – 3x (7 + x)

= 70 + 10x – 21x – 3x²

= – 3x² – 1 1x + 70

2. (11 +2x) (8 – 2y)
Solution :

(11 + 2x) (8 – 2y) = (11+ 2x) × (8 – 2y)

= 11 (8 – 2y) + 2x (8 – 2y)

= 88 – 22y + 1 6x – 4xy

3. (a + by) (4a – 6y)
Solution :

(a + by) (4a – 6y) = (a + by) × (4a – 6y)

= a (4a – 6y) + by (4a – 6y)

= 4a² – 6ay + 4aby – 6by²

4. (2x² y – y² ) (3x-5y)
Solution :

(2x² y – y² ) (3x – 5y) = (2x² y – y² ) × (3 -5y)

= 2x² y (3x – 5y) – y² (3x – 5y)

= 6x³ y – 1 0x² y² – 3xy² + by³

5. \(\left(\frac{x}{2}-\frac{y}{3}\right)\left(\frac{2 x}{3}-\frac{3 y}{5}\right)\)
Solution:

⇒  \(\left(\frac{x}{2}-\frac{y}{3}\right)\left(\frac{2 x}{3}-\frac{3 y}{5}\right)=\left(\frac{x}{2}-\frac{y}{3}\right) \times\left(\frac{2 x}{3}-\frac{3 y}{5}\right)\)

= \(\frac{x}{2}\left(\frac{2 x}{3}-\frac{3 y}{5}\right)-\frac{y}{3}\left(\frac{2 x}{3}-\frac{3 y}{5}\right)\)

= \(\frac{x^2}{3}-\frac{3 x y}{10}-\frac{2 x y}{9}+\frac{y^2}{5}\)

= \(\frac{x^3}{3}-\frac{47 x y}{90}+\frac{y^2}{5}\)

6. \(\left(\frac{2 a^2}{7}-\frac{1}{5}\right)\left(\frac{3 a}{5}-\frac{2}{9}\right)\)
Solution:

⇒ \(\frac{2 a^2}{7}\left(\frac{3 a}{5}-\frac{2}{9}\right)-\frac{1}{5}\left(\frac{3 a}{5}-\frac{2}{9}\right)\)

=  \(\frac{6 a^3}{35}-\frac{4 a^2}{63}-\frac{3 a}{25}+\frac{2}{45}\)

Algebraic Operations Exercise 6.12

1. Let’s find values of the following mentally 

1. 3a × 4b 
Solution:

3a × 4b = 12ab

2. 12ab ÷ 3a 
Solution:

12ab ÷ 3a = 4b

3. 12ab ÷  3 
Solution:

12ab ÷  3 = 4ab or, 12ab ÷ 4ab = 3

4. (-x² ) × x 
Solution:

(-x² ) × x = – x³

5. 9x² – 3x² 
Solution:

9x² – 3x² = 3

6. x² × x² 
Solution:

x² × x² = x⁴

7. \(x^2 \times \frac{1}{x^2}\)
Solution:

\(x^2 \times \frac{1}{x^2}\)= 1

8.  0 ÷ ab = 0
Solution:

0 ÷ ab = 0

9. 4a²b²c² × 0
Solution:

4a²b²c² × 0 = 0

10. 3ab + 3b
Solution:

3ab + 3b = a or, = 3ab ÷ a = 3b.

11. x⁰ xy 
Solution:

x⁰ xy = 1 xy

= y

12. x÷ 0
Solution:

x÷ 0 =.undefined.

Question 2. Let’s multiply 

1. 2x² × (-3y) x 6z
Solution :

2x² × (-3y)× 6z = – 36x²yz

2. 7xy² × 8x²y × xy
Solution :

7xy² × 8x²y × xy= 56x⁴y⁴

3. (- 3a²) × (4a²b) × (-2)
Solution:

(- 3a²) × (4a²b) × (-2) = 24a⁴b

4. \((-2 m n) \times \frac{1}{6} m^2 n^2 \times 13 m^4 n^4\)
Solution: 

\((-2 m n) \times \frac{1}{6} m^2 n^2 \times 13 m^4 n^4\) = \((-2) \times \frac{1}{6} \times 13 m^{1+2+4} n^{1+2+4}\)

= \(\frac{13}{3} m^7 n^7\)

5. \(\frac{2}{3} x^2 y \times \frac{3}{5} x y^2\)
Solution: 

\(\frac{2}{3} x^2 y \times \frac{3}{5} x y^2\)

= \(\frac{2}{3} x \frac{3}{5} x^{2-1} y^{1+2}\)

= \(\frac{2}{5} x^3 y^3\)

6. \(\left(-\frac{18}{5} x^2 z\right) \times\left(-\frac{25}{6} x z^2 y\right)\)
Solution:

⇒ \(\left(-\frac{18}{5} x^2 z\right) \times\left(-\frac{25}{6} x z^2 y\right)\)

= \(\frac{18}{5} \times \frac{25}{6} x^{2+1} z^{1+2} y\)

= 15 x³z³y

7. \(\left(-\frac{3}{5} \mathrm{~s}^2 \mathrm{t}\right) \times\left(\frac{15}{7} \mathrm{st}^2 \mathrm{u}\right) \times\left(\frac{7}{9} \mathrm{su}^2\right)\)

⇒ \(\left(-\frac{3}{5} \mathrm{~s}^2 \mathrm{t}\right) \times\left(\frac{15}{7} \mathrm{st}^2 \mathrm{u}\right) \times\left(\frac{7}{9} \mathrm{su}^2\right)\)

= \(-\frac{3}{5} \times \frac{15}{7} \times \frac{7}{9} s^{2+1+1} t^{1+2} u^{1+2}\)

= – s⁴t³u³

8. \(\left(\frac{4}{3} x^2 y z\right) \times\left(\frac{1}{3} y^2 z x\right) \times\left(-6 x y z^2\right)\)
Solution:

⇒ \(\left(\frac{4}{3} x^2 y z\right) \times\left(\frac{1}{3} y^2 z x\right) \times\left(-6 x y z^2\right)\)

=  \(-\frac{4}{3} \times \frac{1}{3} \times 6 x^{2+1+1} y^{1+2+1} z^{1+1+2}\)

= – \(\frac{8}{3} x^4 y^4 z^4\)

“How to solve Algebraic Operations problems Class 7 WBBSE”

9. 4a (3a + 7b)
Solution :

4a (3a + 7b) = 12a² + 28ab

10. 8a² × (2a + 5b)
Solution :

8a² × (2a + 5b)= 8a² × 2a+ 8a² x 5b

= 16 a³ + 40 a²b

11. – 1 7 x² × (3x – 4)
Solution :

– 1 7 x² × (3x – 4) = – 17x² × 3x – 17 x² (-4)

= – 51x³ + 68x²

12. \(\frac{2}{3} a b c\left(a^2+b^2-3 c^2\right)\)
Solution:

⇒ \(\frac{2}{3} a b c\left(a^2+b^2-3 c^2\right)\)

= \(\frac{2}{3} a b c \times a^2+\frac{2}{3} a b c \times b^2+\frac{2}{3} a b c\left(-3 c^2\right)\)

= \(\frac{2}{3} a^3 b c+\frac{2}{3} a b^3 c-2 a b c^3\)

13.  2 × 5x (10x²y – 100xy²)
Solution :

2 × 5x (10x²y – 100xy²) = 10x x (10x2y – 100xy2)

= 100x³y – 1000x²y²

14. (2x + 3y) (5x-y)
Solution :

(2x + 3y) (5x-y) = 2x (& – y) + 3y (5x – y)

= 10x²– 2xy + 15xy – 3y²

= 10 x² + 13xy – 3y²

15. (a² – b²) (2b – 6a)
Solution :

(a² – b²) (2b – 6a) = a² (2b²– 6a) – b² (2b – 6a)

= 2a²b – 6a³ – 2b³ + 6ab²

16. (x + 2.) (3x + 1 )
Solution :

(x + 2.) (3x + 1 ) = x (3x + 1 ) + 2 (3x + 1 )

= 3x² + x + 6x + 2

= 3x² + 7x + 2

Question 3.

1. Seema planted 3x saplings in a row, In 2x such rows let’s find how many saplings Seema can plant.
Solution :

In each row, Seema planted 3x saplings

∴ In 2x rows she will plant = 3x × 2x = 6x² saplings

2. The length of a rectangle is (4x + 1 ) m and its breadth is 3zm. Let’s calculate the area of the rectangle.
Solution :

Length & Breadth of the rectangle are (4x + 1 ) m & 3x m

∴ Area of the rectangle = (4x + 1) ×  3x sqm

= (12x²+3x) sqm.

3. Presently, the price of a dozen of bananas has increased by Rs. 6. If the previous price of a dozen of bananas was Rs. x, let’s find the cost of 2x dozen of bananas.
Solution :

⇒ Previously the price of one dozen banana was Rs. x.

⇒ Present price of one dozen banana is Rs. (x + 6)

⇒ Price of 2x dozen banana is Rs. (x + 6) × 2x

= Rs. (2x² + 12x).

4. Let’s find the area of a square whose each sode is 7x cm.
Solution :

Area of a square whose each side is 7x cm.

=  7x × 7x sqcm

= 49x² sqcm

5. The area of a rectangle is 8x² sq units. If its length is 4x units, let’s find its breadth.
Solution:

The area of rectangle = 8x² sq unit

And  its length = 4x unit

The breadth of the rectangle = (8x² ÷ 4x) unit

= 2x unit

6. Sushobhan sold 729y4 number of kites in 9y days. Let’s find the number of kites sold in average per day.
Solution:

Susobhan sold in 9y day 729 y⁴ kites

∴ He sold in 1 day = \(\frac{729 y^4}{9 y}\)  kites

= 81 y³ kites

4. Let’s divide the first algebraic expression by the second algebraic expression in the following pairs of algebraic numbers 

1. 8x³b, x²b
Solution :

= 8x³b ÷  x²b

= \(\frac{8 x^3 b}{x^2 b}\)

= 8x

2. 9xy³, xy
Solution:

9xy³, xy = – 9xy³ xy

= \(\frac{-9 x y^3}{x y}\)

= -9y²

3. -15xYz²,-x²yz²
Solution :

-15xyz²,-x²yz²= (- 1 5x²y⁴z²) -4- (- x²yz²)

= \(\frac{-15 x^2 y^4 z^2}{-x^2 y z^2}=15 y^3\)

4. 21 l³m³n³, – 4l⁴mn
Solution:

21 l³m³n³, – 4l⁴mn = (21 l³m³n³) ÷ (4l⁴mn)

= \(\frac{\left.21\right|^3 m^3 n^3}{-\left.4\right|^4 m n}\)

= \(-\frac{21}{4l} m^2 n^2\)

2. (5a² – 7ab²), a
Solution : = (5a² – 7ab²) ÷ a

= \(\frac{5 a^2}{a}-\frac{7 a b^2}{a}\)

= \(5 a-7 b^2\)

6. (- 48x⁹ + 12x⁶), 3x³
Solution :

(- 48x⁹ + 12x⁶), 3x³ = (- 48x⁹ + 1 2x⁶) ÷3x³

= \(\frac{-48 x^9}{3 x^3}+\frac{12 x^6}{3 x^3}\)

= – 16 x⁶ + 4x³

7. 15m²n + 20m²n², 5mn
Solution:

15m²n + 20m²n², 5mn = (15m²n + 20m²n²) ÷ (5mn)

= \(\frac{15 m^2 n}{5 m n}+\frac{20 m^2 n^2}{5 m n}\)

= 3m + 4mn

8. 36a⁵b² – 24a²b5, – 4a²b²
Solution:

36a⁵b² – 24a²b5, – 4a²b² = (36a⁵b² – 24a²b⁵) ÷ (-4a²b²)

= \(\frac{36 a^5 b^2}{-4 a^2 b^2}-\frac{24 a^2 b^5}{-4 a^2 b^2}\)

9. 3pqr + 6p²qr² – 9p³q²r³, – 3pqr
Solution :

3pqr + 6p²qr² – 9p³q²r³, – 3pqr = (3pqr + 6p²qr² – 9p³q²r³) ÷(- 3pqr)

= \(\frac{3 p q r}{-3 p q r}+\frac{6 p^2 q r^2}{-3 p q r}-\frac{9 p^3 q^2 r^3}{-3 p q r}\)

= – 1 – 2pr + 3²2qr²

10. m²n⁴ + m³n³ – m⁴n², – m⁴n⁴
Solution :

m²n⁴ + m³n³ – m⁴n², – m⁴n⁴ = (m²n⁴ + m³n³ – m⁴n²) ÷ (m⁴n⁴)

= \(\frac{m^2 n^4}{-m^4 n^4}+\frac{m^3 n^3}{-m^4 n^4}-\frac{m^4 n^4}{-m^4 n^4}\)

= \(-\frac{1}{m^2}-\frac{1}{m n}+\frac{1}{n^2}\)

5. Let us simplify the following –

1. a (b – c) + b (c – a) + c (a – b)
Solution :

a (b – c) + b (c – a) + c (a – b) a (b – c) + b (c – a) + c (a – b) = ab -ac +bc – ab + ac – bc

= 0

2. a (b – c) – b (c – a) – c (a – b)
Solution :

a (b – c) – b (c – a) – c (a – b) = ab – ac – bc + ab – ac + bc

= 2ab – 2ac .

3.  x (x + 4) + 2x (x – 3) – 3x²
Solution :

x (x + 4) + 2x (x – 3) – 3x² = x² + 4x + 2x² – 6x² – 3x

= x² – 2x – 3x² + 4x – 6x = – 2x

4. 3x² + x (x + 2) – 3x (2x + 1 )
Solution :

3x² + x (x + 2) – 3x (2x + 1 ) = 3x² + x² + 2x – 6x² – 3x

= 4x² – 6x² + 2x – 3x = – 2x² – x

5. (a + b) (a – b) + (b + c) (b – c) + (c + a) (c – a)
.Solution :

(a + b) (a – b) + (b + c) (b – c) + (c + a) (c a)

= a(a – b) + b(a – b) + b(b – c) + c(b – c) + c(c – a) + a (c – a)

= a² – ab + ab – b² + b² – bc + bc – c² + c² – ac + ac – a²

= 0

6. (a² + b²) (a² – b²) + (b² + c²) (b² – c²) + (c² + a²) (c² – a²)
Solution :

(a² + b²) (a² – b²) + (b² + c²) (b² – c²) + (c² + a²) (c² – a²)

= a² (a² – b²) + b² (a² – b²) + b² (b² – c²) + c² (b² – c²) + c² (c² – a²) + a² (c² – a²)

= a⁴ – a²b² + a²b² – b⁴ + b⁴ – b²c² + b²c² – c⁴ + c⁴ – a²c² + a²c² – a⁴

= 0

WBBSE Algebra Chapter 3 Concept Of Index Exercise 3 Solved Problems

Power or Index: The product obtained by multiplying a number several times by itself is called the power or Index of that number.

⇒ 3 x 3 x 3 x 3 =3⁴ (Power or Index of 3 is 4)

⇒ a x a x a x a x a =a⁵ (Power or Index of a is 5)

⇒ Some important formulas on the Index
⇒ [a is non zero integers and also m and n are integers]

1. a^m. aⁿ = a^{m + n}

2. a^m ÷ aⁿ = a^{m – n}

3. (a^m)ⁿ = a^mn

4. \(a^m=\frac{1}{a^{-m}}\)

5. a⁰ = 1

6. (ab)^m = a^m.bⁿ

7. \(\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}\)

“WBBSE Class 7 Maths Algebra Chapter 3 solutions”

Question 1. Choose the correct answer 

Read and Learn More WBBSE Solutions for Class 7 Maths

1. If we express the number 35400000 in an index of 10 we get

1. 354 x 10⁴
2. 354 x 10³
3. 354 x 10⁵
4. 354 x 10⁶

Solution: 35400000
= 354 x 100000
= 354 × 105

So the correct answer is 3. 354 x 10⁵

2. If we express the number 16489 in index form as the power of 10 we get

1. 16.489 x 10²
2. 164.89 x 10²
3. 1.6489 x 10²
4. 1648.9 x 10²

Solution: 16489 = \(\frac{16489}{100}\) x 100

= 164.89 × 10²

So the correct answer is 2. 164.89 x 10²

3. The value of 7 x 10⁶ + 3 x 10⁴+ 4 x 10³ + 6 × 10² + 8 x 10 + 9 is

1. 734689
2. 7346089
3. 7340689
4. 7034689

Solution: 7 x 10⁶ + 3 x 10⁴+ 4 x 10³ + 6 × 10² + 8 x 10 + 9
= 7000000 + 30000 + 4000+ 600 + 80 +9
= 7034689

So the correct answer is 4. 7034689

“Concept of Index Class 7 WBBSE solved problems”

Question 2. Write ‘true’ or ‘false’

1. The simplest form of (a⁷×a^-5)+(a^-3×a⁶) is \(\frac{1}{a}\)

Solution: (a⁷×a^-5)+(a^-3×a⁶)

= \(\frac{a^7 \times a^{-5}}{a^{-3} \times a^6}=\frac{a^{7-5}}{a^{-3+6}}=\frac{a^2}{a^3}=a^{2-3}=a^{-1}=\frac{1}{a}\)

So the statement is true.

2. (-5)⁴× (3)⁴= -50625

Solution: (-5)⁴× (3)⁴

= (-5) × (-5) × (-5) x (-5) × 3 × 3 × 3 × 3
= (+25) × (+25) x 81
= + 625 x 81
= 50625

So the statement is false.

3. The value of \(\frac{3 \times 7^2 \times 2^4}{21 \times 112}\) is 1

Solution: \(\frac{3 \times 7^2 \times 2^4}{21 \times 112}\)

= \(\frac{3 \times 7^2 \times 2^4}{3 \times 7 \times 2^4 \times 7}\)

= 3^1-1 x 7^2-1-1 x 2^4-4
= 3⁰ x 7⁰ x 2⁰
= 1 x 1 x 1
= 1

So the statement is true.

Question 3. Fill in the blanks

1. 20 x 18 x  = [2²×3×5]²

Solution: 20 x 18 x = 2⁴ x 3² x 5²

⇒ 2² × 5 ×3² x 2 x = 2⁴ x 3² x 5²

⇒ 2²⁺¹ x 5 x 3² x = 2⁴ x 3² x 5²

⇒   = \(\frac{2^4 \times 3^2 \times 5^2}{2^3 \times 5 \times 3^2}\)

=2^4-3 x 3^2-2 x 5^2-1
= 2¹x 3⁰ x 5¹

= 2 x 1 x 5
= 10

“Exercise 3 Algebra solutions WBBSE Class 7”

2. \(\left(x^2\right)^3 \times\left(y^{-3}\right)^2=\left(\frac{x}{y}\right)\)

Solution: \(\left(x^2\right)^3 \times\left(y^{-3}\right)^2\)

= x⁶ x y^-6

= \(\frac{x^6}{y^6}=\left(\frac{x}{y}\right)^6\)

So the answer is 6

3. 

Solution: (-7)² x 8²
= 49 x 8²
= 7² x 8²
= (7 x 8)²
= (56)²

⇔ (-7)² x 8²
= (-7 x 8)²= (-56)²

So the answer is 56 or -56.

Question 4. Express 49 x 49 x 49 x 49 as a power 7

Solution: 49 x 49 x 49 x 49
= 7 x 7 x 7 x 7 x 7 x 7 x 7 x 7
= 7⁸

“WBBSE Class 7 Maths Chapter 3 Concept of Index”

Question 5. Express the following numbers in index form as the power of 10 (taking 1, 2, and 3 places of the decimal)

1. 4678
2. 526824

Solution: 1. 4678

= \(\frac{4678}{10}\) x 10 =467.8 x 10

⇒ 4678 = \(\frac{4678}{100}\) x 100 = 46.78 x 10²

⇒ 4678 = \(\frac{4678}{1000}\) x 1000 = 4.678 x 10³

2. 526824

⇒ 526824 = \(\frac{526824}{10}\) x 10 = 52682.4 x 10

⇒ 526824 = \(\frac{526824}{100}\) x 100 = 5268.24 x 10²

⇒ 526824= \(\frac{526824}{1000}\)  x 1000 = 526.824 x 10³

Question 6. From the numbers from their expanded form

1. 9 x 10⁴ + 3 x 10² + 7
2. 3 x 10⁷ + 4 x 10⁶ + 2 x 10⁴ +7 x 10² + 5

Solution:

1. 1. 9 x 10⁴ + 3 x 10² + 7

= 90000 + 300 + 7
= 90307

2. 3 x 10⁷ + 4 x 10⁶ + 2 x 10⁴ +7 x 10² + 5

= 30000000+4000000+ 20000 + 700 + 5
= 34020705

Question 7. Simplify and express each of them in powder form

1. \(\frac{10^3 \times 10^4}{2^5 \times 5^4}\)

2. \(\frac{4^8 \times a^{10} b^4}{4^3 \times a^4 b^2}\) [ a ≠ 0, b ≠ 0]

Solution:

1. \(\frac{10^3 \times 10^4}{2^5 \times 5^4}\)

= \(\frac{(2 \times 5)^3 \times(2 \times 5)^4}{2^5 \times 5^4}\)

= 2^3+4-5 x 5^3+4-4

= 2²x5³
= 4 × 125
= 500

2. \(\frac{4^8 \times a^{10} b^4}{4^3 \times a^4 b^2}\)

=4^8-3 x a^10-4 x b^4-2

= 4⁵ x a⁶ x b²

Question 8. So that \(a^m=\frac{1}{a^{-m}}\)

Solution: a^m = a^0-(-m)

= \(\frac{a^0}{a^{-m}}=\frac{1}{a^{-m}}\)

 

Class 7 Math Solution WBBSE Concept Of Index

Concept Of Index Exercise 5.1

Let’s expand the following in the index of 

Question 1. 8275
Solution:

8275 = 8 × 10³ + 2 × 10² +7 ×10 + 5

Question 2. 90925
Solution:

90925 = 9 × 10⁴ + 0 ×10³+9 ×10² +2 ×10 + 5

Question 3. 12578
Solution:

12578 = 1 × 10⁴ +2 × 10³ + 5 × 10² +7 ×10 + 8

Question  4. 7858
Solution:

7858 = 7 ×  10³ + 8 × 10² + 5×10 + 8

Class Vii Math Solution WBBSE Concept Of Index Exercise 5.2

Question 1. 100 = 10 ^____
Solution:

100  = 10 × 10

= 10²

Question 2. 27 = 3^____
Solution:

27 = 3 × 3 × 3

= 3³

Question 3.125 = 5^____
Solution:

125 = 5 × 5 × 5

= 5³

Question 4. 32 = 2^___
Solution:

32 =  2 × 2 × 2 × 2 × 2

= 2⁵

Question 5. 343 = 7____
Solution:

343 =  7 × 7 × 7

= 7³

“Step-by-step solutions for WBBSE Class 7 Algebra”

Question 6.121 = 11______
Solution:

121=11

121= 11 × 11

=11²

Question 7. 625 = 5______
Solution: =

625 = 5 × 5 × 5 × 5

= 5⁴

Question 8. 2³  = ^____× ^_____×
Solution:

2³  = ^____× ^_____× = 2 × 2 ×2

= 2³

Question 9. 3⁴ = ^_____ ×^_____ ×^_____ ×
Solution: =

3⁴ = ^_____ ×^_____ ×^_____ × = 3 × 3 × 3 × 3=

= 3⁴

= 81

Question 10. 729 = 9^_____
Solution:

729 =  9 × 9 × 9

= 9³

Question 11. 2 × 2× 2 × 2 × 2 = ^_____
Solution:

2 × 2× 2 × 2 × 2 = 2⁵

Question 12. (- 2) × (- 2) × (- 2) = (- 2)_________
Solution:

2 × 2× 2 × 2 × 2 = (-2)³

Question 13. (- 2) ×(- 2) × (- 2) × (- 2) = (- 2) ________________
Solution:

(- 2) ×(- 2) × (- 2) × (- 2) = (- 2)

= (-2)⁴

Class Vii Math Solution WBBSE Concept Of Index Exercise 5.3

Let’s express the following numbers in the product of power from of their prime factors.

Question 1. 24.
Solution:

24= 2 × 2 × 2 × 3 = 2³ × 3

Question 2. 56.
Solution:

56 = 2× 2 × 2 ×7 = 2³ ×7

Question 3. 63.
Solution:

63. = 3 × 3 × 7 = 3² × 7

Question 4. 72
Solution:

72 = 2 × 2 × 2× 3 ×3 = 2³ × 3²

Question 5. 200
Solution:

200 = 2  ×2 × 2× 5× 5 = 2³ × 5²

Class Vii Math Solution WBBSE Concept Of Index Exercise 5.4

Let’s put > or < signs in the respective blank squares

Question 1. 5³ ____________ 3⁵
Solution:

5³ = 5× 5× 5

= 125

3⁵ = 3×3 × 3× 3×3

= 243

∴ 5³ < 3⁵

Question 2. 6²______________2⁶
Solution:

6² = 6 ×6

= 36

2⁶ = 2 × 2 × 2 × 2 × 2 × 2

= 64

∴ 6² < 2⁶

Question 3. 2⁴________4²
Solution:

2⁴ = 2 × 2 × 2 × 2

= 16

4²= 4 × 4 = 16

∴ 2⁴= 4²

Question 4. 7²______2⁷
Solution:

7² = 7x 7

= 49

2⁷ = 2×2 × 2 × 2 × 2 × 2 ×2

= 128

∴ 7²< 2⁷

Question 5. 3⁴_________4³
Solution:

3⁴ = 3 × 3 × 3 × 3

= 81

4³ = 4 × 4 × 4 = 64

∴ 3⁴> 4³

“Class 7 WBBSE Maths Index chapter exercise solutions”

Question 6. 3⁵_______5³
Solution: 

3⁵ = 3 × 3 × 3 × 3 × 3

= 243

5³=5 × 5 × 5

= 125

∴ 3⁵>5³

WB Class 7 Math Solution Concept Of Index Exercise 5.5

Question 1. 2⁵ × 2⁷
Solution:

2⁵ × 2⁷ = 2⁵⁺⁷

= 2¹²

Question 2. (-3)¹⁸× (-3)¹²
Solution:

(-3)¹⁸× (-3)¹²= (3)¹⁸⁺¹²

= (-30)³⁰

Question 3. 10⁸ x 10²
Solution:

10⁸ x 10² = (10)⁸⁺²

= (10)¹⁰

Question 4. 2¹⁵ ÷ 2¹³ 
Solution:

2¹⁵ ÷ 2¹³ 

= (2)^15-13

= 2²

Question 5. 9¹⁵-9¹⁴
Solution:

9¹⁵-9¹⁴ = 9^15-14

= 9

Question 6. 11⁶÷ 11⁴
Solution:

11⁶÷ 11⁴ = 11^6-4

= 11²

Concept Of Index Exercise 5.6

Let’s fill the blank squares given below:

Question 1. 9²÷ 9² = ________________
Solution: =

9²÷ 9² = 9^2-2

= 9⁰

= 1

Question 2. 7³_______0 = 1
Solution:

7³_______0 = 1

7³ ÷7³ = 1

= 7⁰

“Solved problems of Index chapter Class 7 WBBSE”

Question 3. 11⁰= ______________
Solution:

11⁰ = 1

Question 4. 1 =13__________
Solution:

1 =13___

∴ 1= 13⁰= 1

∴ 1-1 = 0

Question 5. 1 =(- 13) _______________
Solution:

1 =(- 13)

= (-13)⁰

WB Class 7 Math Solution Concept Of Index Exercise – 5.7

Question 1. 6⁵÷2⁵ = ________________
Solution:

6⁵÷2⁵

= \(\frac{6^5}{2^5}\)

= \(\left(\frac{6}{2}\right)^5\)

= (3)⁵

Question 2. _____ =  7² ÷ 2²
Solution: 

= \(\frac{7^2}{2^2}\)

= \(\left(\frac{7}{2}\right)^5\)

Question 3. 10² = ______ × ________ 
Solution:

10² = 10 × 10

Question 4. (4)²  × 6² = ________²
Solution:

(4)²  × 6²

(-4² × 6²)  = (-24)²

Question 5. (5)⁰ = ______________
Solution:

(5)⁰ =  1

6.  \(\left(\frac{2}{3}\right)^3\)
Solution:

⇒ \(\left(\frac{2}{3}\right)^3\)

= \(\frac{2^3}{3^3}=\frac{8}{27}\)

Concept Of Index Exercise – 5.8

Question 1. Let us express 8 × 8 × 8 as power of 2.
Solution :

Given

8 × 8 × 8

= 2³ × 2³ × 2³⁺³⁺³

= (2)⁹

Question 2. 25 × 25 × 25 × 25 to be expessed as power of 5.
Solution :

Given

25 × 25 × 25 × 25

= 5² × 5² × 5²× 5²

= (5)^2+2+2+2

= (5)⁸

Question 3. Let’s express 36 × 36 × 36 as the power of 6.
Solution :

Given

36 × 36 ×  36

= (6)² × (6)² × (6)²

= (6)^{2+2 +2}  = 6⁶

Question 4. Let’s express 81 x 81 as power of 3.
Solution :

Given

81 x 81

= 3⁴ × 3⁴ = (3)⁴⁺⁴

= 3⁸

Question 5. Let us find the values of the following

1. \(\frac{2^6 \times 3^5}{(6)^5}\)
Solution:

Given

⇒ \(\frac{2^6 \times 3^5}{(6)^5}\)

= \(\frac{2^6 \times 3^5}{(2 \times 3)^5}\)

= \(\frac{2^6 \times 3^5}{2^5 \times 3^5}\)

= 2¹

= 2

2. \(\frac{10^3 \times 10^4}{2^5 \times 5^4}\)
Solution:

Given

⇒ \(\frac{10^3 \times 10^4}{2^5 \times 5^4}\)

= \(\frac{(2 \times 5)^3 \times(2 \times 5)^4}{2^5 \times 5^4}=\frac{2^3 \times 5^3 \times 2^4 \times 5^4}{2^5 \times 5^4}\)

= \(\frac{(2)^3 \times(2)^4 \times 5^3}{2^5}=2^{3+4-5} \times 5^3\)

2² × 5³ = 4 × 125

= 500

3. \(\frac{5^9 \times 5^6}{5^7}\)
Solution:

Given

⇒ \(\frac{5^9 \times 5^6}{5^7}\)

= \(5^{9+6-7}\)

= 5⁸

4. \(\frac{6^4 \times 3^8}{3^{12}}\)
Solution:

Given

⇒ \(\frac{6^4 \times 3^8}{3^{12}}\)

= \(\frac{(3 \times 2)^4 \times 3^8}{3^{12}}\)

= \(\frac{(3)^{4+8} \times 2^4}{3^{12}}\)

= 2 × 2 × 2 × 2

= 16

5. \(\frac{25^2 \times 25^5}{5^{10}}\)

Given

⇒ \(\frac{25^2 \times 25^5}{5^{10}}\)

= \(\frac{\left(5^2\right)^2 \times\left(5^2\right)^5}{5^{10}}\)

= \(\frac{5^4 \times 5^{10}}{5^{10}}\)

= 5⁴

5 × 5 × 5 × 5 = 625

“Algebra Chapter 3 Concept of Index WBBSE Class 7”

6. \(\frac{2^3 \times 3^9}{3^6 \times 6^3}\)

Given

⇒ \(\frac{2^3 \times 3^9}{3^6 \times 6^3}\)

⇒  \(\frac{2^3 \times 3^9}{3^6 \times(2 \times 3)^3}\)

= \(\frac{2^3 \times 3^9}{3^6 \times 2^3 \times 3^3}\)

=\(\frac{3^9}{3^9}\)=1

7. \(\left(\frac{a^7}{a^5}\right) \times \begin{gathered}
a^2 \\
(a \neq 0)
\end{gathered}\)
Solution:

⇒ \(\left(\frac{a^7}{a^5}\right) \times \begin{gathered}
a^2 \\
(a \neq 0)
\end{gathered}\)

= \(\left(\frac{a^7}{a^5}\right) \times a^2\)

= \(\left(a^{7-5}\right) \times a^2\)

a² × a² = a⁴

8.\(\frac{3 \times 7^2 \times 2^4}{21 \times 112}\)
Solution:

Given

⇒ \(\frac{3 \times 7^2 \times 2^4}{21 \times 112}\)

= \(\frac{3 \times 2^4 \times 7^2}{3 \times 7 \times 7 \times 16}\)

= \(\frac{7^2 \times 2^4}{7^2 \times 2^4}\)

= 1

WB Class 7 Math Solution Concept Of Index Exercise 5.9

Question 1. Let’s express the distances given below in index of 10 and try to get a better idea of the distances Distance of Mercury from Sun is 57900000 km. Distances of Mars and Jupiter from Sun are 227900000 km. and 778300000 km respectively.
Solution :

Given

Distances of Mars and Jupiter from Sun are 227900000 km. and 778300000 km respectively.

1. Distance of Mercury from Sun = 57900000 km.

= 579 × 10⁵ km.

2. Distance of Mars from Sun = 227900000 km

= 2279 ×10⁵ km.

3. Distance of Jupiter from Sun = 778300000 km.

= 7783 × 10⁵ km.

Question 2. Let’s fill in the gaps –

1. The distance between Earth and the Moon is 384,000,000 m. = 384 x____________ m.
Solution :

⇒ Distance between Earth and Moon

⇒ 384,000,000 m = 384 x 10 ⁶

∴  10⁶

2. The speed of light in a vacuum is 3,00,000,000 m/sec. = 3 x_m/sec.
Solution:

⇒ The speed of light in a vacuum

= 3,00,000,000 m/sec. = 3 ×10⁸ m/sec

∴  10⁸

3. Let’s express the following number in index form as power of 10 (taking 1,2 and 3 places of decimal)

1. 978 =
Solution:

= 97. 8 × 10

= 9. 78 × 10²

= 0.978  × 10³

2. 1592170 =
Solution:

= 159217 × 10

= 15921 .7 × 10²

= 1592 .17 × 10²³

“WBBSE Class 7 Maths Exercise 3 Index chapter”

Question 4. Let’s form the numbers from their expanded form given below –

1. 3 ×10³ +2 ×10² +7 × 10+ 2
Solution:

3 ×10³ +2 ×10² +7 × 10+ 2 = 3000 + 200 + 70 + 2

= 3272

2. 2  × 10³+3 ×10+ 5
Solution:

2  × 10³+3 ×10+ 5 = 2000 + 30 + 5

= 2035

3. 8 ×10⁴ + 2 × 10³ + 3 × 10² + 6
Solution:

8 ×10⁴ + 2 × 10³ + 3 × 10² + 6

= 80000 + 2000 + 300 + 6

= 82306

4. 9 ×10⁴ +5 ×10³ +6 × 10² + 6 ×10
Solution:

9 ×10⁴ +5 ×10³ +6 × 10² + 6 ×10

= 90000 + 5000 + 600 + 70

= 95670

Question 5. Let’s simplify and express each of them in powder form

1. \(\frac{2^3 \times 3^5 \times 16}{3 \times 32}\)
Solution:

= \(\frac{2^3 \times 3^5 \times 2^4}{3 \times 2^5}=\frac{2^{3+4} \times 3^5}{2^5 \times 3}\)

= \(2^{7-5} \times 3^{5-1}\)

= \(2^2 \times 3^4\)‘

= 2² × 9²

= (18)²

= 18²

= 324

2. \(\left[\left(6^2\right)^3 \times 6^4\right] \div 6^7\)
Solution:

\(\left[\left(6^2\right)^3 \times 6^4\right] \div 6^7\) = 6⁶ × 6⁴÷ 6⁷

= 6^6+4-7

= 6³

3. \(\frac{3 \times \cdot 7^2 \times 11^0}{21 \times 7}\)
Solution:

⇒ \(\frac{3 \times 7^2 \times 1}{3 \times 7 \times 7}\)

= \(\frac{7^2}{7^2}\)

= 1

4. \(\left(3^0+2^0\right) \times 5^0\)
Solution:

(1+1) ×1

= 2 ×1 = 2

“How to solve Index problems Class 7 WBBSE”

5. \(\frac{4^5 \times a^8 b^3}{4^5 \times a^5 b^2}(b \neq 0)\)
Solution:

6. \(\frac{2^8 x^7}{\left(2^2\right)^3 \times x^3}=\frac{2^8 \times x^7}{2^6 \times x^3}\)

⇒ \(\frac{2^8 x^7}{\left(2^2\right)^3 \times x^3}=\frac{2^8 \times x^7}{2^6 \times x^3}\)

= \(2^{8-6} \times x^{7-3}=2^2 \times x^4\)

= \((2)^2 \times\left(x^2\right)^2=\left(2 x^2\right)^2\)