Haritha

Class 6 Math Solutions WBBSE Chapter 20 Geometrical Concept Of Circle Exercise

Question 1. From the figure, let us answer the following :

1. O is the _____ of the circle.
Solution: O is the centre of the circle

2. Line segment OQ is the ____ of the circle.
Solution: radius of the circle.

“Step-by-step solutions for geometrical concept of circle Class 6 WBBSE”

3. Line segment PQ is the ______ of circle.
Solution: diameter of circle.

“WBBSE Class 6 Maths Chapter 20 geometrical concept of circle solutions”

4. Line segment OP is the _______ of circle.
Solution: Line segment OP is radius of circle.

5. Line segment MN is ______ of the circle.
Solution: Line segment MN is a chord of the circle.

Read and Learn More WBBSE Solutions For Class 6 Maths

6. The points M and N have divided the circle into two _______ parts.
Solution: The points M and N have divided the circle into two arcs.

7. The area bounded by arc SR and radii SO and RO is called ______
Solution: The area bounded by the arc SR and radii SO and RO is called the sector.

8. The diameter PQ has divided the circle into two equal parts which are called _____
Solution: The diameter PQ has divided the circle into two equal parts which are called semi-circles.

Question 2. Let us write (✓) for the correct statement and (X) for the wrong one.

1. All diameters of a circle are chords. [✓]
2. All chords of a circle are its diameter. [X]
3. The length of the diameter of a circle is twice the length of its radius. [✓]
4. Sector of a circle is part of its circular area. [✓]
5. Arc of a circle is part of the circle. [✓]
6. Centre of the circle is a fixed point in the circle. [✓]
7. Two diameters always intersect each of them. [✓]

“WBBSE solutions for Class 6 Maths circle geometry”

Class 6 History	Class 6 Social Science
Class 6 Geography	Class 6 Science
Class 6 Maths	Class 6 Science MCQs
Class 6 General Science	Class 6 Maths Solutions
Class 6 Geography	Class 6 Hindi

Question 3. With the help of a compass and a pencil, a circle of radius 3 cm is drawn. Let’s name its centre, radius, diameter, chord, and arc.
Solution:

1. Centre O
2. Radius OP
3. Diameter AB
4. Chord MN
5. Arc PRB

Question 4. In this figure, let us colour the major segment yellow and minor segment green.

Solution:

1. Major segment — yellow portion
2. Minor segment — green portion

“Class 6 Maths WBBSE Chapter 20 circle geometry exercises”

Question 5. The radii of two circles are 2 cm and 4 cm respectively. Let’s write the lengths of their diameter without measuring.
Solution:

Given

The radii of two circles are 2 cm and 4 cm respectively.

⇒ Radius of 1st circle = 2 cm.

⇒ Diameter of 1st circle = 2×2 cm. = 4 cm.

⇒ Radius of 2nd circle = 4 cm.

⇒ Diameter of 2nd circle = 2 x 4cm. = 8 cm.

Question 6. If the length of the greatest chord of a circle is 10 cm, let’s write what will be the length of its radius.
Solution:

Given

If the length of the greatest chord of a circle is 10 cm

⇒ Length of the greatest chord of a circle = 100 cm.

∴ Diameter of the circle = 10 cm.

∴ Radius of the circle = 10/2 cm = 5 cm.

Class 6 Math Solutions WBBSE Chapter 19 Measurement Of Time Exercise 19

Question 1. Given below is the time chart of Priya on a holiday. Let’s try to get few information.
Solution:

⇒ On that day, Priya woke up: at 6.00 a.m

⇒ Priya studied at: 9.00 a.m

⇒ Priya gossips with friends at: 11.00 a.m

⇒ She had lunch at: 1.00 a.m

⇒ After lunch at: 4.00 p.m

⇒ On that day Priya went to bed: at 10.00 p.m

Question 2. Today I left my house at 10:20 a.m. for school. Returned home at 4:45 p.m. Let’s calculate how long I was out of home.
Solution:

I was out of home = 4:45 p.m – 10:20 am = 6 hrs 25 min

Question 3. Yesterday Deba went to bed at 10: 25 p.m. But today he woke up at 6:10 a.m. Let’s find out how long he slept.
Solution:

Deba slept = 6:10 a.m – 10: 25 p.m = 7 hrs 45 mins

Read and Learn More WBBSE Solutions For Class 6 Maths

Question 4. Today I am going to Puri with my family. The train will leave from Howrah station at 22:35 hours, we will reach Howrah station at 20:44 hrs. Let’s calculate how long we have to wait at the station.
Solution:

Given

Today I am going to Puri with my family. The train will leave from Howrah station at 22:35 hours, we will reach Howrah station at 20:44 hrs.

We have to wait at the station = 22:35 hours – 20:44 hours

= 1 hour 51 mins Hrs min

⇒ \(\begin{array}{lc}
\text { Hrs } & \text { min } \\
22 & 35 \\
20 & 44 \\
\hline 1 & 51 \\
\hline
\end{array}\)

“WBBSE Class 6 Maths Chapter 19 measurement of time solutions”

Question 5. Today I have a history examination. Exam started at 11: 30 a.m. The teacher’s digital wristwatch is showing 13:15. Let me find out how long I had already written the examination. If the exam is to end at 2:30 p.m., how long shall I be able to write the examination, when the exam will be over, let me find out what will the digital watch of the teacher read.
Solution:

Given

Today I have a history examination. Exam started at 11: 30 a.m. The teacher’s digital wristwatch is showing 13:15. Let me find out how long I had already written the examination. If the exam is to end at 2:30 p.m.

⇒ I had already written the examination = 13:15 – 11:30 am = 1 hr 45 min.

⇒ I shalf be able to write the examination = 2:30 p.m. -13:15 = 1 hr 15 min.

⇒ The time of digital watch of the teacher = 13:15 + 1 hr 15 min = 14:30.

Question 6. To reach the father’s office from home, it takes 2 hrs 27 min but while returning home, it takes 2 hrs 51 min. Let me find out how long it takes in total to go to my father’s office and back.
Solution:

Given

To reach father’s office from home, it takes 2 hrs 27 min but while returning home, it takes 2 hrs 51 min.

⇒ The time to go to father’s office and back

= 2 hrs 27 mins + 2 hrs 57 mins

= 5 hrs 24 mins.

⇒ \(\begin{array}{lll}
\text { Hrs } & \min & \\
2 & 27 & \\
2 & 57 & \\
5 & 24 &
\end{array}\)

⇒ \(\begin{array}{rcc}
7 . \mathrm{hr} & \min & \mathrm{sec} \\
8 & 32 & 41 \\
+18 & 42 & 25 \\
& 74 & 66 \\
\hline \mathrm{hrs} & \min & \mathrm{sec} \\
\hline
\end{array}\)

Solution:

1. \(\begin{array}{rrr}
\mathrm{hr} & \mathrm{min} & \mathrm{sec} \\
8 & 32 & 41 \\
+18& 42 & 25 \\
26 & 74 & 66 \\
\hline 27 \mathrm{hrs} & 15 \mathrm{~min} & 6 \mathrm{sec} \\
\hline
\end{array}\)

2. \(\begin{array}{ccc}
\text { 2 hr } & \text { min } & \text { sec } \\
\text { (8) } & 60+11 & 60+37 \\
9 & 12 & 37 \\
-3 & 38 & 41 \\
\hline 5 \mathrm{hr} & 33 \mathrm{~min} & 56 \mathrm{sec} \\
\hline
\end{array}\)

“WBBSE solutions for Class 6 Maths measurement of time”

Class 6 History	Class 6 Social Science
Class 6 Geography	Class 6 Science
Class 6 Maths	Class 6 Science MCQs
Class 6 General Science	Class 6 Maths Solutions
Class 6 Geography	Class 6 Hindi

Question 8. Let’s add and subtract

1. (4 hr 33 min 20 sec) + (9 hr 52 min 25 sec)
Solution:

⇒ \(\begin{array}{ccc}
\text { Hrs } & \text { Mins } & \text { Sec } \\
4 & 33 & 20 \\
+9 & 52 & 25 \\
13 & 85 & 45 \\
\hline=14 \mathrm{Hrs} & 25 \mathrm{~min} & 45 \mathrm{sec} . \\
\hline
\end{array}\)

2. (6 hr 42 min 2 sec) – (2 hr 55 min 42 sec)
Solution:

⇒ \(\begin{array}{ccc}
\text { Hrs } & \text { Min } & \text { Sec } \\
5 & 60+41 & 60+2 \\
6 & 42 & 2 \\
-2 & 55 & 42 \\
\hline 3 & 46 & 20
\end{array}\)

= 3 hrs 46 min 20 Sec

3. (18 hr 19 min 15 sec) + (9 hr 55 min 48 sec)
Solution:

⇒ \(\begin{array}{rcc}
\text { Hrs } & \text { Min } & \text { Sec } \\
18 & 19 & 15 \\
9 & 55 & 48 \\
\hline 27 & 74 & 63 \\
\hline
\end{array}\)

= 28 hrs 15 min 3 sec

4. (23 hr 7 min) – (19 hr 29 min 18 sec)
Solution:

⇒ \(\begin{array}{ccc}
\text { Hrs } & \min& \text { Sec } \\
22 & 60+6 & 60+0 \\
23 & 7 & 00 \\
9 & 55 & 48 \\
\hline 13 & 11 & 12 \\
\hline
\end{array}\)

= 13 hrs 11 min 12 Sec.

Class 6 WBBSE Math Solutions Chapter 19 Measurement Of Time Exercise 19.1

Question 1. Debubabu built a new house. He will paint the two windows of his house. Each window has two panes. If he takes 2 hr 15 min to paint 1 window pane, let’s find how long he will take to colour 2 windows.
Solution:

Given

Debubabu built a new house. He will paint the two windows of his house. Each window has two panes. If he takes 2 hr 15 min to paint 1 window pane,

⇒ Time required to paint 1 window pane = 2 hrs 15 m

∴ Time required to paint 2 window panes = 2 hr 15 min x 2 = 4 hrs 30 min

“Class 6 Maths WBBSE Chapter 19 time measurement exercises”

Question 2. In 11 hrs 36 min, Phani Da can make 4 exactly similar clay statues. Let’s find out how long he will take to make one such statue. He takes equal time to make each statue.
Solution:

Given

In 11 hrs 36 min, Phani Da can make 4 exactly similar clay statues.

⇒ Time required to make 4 clay statues = 11 hrs 36 min

∴ Time required to make 1 clay statue = 11 hr 36 min ± 4

= 2 hrs 54 mins.

Question 3.

1. 3 hrs 26 min x 4 = How many hours and minutes?
Solution:

⇒ 3 hrs 26 min x 4 = 12 hr 104 min = 13 hrs 44 min

2. 7 hrs 13 min x 12 = How many hours and minutes?
Solution:

⇒ 7 hrs 13 min x 12 = 84 hrs 156 min = 86 hr. 36 min

3. 3 hrs 27 min ÷ 9 = How many minutes and seconds?
Solution:

4. 15 hrs ÷ 12 = How many hrs and mins?
Solution:

5. 6 hrs 18 sec ÷ 9 = How many min, sec?
Solution:

6. 5 hrs10 min 4 sec ÷ 4 = How many hours, min, sec?
Solution:

“Step-by-step solutions for measurement of time Class 6 WBBSE”

7. 2 hrs 32 min 41 sec÷ 3 = How many hours, min, secs?
Solution:

⇒ 2 hr 32 min 4 sec x 3

⇒ 2 hr 32 min 4 sec x 3

⇒ 6 hr 96 min 12 sec

= 7 hr 36 min 12 sec

Class 6 Maths Solutions WBBSE Chapter 19 Measurement Of Time Exercise 19.1

Question 1. 1st February of 2010 was Monday. Let’s calculate 1st March 2010 and 1st April 2010 will be which days.
Solution:

⇒ 1 st February 2010 was monday and February has 28 days.

∴ 1st March 2010 was Monday and March has 31 days.

∴ 3 days after Monday is Thursday.

∴ 1st April 2010 is Thursday.

Question 2. 01/02/2012 was Wednesday, then let’s calculate which days were the following dates: 01/03/2012, 01/04/2012, 01/05/2015,04/06/2012
Solution:

⇒ 1st February of 2012 was Wednesday and February 2012 had 29 days.

∴ 1 day after Wednesday is Thursday.

∴ . 1st March 2012 was Thursday.

⇒ March has 31 days.

⇒ 3 days after Thursday is Sunday.

∴ 1st April 2012 was Sunday.

⇒ April has 30 days.

⇒ 2 days after Sunday is Tuesday.

∴ 1st May 2012 was Tuesday.

⇒ May has 31 days, 6 days after Tuesday is Monday.

⇒ 4th June 2012 is Monday.

“WBBSE Class 6 Maths Chapter 19 important questions and answers”

Question 3. 1st January of 1996 was Sunday. Let us calulate which day 1st January of 1997 will be.
Solution:

⇒ 1st January 1996 was Monday.

⇒ Since 1996 was a leap year so 1996 had 366 days.

⇒ 2 days after Monday is Wednesday.

∴ 1st January 1997 was Wednesday.

Question 4. 1st March of 2004 was Monday, which day will be 1st April of 2005.
Solution:

⇒ 1st March 2004 was Monday 2005 year had 365 days.

⇒ 1 day after Monday is Tuesday.

⇒ 1st March 2005 was Tuesday

⇒ 1st March 2005 had 31 days.

⇒ 3days after Tuesday is Friday.

∴ 1st April 2005 was Friday.

Question 5. 1st June of 2008 was Tuesday. Let’s calculate which day was 1st June of 2006.
Solution:

⇒ 1st June 2008 was Sunday. 2008 was a leap year.

⇒ So, 2008 had 366 days.

⇒ 2006 and 2007 had 365 days.

⇒ 4days before Sunday is Thursday.

⇒ 1st June 2006 was Thursday.

Question 6. Independence day of 2013 is Thursday, let’s find which day will be Independence day of 2016.
Solution:

⇒ Independence day of 2013, 15th August 2013 was Thursday.

⇒ 2013 had 365 days and 2016 is a leap year.

⇒ So 2016 has 366 days.

⇒ 4 days after Thursday is Monday.

∴  Independence day of 2016 will be Monday.

Question 7. From the calendar of 2013 let us write which day of the week are the following and without looking into calender, let’s find which days of the week these days were in the year 2011 — Children’s day, Teachers’ day, Gandhi birthday, Republic day, Netaji Jayanti, World environment day (5th of June).
Solution:

1. Children’s Day (14th November)

⇒ 14th November 2013 was Thursday.

⇒ 2014 had 365 days and 2012 had 366 days.

⇒ 3 days before Thursday is Monday.

2. Teachers’ day (5th September)

⇒ 5th September of 2013 was Thursday.

⇒ 2011 had 365 days and 2012 had 366 days.

⇒ 3 days before Thursday is Monday.

∴ Teachers’ day – 5th September 2011 was Monday.

“Solved examples of time measurement WBBSE Class 6”

3. Gandhi birthday – 2nd October.

⇒ 2nd October 2013 was Wednesday.

⇒ 2011 had 365 days and 2012 had 366 days.

3 days before Wednesday Sunday.

∴ Gandhi birthday (2nd October) 2011 was Sunday.

4. Republic day (26th January)

⇒ 26th January of 2013 was Saturday.

⇒ 2011 had 365 days and 2012 had 366 days.

⇒ 3 days before Saturday is Wednesday.

∴ Republic day (26th January, 2011) was Wednesday.

5. Netaji Jayanti (23rd January)

⇒ 23rd January 2013 was Wednesday.

⇒ 2011 had 365 days and 2012 had 366 days.

⇒ 3 days before Wednesday is Sunday.

∴ Netaji Jayanti (23rd January) 2011 was Sunday.

6. World Environment Day (5th June)

⇒ 5th June 2013 was Wednesday.

⇒ 2011 had 365 days and 2012 had 366 days.

⇒ 3 days before Wednesday is Sunday.

⇒ World Environment Day 5th June, 2011 was Sunday.

“Best guide for Class 6 Maths WBBSE measurement of time”

Question 8.

1. Let’s write the leap years between 1895 to 1915.
2. Let’s write the leap years between 2010 to 2030.

Solution:

1. Leap years between 1895 to 1915 = 1896; 1900; 1904,1908, and 1912.
2. Leap years between 2010 to 2030 = 2012, 2016, 2020, 2024, and 2028.

Question 9. I stayed for 4 years in a house at College Ghat Road, from 2010 to 2013. Let’s calculate for how many days I stayed in that house.
Solution:

⇒ In 2010 – Number of days = 365 In 2011 — Number of days = 365

⇒ In 2012 — Number of days = 366 In 2013 — Number of days = 365 Total days in 4 years = 365 + 365 = 366 + 365 = 1461

Question 10. My birthday is on 15th December. In 2013, my birthday was on Sunday. Which day of the week will my birthday be in the years 2014, 2015 and 2016?
Solution:

My birthday is on 15th December 2013, which was Sunday. In 2014 had 365 days

⇒ 1 day after Sunday is Monday.

∴ 15th December 2014 was Monday.

⇒ In 2015 had 365 days.

∴ 1 days after Monday is Tuesday.

∴ 15th December, 2015 is Tuesday.

⇒ 2016 is of 366 days.

∴ 2 days after Tuesday is Thursday.

∴ 15th December 2016 will be Thursday.

Question 11. After independence, how many leap years have passed till 2014?
Solution:

⇒ Independence day was 15th August 1947.

⇒ The leap years that have passed till 2014 are 1948,1952,1956,1960,1964,1968 1972,1976,1980,1984,1988,1992 1996, 2000, 2004,2008 and 2012.

WBBSE Class 6 Maths Solutions Chapter 19 Measurement Of Time Exercise 19.2

Question 1. My date of birth is 19 -11 -1975, i.e., 19th November 1975. Let us find my
Solution:

⇒ \(\begin{array}{lll}
\text { Years } & \text { Months } & \text { Days } \\
& 12+9 & 30+10 \\
2010 & 10 & 10 \\
-1975 & 11 & 19 \\
\hline 24 & 10 & 21 \\
\hline
\end{array}\)

= 24 years 10 months 21 days.

Question 2. The construction of our main road began in summer on 6/6/2010. It took 1 yr 3 months 18 days to finish the work. Let’s calculate to find on which date the road construction work was completed.
Solution:

⇒ \(\begin{array}{ccc}
\text { Years } & \text { Months } & \text { Days } \\
2010 & 06 & 06 \\
+\quad 1 & 03 & 18 \\
\hline 2011 & 09 & 24 \\
\hline
\end{array}\)

⇒ The date of road construction work was completed on 24/09/ 2011.

Question 3. My present age is 11 yrs 7 months 10 days. Let me calculate after how many years I shall be eligible to cast my vote.
Solution:

⇒ \(\begin{array}{cll}
\text { Years } & \text { Months } & \text { Days } \\
17 & 12+0 & 30 \\
(-) 11 & 07 & 10 \\
\hline 6 & 08 & 20 \\
\hline
\end{array}\)

= 6 yrs 04 months 20 days.

“How to measure time accurately Class 6 WBBSE”

Question 4. The age of my father is 52 years 8 months 20 days. My uncle is 3 years 10 months 26 days older than my father. Let me find the age of my uncle.
Solution:

⇒ \(\begin{array}{ccl}
\text { Years } & \text { Months } & \text { Days } \\
52 & 8 & 20 \\
+3 & 10 & 26 \\
\hline 56 & 7 & 16 \\
\hline
\end{array}\)

= Age of my uncle = 56 years 7 months 16 days.

Question 5. Let us find values

1. \(\begin{array}{rcc}
\text { Year } & \text { Months } & \text { Days } \\
9 & 10 & 27 \\
+\quad 5 & 8 & 21 \\
\hline 14 & 18 & 48 \\
\hline
\end{array}\)

Solution: 15 years 7 months 18 days.

2. \(\begin{array}{ccc}
\text { Year } & \text { Month } & \text { Days } \\
29 & 11 & 19 \\
5 & 9 & 25 \\
+6 & 3 & 13 \\
\hline 40 & 23 & 57 \\
\hline
\end{array}\)

Solution: 42 years 0 months 27 days

3. \(\begin{array}{ccl}
\text { Year } & \text { Month } & \text { Days } \\
& 12+3 & 30 \\
11 & 3 & \mathrm{x} \\
-6 & 10 & 28 \\
\hline 5 & 5 & 02 \\
\hline
\end{array}\)

Solution: 5 years 5 months 2 days.

4. \(\begin{array}{|c|c|c|}
\hline \text { Year } & \begin{array}{c}
\text { Month } \\
12+6
\end{array} & \begin{array}{l}
\text { Days } \\
30+19
\end{array} \\
\hline 11 & 6 & 19 \\
\hline-6 & 10 & 21 \\
\hline 5 & 8 & 28 \\
\hline
\end{array}\)

Solution: 5 years 8 months 28 days.

Question 6. (1) 8 yrs 8 months 28 days + 11 yrs 8 months 18 days = How many yr, months and days?
Solution:

⇒ \(\begin{array}{rcl}
\text { Years } & \text { Months } & \text { Days } \\
8 & 8 & 28 \\
+11 & 8 & 18 \\
\hline 19 & 16 & 46 \\
\hline
\end{array}\)

= 20 yrs 5 months 16 days.

2. 20 yrs 11 months and days? -10 yrs 8 months 23 days = How many yr, month
Solution:

⇒\(\begin{array}{ccc}
\text { Years } & \text { Months } & \begin{array}{c}
\text { days } \\
30+0
\end{array} \\
20 & 11 & 0 \\
10 & 89 & 23 . \\
\hline 10 & 2 & 7 \\
\hline
\end{array}\)

= 10 years 2 months 7 days

3. 8 yrs 7 months 21 days x 9 = How many yr, month and days?
Solution:

= 77 yrs.9 months 9 days.

Question 7. My age is □ yrs □ months □ days. The age of my friend is □ yrs □ months □ days. What is the sum of our age and who is older than the two and by how much? Let us find out.
Solution:

My age is 15 years 11 months 26 days. The age of my friend is 14 yrs 9 months 20 days.

⇒ \(\text { Sum }=\begin{array}{ccc}
\text { Year } & \text { Month } & \text { Days } \\
15 & 11 & 26 . \\
14 & 9 & 20 \\
\hline 29 & 20 & 46 \\
\hline
\end{array}\)

= 30 yrs 9 months 16 days.

I am older than my friends.

Difference

⇒ \(\begin{array}{lll}
\text { Years } & \text { Month } & \text { Days } \\
15 & 11 & 26 \\
14 & 9 & 20 \\
\hline 1 & 2 & 6 \\
\hline
\end{array}\)

= 1 yr 2 months 6 days.

“Understanding time conversion and measurement Class 6 WBBSE”

Question 8. My date of birth is □. Today my age is □ yrs □ months □ days.
Solution:

My date of birth is 04.10-2001

Today my age is — (19-11 -2014)

⇒ \(\begin{array}{rll}
\text { Year } & \text { Month } & \text { Days } \\
2014 & 11 & 19 \\
-2001 & 10 & 04 \\
\hline 13 & 1 & 15 \\
\hline
\end{array}\)

Today my age is 13yrs 1month 15days

WBBSE Class 6 Math Solutions WBBSE Chapter 21 Fundamental Concept Of Ratio And Proportion

We take lengths of our pen caps and lengths of pens without caps and try to take their ratio and write them in a table like below.

Let us see the picture count the number ratio in its lowest form and write the ratio.

Class 6 WBBSE Math Solutions Chapter 21 Fundamental Concept Of Ratio And Proportion Exercise 21

Question 1. I have taken measurements of the floor of my living room. I see that the rectangular floor is 8 m in length and 5 m in breadth. Let’s find the ratio of its length to its breadth. Let’s also write if the ratio is the ratio of greater or lesser inequality.

Read and Learn More WBBSE Solutions For Class 6 Maths
Solution:

⇒ Length of the floor = 8 m.

⇒ Breadth of the floor = 5 m.

∴ The ratio of length and breadth = 8:5

It is a ratio of greater inequality.

“WBBSE Class 6 Maths Chapter 21 ratio and proportion solutions”

Class 6 History	Class 6 Social Science
Class 6 Geography	Class 6 Science
Class 6 Maths	Class 6 Science MCQs
Class 6 General Science	Class 6 Maths Solutions
Class 6 Geography	Class 6 Hindi

Question 2. Sabita is making garlands of chinarose and marigold, if she made 12 garlands of chinarose and 15 of marigold, then let’s find what is the ratio of the number of garlands of chinarose and marigold. Let’s also write if the ratio is a ratio of greater or lesser inequality.
Solution:

Given

Sabita is making garlands of chinarose and marigold, if she made 12 garlands of chinarose and 15 of marigold,

⇒ The number of garlands in china rose = 12 and

⇒ Number of garlands of marigolds = 15

∴ Ratio of garlands of qhinarose and marigold = 12:15 = 4:5

It is a ratio of lesser inequality.

Question 3. The ratio of my age to Sutapa’s age is 5: 6. If my age is 10 years, let’s find the age of Sutapa.
Solution:

⇒ The ratio of my age to Sutapa’s age = 5:6

⇒ or, \(\frac{\text { My age }}{\text { Sutapa’s age }}=\frac{5}{6}\)

⇒ or, \(\frac{10 \text { years }}{\text { Sutapa’s age }}=\frac{5}{6}\)

∴ 5 x Sutapa’s age = 6 x 10 years 6×10

∴ Sutapa’s age = \(\frac{6 \times 10}{5}\)  years = 12 years.

“WBBSE solutions for Class 6 Maths fundamental concept of ratio and proportion”

Question 4. My mother gave sweets to me and Raju. Me and Raju ate sweets in the ratio of 1:3. Let’s find how many sweets my mother gave us [Let’s try any 4 values].

⇒ \(\frac{\text { No. of sweets I ate }}{\text { No. of sweets Raju ate }}=\frac{1}{3}=\frac{2}{6}\) i.e., mother gave (2 + 6) = 8 sweets also she can give (1 + 3) = 4 sweets

Solution:

\(\frac{\text { No. of sweets I ate }}{\text { No. of sweets Raju ate }}=\frac{1}{3}=\frac{2}{6}\) = \(\frac{3}{9}=\frac{4}{12}\)

⇒ In 1st case, mother gave us = 1 +3 = 4 sweets

⇒ In 2nd case, mother gave us = 2 + 6 = 8 sweets

⇒ In 3rd case, mother gave us = 3 + 9 = 12 sweets

⇒ In 4th case, mother gave us = 4 +12 = 16 sweets

Question 5. Today 10 of us have come to the park to play. If the ratio of number of girls and boys is 2:3 in our group, let’s find how many girls and boys have come to play.
Solution:

Given

Today 10 of us have come to the park to play. If the ratio of the number of girls and boys is 2:3 in our group,

⇒ Total number of students = 10

⇒ Ratio of number of girls and boys = 2:3

∴ Number of female students = 2/5 parts

∴ Number of boy students = 3/5 parts

⇒ Number of girl students = 2/5 x 10 = 4

⇒ Number of boy students = 3/5 x 10 = 6

“Class 6 Maths WBBSE Chapter 21 ratio and proportion exercises”

Question 6. Father bought 4 pairs of bananas from the market. If brother and sister ate those bananas in the ratio of 1 : 3, let’s find how many bananas each of them ate.
Solution:

Given

Father bought 4 pairs of bananas from the market. If brother and sister ate those bananas in the ratio of 1 : 3,

⇒ Total number of bananas = 4 pairs = 4 x 2 = 8

⇒ Ratio of bananas brother and sister ate = 1:3

∴ Number of bananas brother ate = \(\frac{1}{1+3} \times 8=\frac{1}{4} \times 8=2\)

Number of bananas sister ate = \(\frac{3}{1+3} \times 8=\frac{3}{4} \times 8=6\)

Class 6 Maths Solutions WBBSE Chapter 21 Fundamental Concept Of Ratio And Proportion Exercise 21.1

Question 1. Let’s find if the following numbers are in proportion or not:

1. 13, 52,30,120
Solution:

⇒ 1st term x 4th term = 13 x 120 = 1560

⇒ 2nd term x 3rd term = 52 x 30 = 1560

∴ 13: 52 :: 30:120 (In proportion)

2. 22,11,72, 36
Solution:

⇒ 1 st term x 4th term = 22 x 36 = 792

⇒ 2nd term x 3rd term = 11 x 72 = 792

∴ 22: 11:: 72: 36 (In proportion)

3. 45, 27,15, 25
Solution: 45,27, 15,25

⇒ 1st term x 4th term = 45 x 25 = 1125

⇒ 2nd term x 3rd term = 27 x 15 = 405

∴ 1st term x 4th term  2nd term x 3rd term

∴ 45, 27,15, 25 are not in proportion.

4. 18, 20, 27, 30
Solution:

⇒ 1st term x 4th term = 18 x 30 = 540

⇒ 2nd term x 3rd term = 20 x 27 = 540

∴ 18: 20:: 27: 30 (In proportion)

5. 11,22, 36, 72
Solution:

⇒ 1st term x 4th term = 11 x 72 = 792

⇒ 2nd term x 3rd term = 22 x 36 = 792 (In proportion) .

Question 2. Let’s write if the folowing relations are True/False.

1. 4.5 litre: 13.5 litre:: 4 kg: 12 kg
Solution:

⇒ 4.5 litre: 13.5 litre:: 4kg: 12 kg

⇒ 4.5: 13.5:: 4: 12

⇒ 1st term x 4th term = 4.5 x 12 = 54

⇒ 2nd term x 3rd term = 13.4 x 4 = 54

∴ The relation is True.

“Step-by-step solutions for ratio and proportion Class 6 WBBSE”

2. 12 km: 8 km:: 61 km:: 40 km
Solution:

⇒ 12 km: 8 km:: 60 km: 40 km

⇒ or, 12: 8:: 60: 40

⇒ 1st term x 4th term = 12 x 40 = 480

⇒ 2nd term x 3rd term = 8 x 60 = 480

∴ The relation is True.

3. 20 men: 45 men:: 180 rupee: 270 rupee
Solution:

⇒ 20 men: 45 men:: 180 rupees: 270 rupees

∴ 1 st term x 4th term = 20 x 270 = 5400

⇒ 2nd term x 3rd term = 45 x 180 = 8100

∴ 20:45 ≠ 180:270

∴ The relation is False.

4. 15m: 9m :: 35m: 21m.
Solution:

⇒ 15m: 9 m :: 35 m: 21 m

⇒ or 15:9:: 35:21

⇒ 1st term x 4th term = 15 x 21 = 315

⇒ 2nd term x 3rd term = 9 x 35 = 315

∴ The relation is True.

WBBSE Math Solutions Class 6 Chapter 21 Fundamental Concept Of Ratio And Proportion Exercise 21.2

Question 1. Let’s find if the following numbers are in proportion or not. If they are in proportion, let’s write all the possible proportions with them.

1. 3,15, 4, 20
Solution:

1st case: 3, 15, 4, 20

⇒ 1st term x 4th term = 3 x 20 = 60

⇒ 2nd term x 3rd term = 15 x 4 = 60

∴ 3: 15:: 4.20

2nd case: 3, 4,15, 20

⇒ 1st term x 4th term = 3 x 20 = 60

⇒ 2nd term x 3rd term = 4 x 15 = 60

∴ 3:4:: 15: 20

3rd case: 15, 3, 20, 4

⇒ 1st term x 4th term = 15 x 4 = 60

⇒ 2nd term x 3rd term = 3 x 20 = 60

∴  15: 3:: 20: 4

4th case: 15, 20, 3, 4

⇒ 1st term x 4th term = 15×4 = 60

⇒ 2nd term x 3rd term = 20 x 3 = 60

∴ 15: 20:: 3: 4

“WBBSE Class 6 Maths Chapter 21 important questions and answers”

2. 6,18,7, 21
Solution:

1st case: 6,18, 7, 21

⇒ 1st term x 4th term = 6×21 = 126

⇒ 2nd term x 3rd term = 18 x 7 = 126

∴ 6:18:: 7: 21

2nd case: 6, 7,18, 21

⇒ 1st term x 4th term = 6 x 21 =126

⇒ 2nd term x 3rd term = 7 x 18 = 126

∴ 6: 7:: 18: 21

3rd case: 18, 6, 21: 7

⇒ 1st term x 4th term = 18 x 7 = 126

⇒ 2nd term x 3rd term = 6 x 21 =126

∴ 18: 6:: 21: 7

4th case: 18, 21,6, 7

⇒ 1st term x 4th term = 18 x 7 = 126

⇒ 2nd term x 3rd term = 21 x 6 = 126

∴ 18: 21:: 6: 7

“Solved examples of ratio and proportion WBBSE Class 6”

3. 5, 15, 7, 21
Solution:

1st case: 5,15, 7, 21

⇒ 1st term x 4th term = 5 x 21 =105

⇒ 2nd term x 3rd term = 15×7 = 105

∴ 5: 15:: 7: 21

2nd case: 15, 5, 21,7

⇒ 1st term x 4th term = 15 x 7 = 105

⇒ 2nd term x 3rd term = 5×21 =105

∴ 15: 5 :: 21: 7

3rd case: 5, 7, 15, 21

⇒ 1st term x 4th term = 5 x 21 = 105

⇒ 2nd term x 3rd term = 7 x 15 = 105

∴ 5:7:: 15: 21

4th case: 15, 21,5, 7

⇒ 1st term x 4th term = 15 x 7 = 105

⇒ 2nd term x 3rd term = 21 x 5 = 105

∴ 15:21:: 5: 7

4. 7, 21,4, 12
Solution:

1st case: 7, 21, 4,12

⇒ 1st term x 4th term = 7 x 12 = 84

⇒ 2nd term x 3rd term =21 x 4 = 84

∴ 7:21:: 4: 12

2nd Case: 21, 7,12, 4

⇒ 1st x 4th term = 21 x 4 = 84

⇒ 2nd term 3rd term = 7 x 12 = 84

∴ 21: 7:: 12: 4

3rd case: 7, 4, 21, 12

⇒ 1st term x 4th term = 7 x 12 = 84

⇒ 2nd term x 3rd term = 4×21 =84

∴  7:4:: 21: 12

4th case 21,12, 7, 4

⇒ 1st term x 4th term = 21 x 4 = 84

⇒ 2nd term x 3rd term = 12 x 7 = 84

∴ 21: 12:: 7: 4

“Best guide for Class 6 Maths WBBSE ratio and proportion problems”

5. 3,15,10,50
Solution:

1st case: 3, 15,10, 50

⇒ 1st term x 4th term = 3 x 50 = 150

⇒ 2nd term x 3rd term = 15 x 10 = 150

∴ 3: 15:: 10:50

2nd case: 15, 3, 50,10

⇒ 1st term x 4th term = 15 x 10 = 150

⇒ 2nd term x 3rd term = 3 x 50 = 150

∴ 15:3 :: 50:15

3rd case: 3, 10,15, 50

⇒ 1st term x 4th term = 3 x 50 = 150

⇒ 2nd term x 3rd term = 10×15=150

∴ 3: 10:: 15: 50

4th case: 15, 50, 3,10

⇒ 1st term x 4th term = 15 x 10 = 150

⇒ 2nd term x 3rd term = 50 x 3 = 150

∴ 15: 50:: 3: 10

6. 2, 6, 7, 21.
Solution:

1st case: 2, 6, 7, 21

⇒ 1 st term x 4th term = 2 x 21 = 42

⇒ 2nd term x 3rd term = 6 x 7 = 42

∴ 2:6:: 7: 21

“How to solve ratio and proportion questions Class 6 WBBSE”

2nd case: 6, 2, 21,7

⇒ 1 st term x 4th term = 6 x 7 = 42

⇒ 2nd term x 3rd term = 2 x 21 =42

∴ 6: 2:: 21: 7

3rd case: 2, 7, 6, 21

⇒ 1st term x 4th term = 2 x 21 = 42

⇒ 2nd term x 3rd term = 7 x 6 = 42

∴ 2: 7:: 6: 21

4th case: 6, 21,2, 7

⇒ 1st term x 4th term = 6 x 7 = 42

⇒ 2nd term x 3rd term = 21 x 2 = 42

∴ 6: 21:: 2: 7

Question 2. Let’s take 4 positive whole numbers:

1. 4,6,8,12
Solution:

1st case: 4, 6, 8, 12

⇒ 1st term x 4th term = 4 x 12 = 48

⇒ 2nd term x 3rd term = 6 x 8 = 48

∴ 4: 6:: 8: 12

2nd case: 6, 4,12, 8

⇒ 1st term x 4th term = 6 x 8 = 48

⇒ 2nd term x 3rd term = 4 x 12 = 48

∴ 6:4 :: 12:8

3rd case: 6,12, 4, 8

⇒ 1st term x 4th term = 6 x 8 = 48

⇒ 2nd term x 3rd term = 12 x 4 = 48

∴ 6:12 :: 4:8

4th case: 12, 6, 8, 4

⇒ 1st term x 4th term = 12 x 4 = 48

⇒ 2nd term x 3rd term = 6 x 8 = 48

∴12: 6:: 8:4

Question 3.

1. 3.5, 7, 2, 4
Solution: 3.5, 7, 2, 4

1st case: 3.5, 7, 2, 4

⇒ 1st term x 4th term = 3.5 x 4 = 14

⇒ 2nd term x 3rd term = 7×2 = 14

∴ 3.5: 7:: 2: 4

2nd case: 7, 3.5, 4, 2

⇒ 1st term x 4th term = 7×2 = 14

⇒ 2nd term x 3rd term = 3.5 x 4 = 14

∴ 7: 3.5:: 4:2

3rd case: 3.5, 2, 7, 4

⇒ 1st term x 4th term = 3.5 x 4 = 14

⇒ 2nd term x 3rd term = 2×7 = 14

∴ 3.5: 2:: 7: 4

4th case: 7,4, 3.5, 2

⇒ 1st term x 4th term = 7×2 = 14

⇒ 2nd term x 3rd term = 4 x 3.5 = 15

∴ 7: 4:: 3.5:2

2. 1.5, 4.5, 2.5; 7.5
Solution: 1.5, 4.5, 2.5, 7.5

1st case: 1.5, 4.5, 2.5, 7.5

⇒ 1st term x 4th term = 1.5 x 7.5 = 11.25

⇒ 2nd term x 3rd term = 4.5 x 2.5 = 11.25

∴ 1.5: 4.5:: 2.5: 7.5

2nd case: 4.5, 7.5,1.5, 2.5

⇒ 1st term x 4th term = 4.5 x 2.5 = 11.25

⇒ 2nd term x 3rd term = 7.5 x 1.5 = 11.25

∴ 4.5: 7.5:: 1.5:2.5

3rd case: 1.5, 2.5, 4.5, 7.5

⇒ 1st term x 4th term = 1.5 x 7.5 = 11.25

⇒ 2nd term x 3rd term = 2.5 x 4.5 = 11.25

∴1.5: 2.5:: 4.5 4.5: 4.5

4th case: 4.5,1.5, 7.5, 2.5

⇒ 1st term x 4th term = 4.5 x 2.5 = 11.25

⇒ 2nd term x 3rd term = 1.5 x 7.5 = 11.25

∴ 4.5: 1.5:: 7.5:2.5

3. 0.35,1.05, 0.09, 0.27
Solution: 0.35,1.05, 0.09, 0.27

1st case: 0.35,1.05,0.09, 0.27

⇒ 1st term x 4th term = 0.35 x 0.27 = 0.0945

⇒ 2nd term x 3rd term = 1.05 x 0.09 = 0.0945

∴ 0.35:1.05 :: 0.09:0.27

2nd case: 1.05, 0.35, 0.27, 0.09

⇒ 1st term x 4th term = 1.05 x 0.09 = 0.0945

⇒ 2nd term x 3rd term = 0.35 x 0.27 = 0.0945

∴ 1.05: 0.35:: 0.27: 0.09

3rd case: 0.35,1.05,0.09, 0.27

⇒ 1st term x 4th term = 0.35 x 0.27 = 0.0945

⇒ 2nd term x 3rd term = 1.05 x 0.09 = 0.0945

∴ 0.35:1.05 :: 0.09:0.27

4th case: 1.05,0.27,0.35, 0.09

⇒ 1st term x 4th term = 1.05 x 0.09 = 0.0945;

⇒ 2nd term x 3rd term = 0.27 x 0.35 = 0.0945

∴ 1.05: 0.27 ::0.35:0.09

4. Let’s take 4 positive decimal numbers:
Solution:

1st case: 2.4,4.8,7.5,15

⇒ Here, 1st term x 4th term = 2.4 x 15 = 36

⇒ 2nd term x 3rd term F 4.8 x 7.5 = 36

= 2.4: 4.8:: 7.5:15

2nd case: 4.8, 2.4,15, 7.5

⇒ Here, 1 st term x 4th term = 4.8 x 7.5 = 36

⇒ 2nd term x 3rd term = 2.4 x 15 = 36

∴ 4.8: 2.4:: 15:7.5

3rd case: 15, 7.5, 4.8, 2.4

⇒ Here, 1 st term x 4th term = 15 x 2.4 = 36

⇒ 2nd term x 3rd term = 7.5 x 4.8 = 36

∴ 15:7.5:: 4.8 : 2.4

4th case: 2.4, 7.5, 4.8,15

⇒ Here, 1 st term x 4th term = 2.4 x 15 = 36

⇒ 2nd term x 3rd term = 7.5 x 4.8 = 36

∴ 2.4: 7.5:: 4.8:15

Question 4.

1. \(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{6}\)
Solution:

1st case: \(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{6}\)

⇒ 1st term x 4th term = \(\frac{1}{2} \times \frac{1}{6}=\frac{1}{12}\)

⇒ 2nd term x 3rd term = \(\frac{1}{3} \times \frac{1}{4}=\frac{1}{12}\)

∴ \(\frac{1}{2}: \frac{1}{3}:: \frac{1}{4}: \frac{1}{6}\)

2nd case: \(\frac{1}{3}, \frac{1}{2}, \frac{1}{6}, \frac{1}{4}\)

⇒ 1st term x 4th term = \(\frac{1}{3} \times \frac{1}{4}=\frac{1}{12}\)

⇒ 2nd term x 3 rd term = \(\frac{1}{2} \times \frac{1}{6}=\frac{1}{12}\)

∴ \(\frac{1}{3}: \frac{1}{2}:: \frac{1}{6}: \frac{1}{4}\)

3rd case: \(\frac{1}{2}, \frac{1}{4}, \frac{1}{3}, \frac{1}{6}\)

⇒ 1st term x 4th term = \(\frac{1}{2} \times \frac{1}{6}=\frac{1}{12}\)

⇒ 2nd term x 3rd term = \(\frac{1}{4} \times \frac{1}{3}=\frac{1}{12}\)

∴ \(\frac{1}{2}: \frac{1}{4}:: \frac{1}{3}: \frac{1}{6}\)

4th case: \(\frac{1}{3}, \frac{1}{6}, \frac{1}{2}, \frac{1}{4}\)

⇒ 1st term x 4th term = \(\frac{1}{3} \times \frac{1}{4} \times=\frac{1}{12}\)

⇒ 2nd term x 3rd term = \(\frac{1}{6} \times \frac{1}{2}=\frac{1}{12}\)

∴ \(\frac{1}{3}: \frac{1}{6}:: \frac{1}{2}: \frac{1}{4}\)

2. \(\frac{1}{5}, \frac{1}{10}, \frac{1}{16}, \frac{1}{32}\)

1st case: \(\frac{1}{5}, \frac{1}{10}, \frac{1}{16}, \frac{1}{32}\)

⇒ 1st term  x 4th term = \(\frac{1}{5} \times \frac{1}{32}=\frac{1}{160}\)

⇒ 2nd term x 3rd term = \(\frac{1}{10} \times \frac{1}{16}=\frac{1}{160}\)

∴ \(\frac{1}{5}: \frac{1}{10}:: \frac{1}{16}: \frac{1}{32}\)

2nd case: \(\frac{1}{10}, \frac{1}{5}, \frac{1}{32}, \frac{1}{16}\)

⇒ 1st term x 4th term = \(\frac{1}{10} \times \frac{1}{16}=\frac{1}{160}\)

⇒ 2nd term x 3rd term = \(\frac{1}{5} \times \frac{1}{32}=\frac{1}{160}\)

∴ \(\frac{1}{10}: \frac{1}{5}:: \frac{1}{32}: \frac{1}{16}\)

3rd case: \(\frac{1}{5}, \frac{1}{16}, \frac{1}{10}, \frac{1}{32}\)

⇒ 1st term x 4th term = \(\frac{1}{5} \times \frac{1}{32}=\frac{1}{160}\)

⇒ 2nd term x 3rd term = \(\frac{1}{16} \times \frac{1}{10}=\frac{1}{160}\)

∴ \(\frac{1}{5}: \frac{1}{16}:: \frac{1}{10}: \frac{1}{32}\)

4th case: \(\frac{1}{10}, \frac{1}{32}, \frac{1}{5}, \frac{1}{16}\)

⇒ 1st term x 4th term = \(\frac{1}{10} \times \frac{1}{16}=\frac{1}{160}\)

⇒ 2nd term x 3rd term = \(\frac{1}{32} \times \frac{1}{5}=\frac{1}{160}\)

∴ \(\frac{1}{10}: \frac{1}{32}:: \frac{1}{5}: \frac{1}{16}\)

3. \(\frac{1}{3}, \frac{1}{6}, \frac{1}{9}, \frac{1}{18}\)

1st case: \(\frac{1}{3}, \frac{1}{6}, \frac{1}{9}, \frac{1}{18}\)

⇒ 1st term x 4th term = \(\frac{1}{3} \times \frac{1}{18}=\frac{1}{54}\)

⇒ 2nd term x 3rd term = \(\frac{1}{6} \times \frac{1}{9}=\frac{1}{54}\)

∴ \(\frac{1}{3}: \frac{1}{6}:: \frac{1}{9}: \frac{1}{18}\)

2nd case: \(\frac{1}{6}, \frac{1}{3}, \frac{1}{18}, \frac{1}{9}\)

⇒ 1st term x 4th term = \(\frac{1}{6} \times \frac{1}{9}=\frac{1}{54}\)

⇒ 2nd term x 3rd term = \(\frac{1}{3} \times \frac{1}{18}=\frac{1}{54}\)

∴ \(\frac{1}{6}: \frac{1}{3}:: \frac{1}{18}: \frac{1}{9}\)

3rd case: \(\frac{1}{3}, \frac{1}{9}, \frac{1}{6}, \frac{1}{18}\)

⇒ 1st term x 4th term = \(\frac{1}{3} \times \frac{1}{18}=\frac{1}{54}\)

⇒ 2nd term x 3rd term = \(\frac{1}{9} \times \frac{1}{6}=\frac{1}{54}\)

∴ \(\frac{1}{3}: \frac{1}{9}:: \frac{1}{6}: \frac{1}{18}\)

4th case: \(\frac{1}{6}, \frac{1}{18}, \frac{1}{3}, \frac{1}{9}\)

⇒ 1st term x 4th term = \(\frac{1}{6} \times \frac{1}{9}=\frac{1}{54}\)

⇒ 2nd term x 3rd term = \(\frac{1}{18} \times \frac{1}{3}=\frac{1}{54}\)

∴ \(\frac{1}{6}: \frac{1}{18}:: \frac{1}{3}: \frac{1}{9}\)

4. \(\frac{1}{8}, \frac{1}{24}, \frac{1}{5}, \frac{1}{15}\)
Solution:

1st case: \(\frac{1}{8}, \frac{1}{24}, \frac{1}{5}, \frac{1}{15}\)

⇒ 1st term x 4th term = \(\frac{1}{8} \times \frac{1}{15}=\frac{1}{120}\)

⇒ 2nd term x 3rd term = \(\frac{1}{24} \times \frac{1}{5}=\frac{1}{120}\)

2nd case: \(\frac{1}{24}, \frac{1}{8}, \frac{1}{15}, \frac{1}{5}\)

⇒ 1st term x 4th term = \(\frac{1}{24} \times \frac{1}{5}=\frac{1}{120}\)

⇒ 2nd term x 3rd term = \(\frac{1}{8} \times \frac{1}{15}=\frac{1}{120}\)

∴ \(\frac{1}{24}: \frac{1}{8}:: \frac{1}{15}: \frac{1}{5}\)

3rd case: \(\frac{1}{8}, \frac{1}{5}, \frac{1}{24}, \frac{1}{15}\)

⇒ 1st term x 4th term = \(\frac{1}{8} \times \frac{1}{15}=\frac{1}{120}\)

⇒ 2nd term x 3rd term = \(\frac{1}{5} \times \frac{1}{24}=\frac{1}{120}\)

∴ \(\frac{1}{8}: \frac{1}{5}:: \frac{1}{24}: \frac{1}{15}\)

4th case: \(\frac{1}{24}, \frac{1}{15}, \frac{1}{8}, \frac{1}{5}\)

⇒ 1st term x 4th term = \(\frac{1}{24} \times \frac{1}{5}=\frac{1}{120}\)

⇒ 2nd term x 3rd term = \(\frac{1}{15} \times \frac{1}{8}=\frac{1}{120}\)

∴ \(\frac{1}{24}: \frac{1}{15}:: \frac{1}{8}: \frac{1}{5}\)

5. \(\frac{1}{3}, \frac{1}{5}, \frac{1}{4}, \frac{3}{10}\)
Solution:

1st case: \(\frac{1}{3}, \frac{1}{5}, \frac{1}{4}, \frac{3}{10}\)

⇒ 1st term x 4th term = \(\frac{1}{3} \times \frac{3}{10}=\frac{1}{20}\)

⇒ 2nd term x 3rd term = \(\frac{1}{5} \times \frac{1}{4}=\frac{1}{20}\)

∴ \(\frac{1}{3}: \frac{1}{5} \neq \frac{1}{4}: \frac{3}{10}\)

∴ \(\frac{1}{3} \times \frac{1}{4}=\frac{1}{20}\)

4th case: \(\frac{1}{6}, \frac{1}{18}, \frac{1}{3}, \frac{1}{9}\)

⇒ 1st term x 4th term = \(\frac{1}{6} \times \frac{1}{9}=\frac{1}{54}\)

⇒ 2nd term x 3rd term = \(\frac{1}{18} \times \frac{1}{3}=\frac{1}{54}\)

∴ \(\frac{1}{6}: \frac{1}{18}:: \frac{1}{3}: \frac{1}{9}\)

6. \(\frac{1}{8}, \frac{1}{24}, \frac{1}{5}, \frac{1}{15}\)

1st case: \(\frac{1}{8}, \frac{1}{24}, \frac{1}{5}, \frac{1}{15}\)

⇒ 1st term x 4th term = \(\frac{1}{8} \times \frac{1}{15}=\frac{1}{120}\)

⇒ 2nd term x 3rd term = \(\frac{1}{24} \times \frac{1}{5}=\frac{1}{120}\)

2nd case: \(\frac{1}{24}, \frac{1}{8}, \frac{1}{15}, \frac{1}{5}\)

⇒ 1st term x 4th term = \(\frac{1}{24} \times \frac{1}{5}=\frac{1}{120}\)

⇒ 2nd term x 3rd term = \(\frac{1}{8} \times \frac{1}{15}=\frac{1}{120}\)

∴ \(\frac{1}{24}: \frac{1}{8}:: \frac{1}{15}: \frac{1}{5}\)

3rd case: \(\frac{1}{8}, \frac{1}{5}, \frac{1}{24}, \frac{1}{15}\)

⇒ 1st term x 4th term = \(\frac{1}{8} \times \frac{1}{15}=\frac{1}{120}\)

⇒ 2nd term x 3rd term = \(\frac{1}{5} \times \frac{1}{24}=\frac{1}{120}\)

∴ \(\frac{1}{8}: \frac{1}{5}:: \frac{1}{24}: \frac{1}{15}\)

4th case: \(\frac{1}{24}\), \(\frac{1}{15}\), \(\frac{1}{8}\), \(\frac{1}{5}\)

⇒ 1st term x 4th term = \(\frac{1}{24} \times \frac{1}{5}=\frac{1}{120}\)

⇒ 2nd term x 3rd term = \(\frac{1}{15} \times \frac{1}{8}=\frac{1}{120}\)

∴ \(\frac{1}{24}: \frac{1}{15}:: \frac{1}{8}: \frac{1}{5}\)

7. \(\frac{1}{3}, \frac{1}{5}, \frac{1}{4}, \frac{3}{10}\)

1st case: \(\frac{1}{3}, \frac{1}{5}, \frac{1}{4}, \frac{3}{10}\)

⇒ 1st term x 4th term = \(\frac{1}{3} \times \frac{3}{10}=\frac{1}{20}\)

⇒ 2nd term x 3rd term = \(\frac{1}{5} \times \frac{1}{4}=\frac{1}{20}\)

∴ \(\frac{1}{3}: \frac{1}{5} \neq \frac{1}{4}: \frac{3}{10}\)

∴ \(\frac{1}{3} \times \frac{1}{4}=\frac{1}{20}\)

2nd case: \(\frac{1}{3}, \frac{1}{5}, \frac{3}{10}, \frac{1}{4}\)

= \(\frac{1}{5} \times \frac{3}{10}=\frac{3}{50}\)

∴ \(\frac{1}{3}: \frac{1}{5} \neq \frac{3}{10}: \frac{1}{4}\)

Again, \(\frac{1}{3} \times \frac{1}{4}=\frac{1}{12}, \frac{1}{5} \times \frac{3}{10}=\frac{3}{50}\)

∴ \(\frac{1}{3}, \frac{1}{5}, \frac{1}{4}, \frac{3}{10}\) are not in proportion.

8. \(\frac{8}{3}, 8, \frac{5}{3}, 8\)

1st case: \(\frac{8}{3} \times 8=\frac{64}{3}, 8 \times \frac{5}{3}=\frac{40}{3}\)

2nd case: \(\frac{8}{3} \times \frac{5}{3}=\frac{40}{.3}, 8 \times 8=64\)

∴ \(\frac{8}{3}, 8, \frac{5}{3}, 8\) are not in proportion.

Class 6 Math WBBSE Solutions Chapter 21 Fundamental Concept Of Ratio And Proportion Exercise 21.3

Question 1. Which of the following cases can be represented in ratio? Let’s write.

1. My friend Jatita’s weight and her height.
Solution: My friend Jatita’s weight and her height are not possible.

2. In this month, number of days I went to school and number of days my friend Jahir went to school.
Solution: In this month, the number of days I went to school and number of days my friend Jahir went to school — Possible.

3. The money I had and the money I spent.
Solution: The money I had and the money I spent — Possible.

4. Litres of water in my water bottle and temperature of water.
Solution: Litres of water in my water bottle and temperature of water — Not possible.

5. How long I played today and how long my brother played.
Solution: How long I play today and how long my brother played — Possible.

Question 2. Let’s express the following quantities in ratio and identify the ratio of greater inequality and the ratio of lesser inequality.

1. 10 kg, 15 kg.
Solution: 10 kg: 15 kg

= 10:15

= 2:3

(Ratio of lesser inequality).

2. 27 things and 18 things
Solution: 27 things: 18 things

= 27:18 = 3:2

(Ratio of greater inequality).

3. Rs. 30 and Rs. 22.50
Solution: Rs. 30: Rs. 22.50

= 30: 22.5

= 6 : 4.5 = 4:3

(Ratio of greater inequality).

4. 4.9 litre and 8.4 litre
Solution: 4.9 litre: 8.4 litre

= 4.9: 84

= 49 : 84

= 7:12

(Ratio of lesser inequality).

5. 52 m and 78 m
Solution: 52 m: 78 m = 52: 78

=4:6

= 2:3

(Ratio of lesser inequality).

6. 1 hr 24 mins and 6 hrs 18 min
Solution: 1 hr 24 min: 6 hr 18 min

= (60 + 24) min : (360 + 10) min = 84 min : 378 m

= 12:54

= 2:9

(Ratio of lesser inequality).

Question 3. A bamboo stick is 2 m long. 0.75 m of it is coloured red and rest part is colored white.
Solution:

Length of the bamboo stick = 2 m.

Portion of red colour = 0.75 m.

∴ Portion of white colour = (2 – 0.75) m = 1.25 m.

1. Let us find the ratio between the total length of bamboo and the portion painted red.
Solution: Length of bamboo: Length of red portion

= 2 : 0.75 = 2 : 75/100

= 2 : 3/4 = 8 : 3

2. Let’s find the ratio between the total length of bamboo and the portion painted white.
Solution: Length of bamboo: Length of white portion

= 2: 1.25 = 2: 125/100

= 2 : 5/4 = 8.5

3. Let’s find the ratio between the part coloured red and that coloured white.
Solution: Length of red portion: Length of white portion = 0.75:1.25

= 75:125

= 3:5

Question 4. The ratio of length and breadth of my room is 7: 5. Let’s write 4 possible perimeters of the room in the above ratio.
Solution: Length: Breadth = 7:5

∴ \(\frac{\text { Length }}{\text { Breadth }}=\frac{7}{5}=\frac{14}{10}=\frac{28}{20}=\frac{56}{40}\)

1. Perimeter = 2(7 + 5) = 2 x 12 = 24 units
2. Perimeter = 2(14 + 10)= 2 x 24 = 48 units
3. Perimeter = 2(28 + 20) = 2 x 48 = 96 units
4. Perimeter = 2(56 + 40) = 2 x 96 = 192 units.

Question 5. I have 26 stamps. Myself and Mita shall divide the stamps in the ratio of 8: 5. Let’s find how many stamps each of us will have.
Solution:

Total number of stamps = 26

∴ \(\frac{\text { No. of Stamps of me }}{\text { No. of Stamps of Mita }}=\frac{8}{5}=8: 5\)

No. of Stamps of mine = 8/13 x 26 = 16

No. of stamps of Mita = 5/13 x 26 = 10

Question 6. The ratio of my textbooks and story books is 4:3. If number of textbooks are 28, let’s find the number of story books and a total number of books.
Solution:

The ratio of my textbooks and story books = 4:3

⇒ Total number of textbooks = 28.

∴ \(\frac{\text { No. of my text books }}{\text { No. of story books }}=\frac{4}{3}\)

⇒ or, \(\frac{28}{\text { No. of story books }}=\frac{4}{3}\)

⇒ 4 x No. of story books = 3 x 28

⇒ No. of story books = \(\frac{3 \times 28}{4}\) = 21.

∴ Total number of books = No. of text books + No. of storybooks

= 28 + 21 = 49

Question 7. In a particular type of jewellery the ratio of gold and silver is 4: 7. In such type of jewelry ornament, how many milligrams of gold has been mixed with 357 milligrams of silver, it is to be calculated.
Solution:

In a particular type of jewellery the ratio of gold and silver is 4: 7. In such type of jewelry ornament,

Ratio of Gold and Silver = 4:7

∴ \(\frac{\text { Gold }}{\text { Silver }}=\frac{4}{7}\)

⇒ \(\frac{\text { Weight of Gold }}{\text { Weight of Silver }}=\frac{4}{7}\)

⇒ \(\frac{\text { Weight of Gold }}{357 \mathrm{mg}}=\frac{4}{7}\)

7 x weight of Gold = \(4 \times 357 \mathrm{mg}\)

Weight of Gold = \(\frac{4 \times 357}{7} \mathrm{mg} . \)

∴ Weight of Gold = 204 mg

Question 8. Let’s write the ratio of the three angles of an equilateral triangle.
Solution:

Each angle of an equilateral triangle = 60°

∴ 1st angle : 2nd angle : 3rd angle = 60°: 60°: 60°

= 1:1:1.

Question 9. Let’s write the ratio of the angles of a right-angled isosceles triangle.
Solution:

The three angles of the right-angled isosceles triangle are 90°, 45°, and 45°.

∴ 1st angle : 2nd angle : 3rd angle = 90° : 45°, 45°

= 2:1:1 .

Question 10. Let me divide Rs. 210 among Fatema and Sakira in the ratio of 3:4. Let me find how much each would get.
Solution:

Total Amount = Rs. 210.

Amount of Fatema: Amount of Sakira = 3:4

∴ Amount of money Fatema received

Amount of money Sakira received = Rs. (210 – 90) = Rs. 120.

Question 11. Mohit bought 6 bananas for Rs. 18 from a vendor and Raju bought 2 dozen of bananas for Rs. 72 from another vendor. Let’s express in ratio to find who paid more for bananas.
Solution:

⇒ Mohit bought 6 bananas for Rs. 18.

∴ Mohit bought 1 banana for Rs. 18/6 = Rs. 3.

⇒ Raju bought 2 dozen (24) bananas for Rs. 72.

∴ Raju bought 1 banana for Rs. 72/24 = Rs. 3.

∴ Ratio of prices = 3:3 = 1 :1 (Equal amount)

Question 12. The distances of Ayesha and Kamal’s house from our school are 1 km and 600 m respectively. Today Ayesha and Kamal came to school in 20 min and 12 min respectively. Let’s find, expressed in ratios, if they reached school at the same time or one of them has reached earlier.
Solution:

Given

The distances of Ayesha and Kamal’s house from our school are 1 km and 600 m respectively. Today Ayesha and Kamal came to school in 20 min and 12 min respectively.

Distance of Ayesha’s house from school: Distance of Kamal’s house from school

= 1 km : 600 m

= 1000 m: 600 m = 5:3

⇒ Again, time taken by Ayesha: time taken by Kamal = 20 min: 12 min

= 5:3

∴ Their speeds are equal.

Question 13. Let’s find which of the following ratios are equal ratios

1. 3:3 and 5:5
Solution: = 1: 1 and 1: 1

∴ Equal ratio

2. 20 : 24 and 25 : 30
Solution: 20: 24 and 25: 30

= 5: 6 and 5: 6.

∴ Equal ratio.

3. 1: 9 and 9:18
Solution: 1:9 and 9: 18

= 1: 9 and 1: 2

∴ Not equal ratio

4. 28 : 21 and 20:15
Solution: 28:21 and 20:15

= 4 : 3 and 4 : 3

∴ Equal ratio

5. 1.4: 0.6 and 6.3 : 2.7
Solution: 1.4 : 0.6 and 6.3 : 2.7

= 14 : 6 and 63 : 27

= 7 : 3 and 7 : 3

∴ Equal ratio

6. 52 : 39 and 44 : 33
Solution: 52: 39 and 44: 33

= 4 : 3 and 4 : 3

∴ Equal ratio

Question 14. Let’s verify which of the following numbers are in proportion:

1. 9, 7, 36, 28
Solution: 9, 7, 36, 28

1st term x 4th term = 9 x 28 = 252

2nd term x 3rd term = 7 x 36 = 252

∴ 9, 7, 36, and 28 are in proportion.

2. 12, 30,14, 70
Solution: 12, 30, 14, 70

1st term 4th term = 1 2 x 70 = 840

2nd term x 3rd term = 30 x 1 4 = 420

∴ 12, 30, 14, 70 are not in proportion.

3. 24, 6, 108, 27
Solution: 24, 6, 108, 27

⇒ 1st term x 4th term = 24 x 27 = 648

⇒ 2nd term x 3rd term = 6 x 1 08 = 648

∴ 24, 6, 108, 27 are in proportion.

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

= \(\frac{1}{2}, 1, \frac{3}{5}, \frac{6}{5}\)

⇒ 1st term x 4th term = \(\frac{1}{2} \times \frac{6}{5}=\frac{3}{5}\)

⇒ 2nd term x 3rd term = \(1 \times \frac{3}{5}=\frac{3}{5}\)

∴ \(\frac{1}{2}, 1, \frac{3}{5}, 1 \frac{1}{5}\) are in proportion.

5. 72, 45, 70, 25
Solution: 72, 45, 70, 25

⇒ 1st term x 4th term = 72 x 25 = 1800

⇒ 2nd term x 3rd term = 45 x 70 = 31 50

∴ 72, 45, 70, 25 are not in proportion

6. _______ Let’s take + positive whole numbers.
Solution: 7, 8, 35, 40

1 st term x 4th term = 7 x 40 = 280

2nd term x 3rd term = 8 x 35 = 280

∴ 7, 8, 35, 40 are in proportion.

Question 15. Let’s first try to find, if the following numbers are in proportion, let’s find all other possible proportions for each of them.

1. 60, 2,10,12
Solution: 60, 2,10,12

Case 1: Here, 1st term x 4th term = 60 x 12 = 720

⇒ 2nd term x 3rd term = 2 x 10 = 20

∴ 60: 2 ≠10: 12

Case 2: 60, 12, 10,2

⇒ Here, 1st term x 4th term = 60 x 2 = 120

⇒ 2nd term x 3rd term = 12 x 10 = 120

∴ 60: 12 :: 10: 2

Case 3: 60, 12,12, 2

⇒ Here, 1 st term x 4th term = 60 x 2 = 120

⇒ 2nd term x 3rd term = 10 x 12 = 120

∴ 60 : 10 :: 12 : 2

Case 4: 10,2,60, 12

⇒ Here, 1 st term x 4th term = 10 x 12 = 120

⇒ 2nd term x 3rd term = 2 x 60 = 120

∴ 10: 2:: 60: 12

2. 4,10, 6,15
Solution: 4,10, 6,15

⇒ Case 1: 4, 10, 6,15

⇒ Here, 1 st term x 4th term = 4 x 15 = 60

⇒ 2nd term x 3rd term = 6 x 10 = 60

∴ 4 : 10 :: 6: 15

Case 2: 4, 6,10, 15

⇒ Here, 1 st term x 4th term = 4 x 15 = 60

⇒ 2nd term x 3rd term = 6 x 10 = 60

∴ 4 : 6:: 10 : 15

Case 3: 60, 10, 12,2

⇒ Here, 1st term x 4th term = 10 x 6

⇒ 60 2nd term x 3rd term = 4 x 15 = 60

∴ 10 : 4:: 15 : 6

Case 4: 10, 15, 4,6

⇒ Here, 1st term x 4th term = 10 x6 = 60

⇒ 2nd term x 3rd term = 15 x 4 = 60

∴ 10: 15:: 4: 6

3. 8, 9, 24, 2
Solution: 8, 9, 24, 2

Case 1: 8, 24, 2, 9

⇒ Here, 1 st term X 4th term = 8 x 9 = 72

⇒ 2nd term x 3rd term = 24 x 2 = 48

∴ 8: 24 ≠ 2: 9

Case 2: 8, 9, 2, 24

⇒ Here, 1st term x 4th term = 8 x 24 = 192

⇒ 2nd term x 3rd term = 9×2 = 18

∴ 8: 9 ≠ 2: 24

Case 3: 8, 24, 9, 2

⇒ Here, 1st term x 4th term = 8×2 = 16

⇒ 2nd term x 3rd term = 24 x 9 = 216

∴ 8 : 24 ≠ 9:2

∴ 8, 9, 24, 2 are not in proportion

4. 3, 5,15, 25
Solution: 3, 5, 15, 25

Case 1: 5, 3, 25, 15

⇒ Here, 1 st term x 4th term = 5 x 15 = 75

⇒ 2nd term x 3rd term = 3 x 25 = 75

∴ 5 : 3:: 25 :15

Case 2: 5, 25, 3, 15

⇒ Here, 1 st term x 4th term = 5 x 15 = 75

⇒ 2nd term x 3rd term = 25 x 3 = 75

∴ 5: 25 :: 3:15

Case 3: 25, 5, 15, 3

⇒ Here, 1st term x 4th term = 25 x 3 = 75

⇒ 2nd term x 3rd term = 5×15 = 15

∴ 10: 4:: 15:6

Case 4: 25,15, 5, 3

⇒ Here, 1st term x 4th term = 25 x 3 = 75

⇒ 2nd term x 3rd term = 15 x 5 = 75

∴ 25 : 15 :: 5 : 3

5. 45, 5,75,5

Solution: 45, 5, 75: 5

⇒Here, 45 x 5 = 225 75 X 5 = 375

∴ 45:5 ≠ 75.5

⇒ Again, 45 x 75 5 x 5 = 25

∴ 45 : 5 ≠ 5: 75

∴ 45, 5, 75, and 5 are not in proportion.

6. 24, 4, 36, 6
Solution: 24, 4, 36, 6

Case 1: 24, 4, 36, 6

⇒ Here, 1 st term x 4th term = 24 x 6 = 144

⇒ 2nd term x 3rd term = 4 x 36 = 144

∴ 24 : 4:: 36 : 6

Case 2: 24, 36, 4, 6

⇒ Here, 1st term x 4th term = 24 x 6 = 144

⇒ 2nd term x 3rd term = 36 x 4 = 144

∴ 24 : 36 :: 4: 6

Case 3: 4, 24, 6, 36

⇒ Here,. 1 st term x 4th term = 4 x 36 = 144

⇒ 2nd term x 3rd term = 24 x 6 = 144

∴ 4: 24 6: 36

Case 4: 4, 6, 24, 36

⇒ Here, 1 st term x 4th term = 4 x 36 = 144

⇒ 2nd term x 3rd term = 6 x 24 = 144

∴ 4:6:: 24 : 36

Question 16. The height of my friend Priya is 160 cm and the height of her mother is 170 cm. Again, Priya’s weight is 40 kg and her mother’s weight is 42.5 kg. Let’s find if their weights are in proportion to their heights.
Solution:

Given

The height of my friend Priya is 160 cm and the height of her mother is 170 cm. Again, Priya’s weight is 40 kg and her mother’s weight is 42.5 kg.

⇒ Height of Priya = 160 m.

⇒ Height of Priya’s mother = 170 m.

∴  Ratio of their heights = 160 m: 170 m.

= 16:17

⇒ Again, weight of Priya = 40 kg.

⇒ Weight of Priya’s mother = 42.5 kg.

∴ Ratio of their weights = 40 kg: 42.5 kg.

= 40: 425/10

= 16:17

∴ Their weights are in proportion to their heights.

WBBSE Chapter 18 Square Root Exercise 18

Question 1. 441 oranges were picked from Nisar’s fruit garden. Oranges are to be kept in a basket. Each basket will contain as many oranges as there are many baskets. Let’s calculate to find the number of baskets.
Solution:

⇒ Total number of oranges = 441

⇒ No. of baskets = \(\sqrt{441}=\sqrt{21 \times 21}=21\)

Question 2. Today morning I arranged the books in the almirah of my room. On each shelf, I arranged books equal to the number of shelves of the almirah. But 5 books remained out of the almirah. If the total number of books is 86, let’s find the number of shelves of the almirah.
Solution:

Given

Today morning I arranged the books in the almirah of my room. On each shelf, I arranged books equal to the number of shelves of the almirah. But 5 books remained out of the almirah. If the total number of books are 86,

⇒ Total number of books = 86.

⇒ But 5 books remained out of the almirah.

∴ The total number of books in the almirah = 86 – 5 = 81.

∴ Number of shelves of the almirah = \(\sqrt{81}=\sqrt{9 \times 9}=9\)

Read and Learn More WBBSE Solutions For Class 6 Maths

Question 3. In the playground, we decided to arrange ourselves and stand square. After some have formed this square, it’s found that 4 friends are staying apart. If they join the square, it will not remain square anymore. 40 students are present today in our class. Let’s calculate in this arrangement how many students are there in each row.
Solution:

Given

In the playground, we decided to arrange ourselves and stand square. After some have formed this square, it’s found that 4 friends are staying apart. If they join the square, it will not remain square anymore. 40 students are present today in our class.

⇒ Number of students present today = 40.

⇒ To make a square form 4 students will be excess.

∴ Number of students required to make a square form = 40 – 4 = 36.

∴ Number of students is each row = \(\sqrt{36}=\sqrt{6 \times 6}=6\)

“WBBSE Class 6 Maths Chapter 18 square root solutions”

Question 4. In the Sukanta Memorial Library of our locality, the members contributed as many rupees as there are members. If the total collection is Rs. 729 let’s find the number of members of the library.
Solution:

⇒ Total collections from the members = Rs. 729

⇒ As each members contribute as many rupees as there are members.

∴ Number of members = \(\sqrt{729}=\sqrt{3 \times 3 \times 3 \times 3 \times 3 \times 3}\)

Question 5. To dig a pond in Raghunathpur village, the number of people employed worked for as many numbers of days as there were number of people working, and got as payment Rs. 12375.
Solution:

⇒ Number of people working and get as payment Rs 12375.

⇒ If each, gets daily wage of Rs. 55.

⇒ Number of people joined the work  =\(\sqrt{\frac{12375}{55}}=\sqrt{225}=\sqrt{15 \times 15}=15\)

Question 6. If each gets a daily wage of Rs. 55 then let’s find how many pepole joined the work.
Solution:

⇒ The total money collected = Rs. 4096.

⇒ As each member paid an amount 4 times the number of members.

∴ Number of members went for the trip

= \(\frac{\sqrt{4096}}{4}=\sqrt{1024}=\sqrt{4 \times 4 \times 4 \times 4 \times 2 \times 2}\)

= 4x4x2 = 32

Question 7. Today is Children’s Day. Few of us distributed toffee and biscuits among the other students of the school. But 800 toffees were left. So each took twice the number of toffees as we are in number. Let’s calculate among how many of us those 800 toffees were divided.
Solution:

Given

Today is Children’s Day. Few of us distributed toffee and biscuits among the other students of the school. But 800 toffees were left. So each took twice the number of toffees as we are in number.

⇒ Each took twice the number of toffees as we are in number.

⇒ 800 toffee were left.

∴ Number of children = \(\sqrt{\frac{800}{2}}=\sqrt{400}=\sqrt{20 \times 20}=20\)

“WBBSE solutions for Class 6 Maths square root chapter”

Class 6 History	Class 6 Social Science
Class 6 Geography	Class 6 Science
Class 6 Maths	Class 6 Science MCQs
Class 6 General Science	Class 6 Maths Solutions
Class 6 Geography	Class 6 Hindi

Question 8. Safikul uncle of Atla village brought 780 saplings of papaya tree. He decided to plant them in as many rows as there are number of plants in each row. But by trying to do, he fell short of 4 saplings. In how many rows did Rafikul’s uncle decide to plant the saplings, let’s find.

Solution:

Safikul uncle of Atla village brought 780 saplings of papaya tree. He decided to plant them in as many rows as there are number of plants in each row. But by trying to do, he fell short of 4 saplings.

⇒ The number of saplings of the papaya tree = 780 He fell short of 4 saplings.

∴ Total number of saplings required = 780 + 4 = 784

∴ Number of rows = \(\sqrt{784}=\sqrt{4 \times 4 \times 7 \times 7}\) = 4 x 7 = 28.

Question 9. I made a square cardboard box which has many square cells. Each row has many cells as there are a number of rows. My brother kept one 5 rupee, one two rupee, and a one rupee coin in each cell. If total money kept is Rs. 1152, then let’s find how many cells there are in each row of may cardboard box.
Solution:

Given

I made a square cardboard box which has many square cells. Each row has many cells as there are a number of rows. My brother kept one 5 rupee, one two rupee, and a one rupee coin in each cell. If total money kept is Rs. 1152

⇒ The total money kept is Rs. 1152.

⇒ My brother kept one 5 rupee, a two rupee and a one rupee coin in each cell.

∴ Number of cells = \(\sqrt{\frac{1152}{5+2+1}}=\sqrt{\frac{1152}{8}}\)

= \(\sqrt{144}=\sqrt{12 \times 12}=12\)

“Class 6 Maths WBBSE Chapter 18 square root exercises”

Question 10. Let’s find the value of the following

1. Square of 7 = ____
Solution: Square of 7 = 7 x 7 = 49

2. Square root of 121 = _______
Solution: Square root of 121 = √121

= \(\sqrt{11 \times 11}=11\)

3. 9² = _______
Solution: 92 = 9 x 9 = 81

4. √100 = ______
Solution: \(\sqrt{100}=\sqrt{10 \times 10}=10\)

5. √49 = _________
Solution: \(\sqrt{49}=\sqrt{7 \times 7}=7\)

6. √144 = ______
Solution: \(\sqrt{144}=\sqrt{12 \times 12}=12\)

7. \(\sqrt{3^2 \times 2^2}\) = ________
Solution: \(\sqrt{3^2 \times 2^2}=\sqrt{9 \times 4}\)

= \(\sqrt{36}=\sqrt{6 \times 6}=6\)

8. \(\sqrt{5 \times 7 \times 5 \times 7}\) = ________
Solution: \(\sqrt{5 \times 7 \times 5 \times 7}=\sqrt{5^2 \times 7^2}\) = 5 x 7 = 35

9. \(\sqrt{13 \times 13}\) = _____
Solution: \(\sqrt{13 \times 13}\) = 13

“Step-by-step solutions for square root Class 6 WBBSE”

Question 11. Let’s find the square root of the following by finding factors

1. 169
Solution: \(\sqrt{169}=\sqrt{13 \times 13}=13\)

2. 225
Solution: \(\sqrt{225}=\sqrt{15 \times 15}=15\)

3. 4² + 3²
Solution: \(\sqrt{4^2+3^2}=\sqrt{16+9}\)

= \(\sqrt{25}=\sqrt{5 \times 5}=5\)

4. √144
Solution: \(\sqrt{144}=\sqrt{12 \times 12}=12\)

5. 576
Solution: \(\sqrt{576}=\sqrt{4 \times 4 \times 6 \times 6}\) = 4 x 6 = 24

6. 15² + 20²
Solution: \(\sqrt{15^2+20^2}=\sqrt{225+400}\)

= \(\sqrt{625}=\sqrt{5 \times 5 \times 5 \times 5}\)

= \(5 \times 5=25\)

7. 900
Solution: \(\sqrt{900}=\sqrt{3 \times 3 \times 10 \times 10}\)

Class 6 WBBSE Math Solutions Chapter 18 Square Root Exercise 18.1

Question 1. Let’s find if 108, 64,162, 81 are square numbers or not let us find the smallest whole number other than ‘0’ with which these should be multiplied or divided to give a square number.

1. 108
Solution:

⇒ 108 = 2x2x3x3x3

∴ 108 is not a square number.

⇒ If 3 is multiplied or divided with 108, it will be a square number.

∴ 108 x 3 = 324, a square number or, 108 ÷ 3 = 36, a square number

“WBBSE Class 6 Maths Chapter 18 important questions and answers”

2. √64
Solution:

= \(\sqrt{2 \times 2 \times 2 \times 2 \times 2 \times 2}\)

=2x2x2=8

∴ 64 is a square number.

3. 162
Solution:

∴162 = 3x3x3x3x2

∴ 162 is not a square number.

⇒ If 2 is multiplied or divided with 162, it will be a square number.

∴ 162 x 2 = 324 = 18 x 18, a square number,

or, 162 ÷ 2 = 81 = 9 x 9, a square number.

4. √81
Solution:

= \(\sqrt{3 \times 3 \times 3 \times 3}\) =3×3 = 9

∴ 81 is a square number.

Class 6 Maths Solutions WBBSE Chapter 18 Square Root Exercise 18.2

Question 1. Apart from 0, by which least whole number the following numbers must be divided to get square numbers

1. 845
2. 450
3. 18 x 6
4. 25 x 35

1. 845
Solution:

845 = 13x13x5

⇒ Hence, it is not a square number.

⇒ If we divide the number by 5, we will get a square number.

∴ The required number is 5.

2. 450
Solution:

∴ 450 = 5x5x3x3x2

⇒ Hence, it is not a square number, if we divide the number by 2, it will be a square number.

∴ The required number is 2.

“Solved examples of square root WBBSE Class 6”

3. 18×6
Solution:

= (3 x 3 x 2) x (2 x 3)

=3x3x2x2x3

⇒ Hence, it is not a square number, if we divide the number by 3, it will be a square number.

∴ The required number is 3.

4. 25 x 35 = (5 x 5) x (5 x 7)
Solution:

⇒ Hence, it is not a square number, if we divide the number by 35 (5 x 7), it will be a square number.

∴ The required number = 35.

Question 2. By which least whole number apart from 0, the following numbers should be multiplied to give a square number:

1. 432
Solution:

=2x2x3x7x7x3

⇒ Hence, 588 is not a square number, if we multiply 3 with it, it will be a square number.

∴ The required number is 3.

Class 6 Maths Solutions WBBSE

2. 588
Solution:

= 2x2x3x7x7x3

⇒ Hence 588 is not a square number, if we multiply 3 with it, it will be a square number.

∴ The required number is 3

3. 25×20
Solution:

= 5x5x2x2x5

⇒ Hence, it is not a square number, if we multiply 5 with it, it will be a square number.

∴ The required number is 5.

4. 24x 28
Solution:

=2x2x2x3x2x2x7

⇒ Hence, it is not a square number, if we multiply 2x3x7 with it, it will be a square number.

∴ The required number is 2 x 3 x 7 = 42.

WBBSE Math Solutions Class 6 Chapter 18 Square Root Exercise 18.3

Question 1. Let’s work out to find the least square number other than 0, which is divisible by 12,16, 20, and 24.
Solution:

∴ L.C.M.of 12, 16, 20 and 24 = 2x2x2x2x3x5 = 240

⇒ Hence, 240 is not a least square number

⇒ If we multiplied 240 by 3 x 5, then it will be the least square number.

∴ The required least square number.

= 240 x 3 x 5 = 3600.

Question 2. The product of two positive whole numbers is 98 and the greater one is twice the smaller number. Let’s find the two numbers.
Solution:

⇒ Greater number = 2x smaller number

⇒ Greater number x smaller number = 98

⇒ or, (2 x smaller number) x smaller number = 98

⇒ or, 2 x (smaller number)² = 98

∴ (Smaller number)² = 98/2 = 49

∴ Smaller number = √49 =7

∴ Greater number = 2 x smaller number = 2×7 = 14

∴ The required numbers are 7 and 14.

Question 3. Let’s find which least square number has factor 17.
Solution: The least square number which has a factor =17×17 = 289

Question 4. Let’s find the least square number other than 0 which is divisible by 10,15,20 and 30. Also, let’s write the next square number which will be divisible by these numbers.
Solution:

∴ L.C.M. = 2x2x3x5 = 60

⇒ It is not a square number. If we multiplily by 3 x 5, we will get a square number.

∴ Least square number = 60 x 3 x5 = 900.

∴ The next square number = 900 x 4 = 3600.

WBBSE Math Solutions Class 6

Question 5. Let’s analyse the given numbers and place them in their respective columns-20, 27, 50, 75, 100, 108,144, 169,180, 256
Solution:

⇒ 20 = 2 x 2 x 5 – it is not a square number

⇒ 27 = 3 x 3 x 3 – it is not a square number

⇒ 50 = 5 x 5 x 2 – it is not a square number

⇒ 75 = 5 x 5 x 3 – it is not a square number

⇒ 100 = 2 x 2 x 5 x 5 – it is a square number

⇒ 108 = 6 x6 x 3 – it is not a square number

⇒ 144 = 2 x 2 x 6 x 6 – it is a square number

⇒ 169 = 13 x 13 – it is a square number

⇒ 180 = 2 x 2 x 3 x 3 x 5 – it is not a square number

256 = 4 x 4 x 4 x 4 – it is a square number

Question 6. This year on Netaji’s birthday, the students present in Physical Education class were made to stand in rows of 18,24,27, at different points of time, to march past At one point of time we formed a solid square. Let us find the least number of students present in school on that day.
Solution: L.C.M of 18, 24, 27

= 2 x 2 x 2 x 3 x 3 x 3

= 216

⇒ It is not a square number, if we multiply 2×3 with it, it will be a square number.

∴ Least number of students = 216x2x3 = 1296

WBBSE Math Solutions Class 6

Question 7. The product of two positive numbers is 147. The greater number is 3 times the smaller number. Let’s find the numbers.
Solution:

⇒ Greater number = 3 x smaller number .

⇒ Greater number x smaller number = 147

⇒ or, 3 x smaller number x smaller number = 147

⇒ or, (Smaller number)² = 147/3 = 49

∴ Smaller number = √49 = 7

⇒ Greater number = 3×7 = 21.

“Best guide for Class 6 Maths WBBSE square root problems”

Question 8. Arrange the following in ascending order of their value:

⇒ \((\sqrt{36}+\sqrt{25}),(\sqrt{49}+\sqrt{9}) ;(\sqrt{25}+\sqrt{100}),(\sqrt{4}+\sqrt{16})\)

Solution:

⇒ \((\sqrt{36}+\sqrt{25})\)

= 6 + 5 = 11

⇒ \((\sqrt{49}+\sqrt{9})\)

= 7 + 3 = 10

⇒ \((\sqrt{25}+\sqrt{100})\)

= 5 + 10 = 125

⇒ \(\sqrt{4}+\sqrt{16}\)

= 2 + 4 = 6

⇒ In ascending order

6 ∠ 10 ∠ 11 ∠ 15

i.e., \(\sqrt{4}+\sqrt{16} ; \sqrt{49}+\sqrt{9} ; \sqrt{36}+\sqrt{25} ; \sqrt{25}+\sqrt{100}\)

WBBSE Class 6 Maths Solutions

Question 9. Of three positive numbers, the product of first and second is 24 second and third is 48 and that of first and third is 32. Let’s find the three numbers.
Solution:

⇒ 1st number x 2nd number = 24

⇒ 2nd number x 3rd number = 48

⇒ 1st number x 3rd number = 32

⇒ (1st number x 2nd number) x (1st number x 3rd number)/ (2nd number x 3rd number)

∴ (1st number)² = 16

∴ 1st number = √16 = 4

⇒ Again, 1st number x 2nd number = 24

⇒ or, 4 x 2nd number = 24

∴ 2nd number = 24/4 = 6

⇒ Again, 2nd number x 3rd number = 48

⇒ 6 x 3rd number = 48

∴ 3rd number = 48/6 = 8

∴ Required numbers are 4, 6, and 8.

Question 10. On Republic Day, our Physical Education teacher made all students stand in rows having 12,15, and 20 students, at different points of time, for march past. At one point of time, the students were arranged in solid square. Let’s find out at least how many students were present in school on that day.
Solution:

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

= 2 x 2 x 3 x 5

⇒ But 60 is not a square number, if we multiply 3 and 5 with it, it will be square number.

∴ The required number = 60 x 3 x 5 = 900.

WBBSE Class 6 Maths Solutions Chapter 18 Square Root Exercise 18.4

Question 1. Let’s find the square number nearest to 1000.
Solution:

∴ 1000 – 39 = 961 is a square number.

∴ √961 = 31

⇒ Next square number is (32)² = 1024

⇒ 1024- 1000 = 24 ∠ 39

∴ Square number nearest to 1 000 is 1024.

Question 2. Let’s find the least number that will be subtracted from 9585 to give a square number.
Solution:

∴ The required number is 176.

Question 3. Let’s find what least number must be added to 5320 to make it a perfect square.
Solution:

Next square number is (73)² = 5329.

∴ The required number that is to be added to 5320 to make a square number is 5329-5320 = 9.

Question 4. Let’s find the least square number other than 0, which is divisible by 15, 25, 35, 45.
Solution:

L.C.M.of 15,25, 35 and 45

= 3x5x5x7x3 = 1575.

⇒ It is not a square number, if we multiply 7 with it, it will be a square number.

∴ The required least square number = 1575 x 7 = 11025.

Question 5. Let’s find the least square number of 4 digits which are divisible by 8, 15, 20, and 25.
Solution:

L.C.M. of 8, 15, 20 and 25

= 2 x 2 x 5 x 2 x 3 x 5 = 600

⇒ If we multiply 600 by 2 and 3 then we get a square number.

∴ The least-square number of 4 digits = 600 x 6 = 3600.

Question 6. Let’s find the least square number of 4 digits.
Solution:

The least 4 digit number is 1000.

⇒ 1000- 39 = 961 is a square number.

⇒ √961 =31

⇒ Next square number = (32)² = 1 024.

∴ Least square number of 4 digits = 1 024.

“How to calculate square root easily Class 6 WBBSE”

Question 7. Let’s find the greatest square number of 4 digits.
Solution:

⇒ The greatest number of 4 digits = 9999.

⇒ The greatest square number of 4 digits = 9999 -198 = 9801.

Question 8. Let us find the square root of the following by the method of division:
Solution:

Question 9. Without finding square root, let us find the possible digits in unit’s place and number of digits in the square root number:

1. 784
Solution:

⇒ Possible digits in unit’s place = 2, 4

∴ Number of digits in the square root number = 2

Class 6 Math WBBSE Solutions

2. 3676
Solution:

⇒ Possible digits in 4 unit’s place = 6

∴ Number of digits in the square root number = 2

3. 160000
Solution:

⇒ Possible digits in unit’s place = 0

⇒ Number of digits in the square root number = 3

4. 1225
Solution:

Possible digits in unit’s place = 5

∴ Number of digits in the square root number = 2

5. 2401
Solution:

⇒ Possible digits in unit’s place = 1,9

∴ Number of digits in the square root number = 2

6. 10201
Solution:

⇒ Possible digits in unit’s place =,1,9.

∴ Number of digits in the square root number = 3

“Understanding square root concepts for Class 6 WBBSE Maths”

Question 10. Let’s find two square numbers nearest to 5000.
Solution: 5000

⇒ Nearest to 5000 = (70)² and (71)²

= 4900 and 5041

Question 11. The product of two numbers is 1576 and their quotient is 9/7, let’s find the numbers.
Solution:

⇒ 1st number x 2nd number = 1575

and \(\frac{1 \text { st number }}{2 \text { nd number }}=\frac{9}{7}\)

∴ \((1 \text { st number } \times 2 \text { nd number }) \times \frac{1st number }{2nd num ber}=1575 \times \frac{9}{7}\)

∴ (1st number)² = 225 x

⇒ 1st number = \(\sqrt{225 \times 9}\) = 15 x 3 = 45

∴ 2nd number = 1575 ÷ 45 = 35

∴ The numbers are 45, 35.

Question 12. Let us find what digit will be placed in * of 202*, so that it becomes a square number.
Solution: 202 *

⇒ 1 st if we put 1, then 2021 is not a square number.

⇒ 2nd if we put 2, then 2022 is not a square number.

⇒ 3rd if we put 3, then 2023 is not a square number.

⇒ 4th, if we put 4, then 2024 is not a square number.

⇒ 5th if we put 5, then 2025 is a square number.

∴ The required digit is 5.

Class 6 Math Solutions WBBSE Chapter 17 Geometrical Concepts Based On Different Instruments Of Geometry Box Exercise 17

Question 1. I myself drew a quadrilateral and a triangle with the help of scale. I also measured their sides with the scale.
Solution:

ABCD is a quadrilateral

⇒ ABCD is a quadrilateral whose AB = 7 cm, BC = 1.8 cm, CD = 5 cm. and DA = 4 cm.

PQR is a triangle

⇒  PQR is a triangle whose PQ = 4.5 cm, QR = 6 cm, and RP = 5.4 cm.

Read and Learn More WBBSE Solutions For Class 6 Maths

Question 2. I have drawn two line segments of length 2.8 cm, and 5.3 cm, and named them.
Solution:

⇒ MN is a line segment whose length = 2.8 cm and EF is another line segment whose length = 5.3 cm.

Class 6 WBBSE Math Solutions Chapter 17 Geometrical Concepts Based On Different Instruments Of Geometry Box Exercise 17.1

Question 1. Let’s draw a line segment of length 4.5 cm with scale and divider.
Solution:

⇒ PR is a line segment whose length is 4.5 cm.

“WBBSE Class 6 Maths Chapter 17 geometrical concepts solutions”

Question 2. From a straight line let’s mark out a length – XY equal to 6 cm.
Solution:

Question 3. Using scale and divider, let’s draw two line segments of length 6 cm and 5.7 cm, and name them.
Solution:

Question 4. Using scale and divider let us draw two line segments AB and CD of lengths 3.6 cm and 2.2 cm.
Solution:

“WBBSE solutions for Class 6 Maths geometry box instruments”

Question 5. On a straight line, let’s draw a line segment XY which is the sum of the line segments AB and CD marked with a divider separately on the line. On another straight line, the segments AB and CD are separately marked in a way with a divider such that the line segment EF is the difference of the line segments AB and CD.
Solution:

let’s draw a line segment

Class 6 History	Class 6 Social Science
Class 6 Geography	Class 6 Science
Class 6 Maths	Class 6 Science MCQs
Class 6 General Science	Class 6 Maths Solutions
Class 6 Geography	Class 6 Hindi

Question 6. Using scale and compass and taking two separate points as centres let is draw two circles having radii of lengths 3.2 cm and 2.9 cm
Solution:

Question 7. Let’s measure and write the values of the angles given below with a protractor. Let’s identify the angles as acute angle or right angle obtuse angle straight angle or reflex angle and arrange them in the increasing order of their values.
Solution:

1. ∠ACB = 40°; (Acute angle)
2. ∠CDE = 180° (Straight angle)
3. ∠JLK = 120° (Obtuse angle)
4. ∠PQR = 30° (Acute angle)

Reflex Z PQR = 360°-30° = 330°

1. ∠IJH = 120° (Obtuse angle)
2. ∠FEG = 90° (Right angle)
3. ∠YXZ = 30° (Acute angle)
4. ∠XYZ = 140° (Obtuse angle)

Question 8. Let’s write the sides and the vertex for each of the following angles.
Solution:

1. Arms: \(\overrightarrow{\mathrm{OA}}\); \(\overrightarrow{\mathrm{OB}}\)

Sides: \(\overrightarrow{\mathrm{OA}} ; \overrightarrow{\mathrm{OB}}\)

Vertex: O

“Class 6 Maths WBBSE Chapter 17 geometrical concepts using instruments”

2. Arms: \(\overrightarrow{\mathrm{OC}}, \overrightarrow{\mathrm{OD}}\)

Sides: \(\overrightarrow{O C}, \overrightarrow{O D}\)

Vertex: O

3. Arms: \(\overrightarrow{X Y}, \overrightarrow{Y Z}\)

⇒ Sides: \(\overrightarrow{X Y}, \overrightarrow{Y Z}\)

Vertex: Q

4. Arms: \(\overrightarrow{P Q}, \overrightarrow{Q R}\)

⇒ Sides: \(\overrightarrow{\mathrm{PQ}}, \overrightarrow{\mathrm{QR}}\)

⇒ Vertex: Y

Question 9. Let us draw the following angles with a protractor. Let’s name the angle and write their sides and vertex. Also mention the type of the angles – 38°, 75°, 90°, 145°, 180°, 200°, 270°.
Solution:

1. ∠ABC =38°(Acute angle)

Sides: \(\overline{\mathrm{AB}}, \overline{\mathrm{BC}}\)

Vertex: B

2. ∠PQR = 75°(Acute angle)

Sides: \(\overline{\mathrm{PQ}}, \overline{\mathrm{QR}}\)

Vertex: Q

3. ∠DEF = 90°(Rigth angle)

Sides: \(\overline{\mathrm{DE}}, \overline{\mathrm{EF}}\)

Vertex: E

4. ∠MNR = 145°(Obtuse angle)

Sides: \(\overline{\mathrm{MN}}, \overline{\mathrm{NR}}\)

Vertex: N

“Step-by-step solutions for geometrical concepts Class 6 WBBSE”

5. ∠COD = 180°(Strigth angle)

Sides: \(\overline{\mathrm{CO}} ; \overline{\mathrm{OD}}\)

Vertex: O

6. ∠IJK = 200°, (Reflex angle)

Sides: \(\overline{\mathrm{IJ}}, \overline{\mathrm{JK}}\)

Vertex: J

7. ∠APC = 270°(Reflex angle)

Sides: \(\overline{\mathrm{AP}}, \overline{\mathrm{PC}}\)

Vertex: P

Question 10. Let Do It
Solution:

∠BAD = 30°; ∠CAD = 60° and ∠BAC = 90°

∠BAD + ∠CAD = ∠BAC

Again, ∠BAC-∠BAD =∠DAC

∠ADC + ∠ADB = 180° = ∠BDC

∠BDC – ∠ADB = ∠ADC = 90°

1. ∠EAG + ∠GAF = 90°;
2. ∠FAE – ∠FAG = 40°
3. ∠ADC+ ∠ADB = 180°;
4. 180°-∠ADC = 60°
5. ∠SPR + ∠RPQ = ∠SPQ
6. ∠SPQ – ∠SPR = ∠RPQ

Class 6 Math WBBSE Solutions Chapter 17 Geometrical Concepts Based On Different Instruments Of Geometry Box Exercise 17.2

Question 1. Using the outer and inner scales of the protractor, let’s draw the following angles

1. 54°
Solution: Angle POQ = 54°

“WBBSE Class 6 Maths Chapter 17 important questions and answers”

2. 67°
Solution: ∠CDE = 67°

3. 85°
Solution: ∠ABC = 85

4. 95°
Solution: ∠MNK = 95°

5. 120°
Solution: ∠RPK = 120°

Question 2. Let’s draw 187°, 235°, 310° and 325° angles using protractor:
Solution: 187°

Reflex ∠ABC = 187°

Reflex ∠PQR = 235°

Reflex ∠MKL = 310

Reflex ∠CDE = 325°

Question 3. Let’s observe the figure and comprehend and write:

1. Let’s measure angle number 1, name the angle, type of angle. [Measure with protractor]
2. Let angle number 2 be measured with a protractor, name the angle, and write the type of angle.
3. The measures of angles 1 and 2 are added and the angle formed.

Solution:

1. ∠1 =40° (Acute angle)

∠BOC= 40°

2. ∠2 = 50° (Acute angle)

∠AOB = 50°

3. ∠1 + ∠2 = 40° + 50° = 90° (Right angle)

∠AOC = 90°

Question 4. In the figure, let’s take angle number 1 = ∠AOB; angle number 2 = ∠BOC; angle number 3 = ∠COD

1. Let’s measure angle number 1 with a protractor and note its value.
2. Let’s measure angle number 2 with a protractor and note its value.
3. Let’s measure angle number 3 with a protractor and note its value.
4. By adding angle number 1 and angle number 2, we get an angle, name that angle from the figure. ‘
5. By adding angle number 2 and angle number 3, the angle that is formed is named from the figure.
6. The angle formed by adding angle number 1, number 2, and number 3 is named from the figure and let’s express it as the sum of three angles.

Solution:

1. ∠1 = ∠AOB = 60°
2. ∠2 = ∠BOC = 90°
3. ∠3 =∠COD = 30°
4. ∠1 + ∠2 = ∠AOB + ∠BOC = 60° + 90° = 150°
5. ∠2+ ∠3 = ∠BOC + ∠COD.= 90° + 30° = 120°
6. ∠1 + ∠2 + ∠3 = ∠AOB+ ∠BOC+ ∠COD= 60° + 90° + 30° = 180°

Question 5. Let’s answer from the figure:

1. Let’s name one acute angle: ∠AOB
2. Let’s name one obtuse angle:∠BOD
3. Let’s name one right angle: ∠COD
4. Let’s name one straight angle: ∠AOD
5. Let’s name a reflex angle: ∠AOE
6. ∠AOB + ∠BOC =∠AOC
7. ∠BOC + ∠COD = ∠BOD
8. ∠AOC – ∠BOC = ∠AOB
9. ∠BOD – ∠BOC = ∠COD

Class 6 Maths Solutions WBBSE Chapter 17 Geometrical Concepts Based On Different Instruments Of Geometry Box Exercise 17.3

Question 1. Lengths of the sides of triangles are given. Let’s identify the types, of triangles according to angles:

1. 18 cm 18 cm 10 cm;
2. 5.2 cm, 5.2 cm, 5.2 cm;
3. 8 cm, 2 cm, 9 cm

Solution:

1. 18 cm, 18 cm, 10 cm: Here the lengths of two sides of a triangle are equal, then it is an isosodes, triangle.

2. 5.2 cm, 5.2 cm, 5.2 cm: Here the lengths of three sides of a triangle are equal, then it is an equilateral triangle.

3. 8 cm, 8 cm, 9 cm: Here the lengths of the three sides of a triangle are not equal, then it is a scalene triangle.

“Solved exercises for geometry box instruments WBBSE Class 6”

Question 2. The measures of the three angles of the triangles are given below. Let’s identify the types according to angles:

1. 90°, 45°, 45°
2. 90°, 30°, 60°
3. 75°, 70°, 35°
4. 60°, 60°, 60″
5. 120°, 30°, 60°

Solution:

1. 90°, 45°, 45°: Here one angle of a triangle is 90° and the other two angles are equal. So it is a right-angled isosceles triangle.

2. 90°, 30°, 60°: Here one angle is 90°; so, it is a right-angled triangle.

3. 75°, 70°, 35°: Here three angles of a triangle are not equal so it is a scalono trianglo.

4. 60°, 60°, 60°: Here three angles of a triangle are equal; so, it is an equilateral triangle.

5. 120°, 30°, 60°: Here one angle is 120°, greater than 90°- it is an obtuse angle. So tho triangle is an obtuse, angled triangle.

Question 3. A, B, and C are non-collinear points. Let’s join A, B; B, C, and C, A, and try to find answers of the following.

1. Let’s identify the geometric figure formed by joining the line segments.
2. Let’s write the angle opposite to side BC.
3. Let’s write the angle opposite to side AC.
4. Let’s write the side opposite to ∠BAC.
5. Let’s write the side opposite to ∠ACB.

Solution:

1. It is a triangle.
2. ∠BAC
3. ∠ABC
4. Side BC
5. Side AB

Question 4. Let’s find if the statements given below are correct or not
Solution:

1. Hypotenuse is the smallest side of a right-angled triangle.(x)
2. One angle of a right-angled triangle is 90°. (√)
3. There are at least two acute angles in a triangle. (√)
4. Each equilateral triangle is called an isosceles triangle. (√)
5. The sum of three angles of a triangle is 360°. (X)
6. The right-angled triangle can never be an equilateral triangle.(√)
7. The right-angled triangle can not be an isosceles triangle. (X)

Question 5. Let’s try to draw 120°, 135°, 150° angles using set squares.
Solution:

⇒ ∠ABC = 90° + 45° = 135°

To draw 120°

⇒  ∠ABC = 90°+ 60° = 150°

⇒ To draw 150°

⇒ ∠ABC = 90°+ 60° = 150°

6. Let’s write the observations in a tabular form.
Solution:

Question 7. Let’s write a logical statement for the set of words given below.

1. Square, Rectangle, Parallelogram — are all quadrilateral.
Solution: Square, rectangle, parallelogram — all have 4 sides, so they are quadrilateral.

2. A rectangle is a special type of parallelogram.
Solution: A rectangle is a special type of parallelogram as opposite sides of the rectangle are equal and parallel.

“Best guide for Class 6 Maths WBBSE geometrical concepts”

3. Parallelogram is a special trapezium.
Solution: Parallelogram is a special type of trapezium as its two sides are parallel.

4. Rhombus is a special type of parallelogram.
Solution: Rhombus is a special type of parallelogram as its opposite sides are equal and parallel.

Question 8. Let’s mark (√) and (X) the incorrect and correct statements respectively given below:

1. A rectangular figure has all the sides equal. (X)
2. Each angle of a square is a right angle. (√)
3. Opposite sides of a parallelogram are parallel. (√)
4. For any trapezium, all its sides are equal. (X)
5. None of the angles of a rhombus are right angles. (√)

Class 6 Math Solutions WBBSE Chapter 16 Concept Of Directed Numbers And Number Line  Exercise 16

Question 1. Let us draw a number line put the following numbers on the line and name them.

1. Name + 5;- 2, + 3;- 6, + 2,- 5 respectively as A, B, C, D, E, and F.
Solution: + 5;-2; + 3;-6; + 2;-5,

name as A, B, C, D, E, F respectively.

2. Let’s measure how many units E is from B.
Solution: Measure of E from B is units.

3. How many units is A from B towards the right?
Solution: Measure of A from B towards the right is 7 units.

4. How many units is D from E towards the left?
Solution: Measure of D from E towards the left is 8 units.

5. How many units is D from F towards the left?
Solution: Measure of D from F towards the left is 1 unit.

Read and Learn More WBBSE Solutions For Class 6 Maths

6. What is the relation between the numbers which are at A and F?
Solution: Relation between the numbers which are at A and F – Mutually opposite numbers.

7. What are the absolute values of the numbers of B and E?
Solution: Absolute values of the numbers of B and E = 2.

“WBBSE Class 6 Maths Chapter 16 concept of directed numbers and number line solutions”

Question 2. What do the following mean?

1. Profit of -10 rupees
Solution: Profit of -10 rupees = Loss of Rs. 10.

2. -15 m above
Solution: -15 m above = 15 m down

3. -36 g less
Solution: -36 gm less = 36 gm more.

4. -18 meter towards East
Solution: -18 meters towards East = 18 meters toward West.

5. Saved -23 rupees
Solution: Saved -23 rupees = Spent Rs. 23.

6. -5 km towards South.
Solution: -5 km towards south = 5 km towards North.

Class 6 History	Class 6 Social Science
Class 6 Geography	Class 6 Science
Class 6 Maths	Class 6 Science MCQs
Class 6 General Science	Class 6 Maths Solutions
Class 6 Geography	Class 6 Hindi

Question 3. Let’s write the absolute values of the following numbers

1. -12
2. + 13
3. – 22
4. – 61
5. + 17

Solution: Absolute value of the following

1. -12 → 12
2. +13 → 13
3. -22 → 22
4. -61 → 61
5. +17 → 17

“WBBSE solutions for Class 6 Maths directed numbers and number line”

Question 4. Let’s find the opposite of the following:

1. Spent Rs.-10
Solution: Spent Rs. -10 Rs. 10 saved

2. Climbed up -15 m.
Solution: Climbed up -15m 15m going up

3. Profit of 81
Solution: Profit of 81 :- Loss of Rs. – 81

4. move – 35m down
Solution: move – 35 m down: and 35 m up

5. – 24 kg increase in weight.
Solution: – 24 kg increase in weight = 24 kg weight decreased

6. 28 m towards the right
Solution: 28 m towards right = 28 m towards right

7. 9 kg decrease in weight.
Solution: 9 kg decrease in weight = 9 kg decrease in weight.

Question 5. Using a number line, put < or > in the blank spaces.

“Class 6 Maths WBBSE Chapter 16 directed numbers and number line exercises”

1. 0 ______ 5
Solution: 0 < 5

2. 0 ________ -6
Solution: 0 > -6

3. 6 ________ -6
Solution: 6 > -6

4. – 8
Solution: -8 > -10

5. -1 ______ -11
Solution: -1 > -11

6. ______ 15
Solution: 11 < 15

7. —10 _____ 12
Solution: -10 < 2

8. -100 _____ -50
Solution: -100 < -50

Question 6. (1) Let’s write 4 negative whole numbers less than -12.
Solution: —15; —18; — 20; – 25;

(2) Let’s write 4 negative whole numbers greater than – 8.
Solution: -6; -4; -3; -1

Class 6 WBBSE Math Solutions Chapter 16 Concept Of Directed Numbers And Number Line Exercise 16.1

Question 1. Let us add on a number line

1. (+7) – (+2)
Solution: (+7) – (+2)

∴ (+7) + (+2) = +9.

2. (+2)- (-4)
Solution: (+2)- (-4)

∴ (+2) + (—4) = -2

3. (+6) + (-11)
Solution: (+6) + (-11)

∴ (+6) + (-11) = -5

4. (-5), (-7)
Solution: (-5), (-7)

∴ (-5), (-7) = -12

5. (+8) + (-8)
Solution: (+ 8), (-8)

∴ (+8) + (-8) = 0

6. (+7), (-7)
Solution: (+7), (-7)

∴ (+ 7) + (-7) = 0

7. (+9), (-17)
Solution: (+9), (-17)

∴ + 9 +(-17)=-8.

Class 6 Maths Solutions WBBSE

8. (-11), (-9)
Solution: (-11), (-9)

∴ (-11) +(-9) = -20

Question 2. Let’s add the following

1. (+9) + (+2)
Solution:

(+9) + (+2) = + 11

2. (+11) – (+ 5)
Solution: (+11) +(+5) = +16

3. (+27) + (-11)
Solution: (+27)+ (-11) = +16

4. (-25) + (+6)
Solution: (-25) + (+6) = -19;

5. (-5) + (+9)
Solution: (—5) + (+9) = +4

6. (+13)+ (-13)
Solution: (+13) + (-13) = 0

“Step-by-step solutions for directed numbers and number line Class 6 WBBSE”
“WBBSE Class 6 Maths Chapter 16 important questions and answers”
“Solved examples of directed numbers and number line WBBSE Class 6”
“Best guide for understanding directed numbers and number line Class 6 WBBSE”
“How to understand directed numbers and number line Class 6 WBBSE”

Question 3. Let us simplify

1. (+13) + (+12) + (-10)
Solution: (+13) + (+12) + (-10) = +25 – 10 = (+15)

2. (+31) + (+13) + (-16)
Solution: (+31)+ (+13)+ (-16)

= +44 + (-16)

= 44 – 16 = (+28)

“Step-by-step solutions for directed numbers and number line Class 6 WBBSE”
“WBBSE Class 6 Maths Chapter 16 important questions and answers”
“Solved examples of directed numbers and number line WBBSE Class 6”
“Best guide for understanding directed numbers and number line Class 6 WBBSE”
“How to understand directed numbers and number line Class 6 WBBSE”

3. (+25) + (-16) + (+ 2)
Solution: (+25) +(-16) +(+2)

= (+9) + (+2)

= (+11)

4. (-11) + (+18) + (-16)
Solution: (-11)+ (+18)+ (-16)

= (+7)+ (-16)

= (-9)

Question 4. Let’s add the following

1. (-2), (-10), (+21)
Solution: (-2)+ (-10)+ (+21)

= (-12) + (+21)

= (+9)

2. (-18), (-7), (-2)
Solution: (-18) + (-7) + (-2)

=(-25)+ (-2)

= (-27)

Class 6 Maths Solutions WBBSE

3. (+10), (-8), (-10)
Solution: (+10) + (-8) + (-10)

= (+2) + (-10)

= (-8)

4. (-8), (-10), (+2)
Solution: (-8) + (-10) + (+2)

= (-18) + (+2)

= (-16)

5. (-19),(-9),(+5)
Solution: (-19) + (-9) + (+5)

= (-28) + (+5)

= (-23)

6. (+20), (-9), (-6)
Solution: (+ 20) + (-9) + (-6)

= (+11) + (-6)

= (+5)

7. (-14), (-12), (+ 21)
Solution: (-14) + (-12) + (+21)

= (-26) + (+21)

= (-5)

8. (-13), (+7), (-2)
Solution: (-13) + (+7) + (-2)

= (-6) + (-2)

= (-8)

9. (+15), (-9), (-5)
Solution: (+15) + (-9) + (-6)

= (+6) + (-6)

= 0

WBBSE Math Solutions Class 6 Chapter 16 Concept Of Directed Numbers And Number Line  Exercise 16.2

Question 1. Let’s do it

1. (- 6) – (+2) = _____
Solution: (-6) – (+2)

= -6-2 = —8

2. (-12) — (+12) = ______
Solution: (-12) – (+12)

= -12-12 = -24

3. (+11)-(+3) = ______
Solution: (+11)-(+3)

= 11-3 = +8

4. (-7) – (+8) = _____
Solution: (-7) – (+8)

= -7-8 = -15

5. (+20) – (-7) = ______
Solution: (+20) – (-7)

= 20 + 7 = 27

6. (-18)-(-8) = _________
Solution: (-18)-(-8)

= -18 + 8

= -10

7. (- 9) – (-9) = _______
Solution: (- 9) – (- 9)

= -9+9

= 0

WBBSE Math Solutions Class 6

8. (+ 13)- (-7) = ________
Solution: (+ 13)-(-7)

= 13 + 7 = 20

Question 2. Let’s complete the table given below

Solution:

Question 3. Let us mark (√) for the correct statement and (x) for wrong statement:

1. There are definite numbers of positive whole numbers.
Solution: There are definite numbers of positive whole numbers. [√]

2. 5.3 is a natural number.
Solution: 5.3 is a natural number. [X]

3. – 2.1 is a negative whole number.
Solution: -2.1 is a negative whole number. [x]

4. There is no existence of biggest whole number.
Solution: There is no existence of biggest whole number. [√]

“Step-by-step solutions for directed numbers and number line Class 6 WBBSE”
“WBBSE Class 6 Maths Chapter 16 important questions and answers”
“Solved examples of directed numbers and number line WBBSE Class 6”
“Best guide for understanding directed numbers and number line Class 6 WBBSE”
“How to understand directed numbers and number line Class 6 WBBSE”

Question 4. 1. (+14) — (+16)
Solution: (+14)-(+16)

= (+14)+ (-16)

= 14-16 = – 2

2. (+25) – (+21)
Solution: (+ 25)-(+21)

= (+ 25) + (-21)

= 25-21 = 4

3. (+ 34) – (-19)
Solution: (+ 34) – (-19)

= (+34)+ (+19)

= 34 + 19

= +53

4. (-15)-(-27)
Solution: (-15)-(-27)

=(-15)+(+27)

= -15 + 27

= +12

“Step-by-step solutions for directed numbers and number line Class 6 WBBSE”
“WBBSE Class 6 Maths Chapter 16 important questions and answers”
“Solved examples of directed numbers and number line WBBSE Class 6”
“Best guide for understanding directed numbers and number line Class 6 WBBSE”
“How to understand directed numbers and number line Class 6 WBBSE”

5. (-25) -(+13)
Solution: (-25) -(+13)

= (-25) + (-13)

= -25-13

= -38

6. (-16)-(-10)
Solution: (-16)-(-10)

= (-16)+ (+10)

= -16 + 10

= – 6

7. (+ 31) — (— 12)
Solution: (+31)-(-12)

= (+31)+ (+12)

= +31+12 = +43

8. (—31) — (—45)
Solution: (-31)-(-45)

= (-31)+ (+45)

= – 31 + 45 = +14

9. (-21)-(+21)
Solution: (-21)-(+21)

= (-21)+ (-21)

= -21-21

= -42

“Step-by-step solutions for directed numbers and number line Class 6 WBBSE”

Question 5. Let’s put >, < or = in respective blank spaces:

1. (+13) + (- 8) _______ (+3) – (-2)
Solution: L.H.S. = (+13) + (-8)

= 13-8 = 5

⇒ R.H.S. = (+ 3) – (-2)

= 3 + 2 = 5

∴ (+13) + (-8) [ = ] (+3) – (-2)

2. (-12)-(-10) _______ (-9) +(+3)
Solution: L.H.S. = (-12)-(-10)

= -12 + 10 = – 2

⇒ R.H.S. = (-9) + (+3)

= -9 + 3

= -6

∴ (-12) – (-10) [ > ] (-9) + (+3)

3. (+35) – (-5) _____ (- 24) – (-64)
Solution: L.H.S. = (+ 35) – (-5)

= 35 + 5 = 40

⇒ R.H.S. = (-24)-(-64)

= – 24 + 64 = 40

∴ (+35) – ( -5) [ = ] (-24) – (-64)

Class 6 Math WBBSE Solutions

4. (-18) – (+6) ______ (-18) – (-6)
Solution: L.H.S. = (-18) – (+6)

= -18-6 = -24

⇒ R.H.S. = (-18) -(-6)

= -18 + 6 = -12

∴ (-18) (+ 6) [ < ] (—18) — (—6)

5. (-45) – (-52) _______ (-52) – (-45)
Solution: L.H.S. = (-45) – (-52)

= -45 + 52 = + 7

⇒ R.H.S. = (-52)-(- 45)

= – 52 + 45 = —7

∴ (-45) – (—52) [ > ] (-52) – (-45)

6. (+25) -(-19) _______ (-25) -(+19)
Solution: L.H.S. = (+25)-(-19)

= +25 + 19 = + 44

⇒ R.H.S. = (-25)-(+19)

= -25-19 = -44

∴ (+25) — (—19) [ > ] (—25) – (+19)

Question 6. Let’s put number in blank spaces:

1. (-3) + ________ = 0
Solution: (—3) + [+3] = 0

2. (+16) + ________ = 0
Solution: (+16) + [-16] = 0

3. (-9) + _______ = (-15)
Solution: (-9) + [-6] = – 15

4. _____ + (-7) = (-10)
Solution: [-3] +(-7) = (-10)

Question 7. Let’s simplify:

1. (-5) + (opposite number of -7) – (-5)
Solution: (-5) + (+7) + (5)

= + 7

2. 12 – (-3) + (opposite number of +6)
Solution: 12 + 3 + (-6)

= 15-6 = 9

3. 15 – (+4) + (opposite number of +9)
Solution: 15 – 4 + (-9)

= 15-4-9 = 15-13 = 2

4. (opposite number of +20) – (opposite number of -7) – (-8)
Solution: (-20) – (+7) + 8

= -20 – 7 + 8 = – 27 + 8

= -19

Question 8. Let’s find what must be added to the first to get the second:

1. -7,-12
Solution: -12-(-7)

= -12 + 7 = – 5

2. 24, – 32
Solution: -32 – (+24)

= – 32 – 24

= -56

3. -17,12
Solution: 12-(-17)

= 12+17

= 29

Class 6 Math WBBSE Solutions

4. 16, 0
Solution: 0 – (+ 16)

= 0-16

= -16

5. 25,-42
Solution: -42 – (+25)

= -42 -25 = -67

Question 9. Let’s find what must be added to the first to get the second:

1. (+7), (+2)
Solution: (+2) – (+7)

= 2-7 = -5

.2. (+7), (-2)
Solution: (-2)-(+7)

= -2-7 = —9

3. (-7), (+2)
Solution: (+2)-(-7)

= 2 + 7 = 9

4. (-7), (-2)
Solution: (-2)-(-7)

= -2 + 7

= + 5

Question 10. Let’s verify the commutative property of addition for the following:

1. (+5), (+3)
Solution: (+5) + (+3) = +8,

⇒ Again, (+3) + (+5) = +8 (+5) + (+3) = (+3) + (+5)

2. (+ 5), (-3)
Solution: (+5) + (-3) = +2,

⇒ Again, (-3) + (+5) = + 2 (+5) + (-3) = (-3) + (+5)

3. (-5), (+3)
Solution: (-5) + (+3) = -2,

⇒ Again, (+3) + (-5) = – 2 (-5) + (+3) = (+3) + (-5)

4. (-5), (-3)
Solution:

⇒ (-5) + (-3) = -8,

⇒ Again (-3) + (-5) = -8

⇒ (-5) + (-3) = (-3) + (-5)

“WBBSE Class 6 Maths Chapter 16 important questions and answers”

Question 11. Let’s verify the associative property of addition for the following:

1. (+5), (+3), (+2)
Solution: (+5), (+3), (+2)

= {(+5) + (+3)} + (+2)

= (+8) + (+2)

= (+10)

⇒ and, = (+5) + {(+3) + (+2)}

= +5 + 5 = (+10)

{(+5) + (+3)} + (+2)

= (+5) + {(+3) + (+2)}

(+5), (+3), (+2) = (+5) + {(+3) + (+2)}

2. (+5), (-3), (+2)
Solution: (+5), (-3), (+2)

= {(+5) + (-3)) + (+2)

= (+2) + (+2) = (+4)

⇒ and, (+5) + {(-3) + (+2)}

= (+5) + (-1) = (+4)

= {(+ 5) + (-3)} + (+2)

= (+5) + {(-3) + (+2)}

(+5), (-3), (+2) = (+5) + {(-3) + (+2)}

3. (-5), (-3), (+2)
Solution: (-5), (-3), (+2)

⇒ {(-5) + (-3)} + (+2)

= (-8) + (+2) = -6 and, (—5) + {(—3) + (+2)}

= (—5) + (—1) = -6

⇒ {(-5) + (-3)} + (+2)

= -5 + {(-3) + (+2)}

{(-5) + (-3)} + (+2) = -5 + {(-3) + (+2)}

4. (—5), (—3), (-2)
Solution: (-5), (-3), (-2)

⇒ {(- 5) + (-3)} + (-2)

= (-8) + (-2) = (-10)

⇒ and, (-5) + {(-3) + (-2)}

= (- 5) + (- 5) = (-10)

⇒ {(-5) + (-3)} + (-2)

= (-5) + {(-3) + (-2)}

(-5), (-3), (-2) = (-5) + {(-3) + (-2)}

Class 6 Math Solutions WBBSE Chapter 15 Determination Of Area And Perimeter Exercise 15

Let’s draw, observe, and write what I know

Question 1. I found my brother’s first geometric shape model is a triangle.
Solution:

⇒ Its perimeter is = _____ cm + _______ cm + ______ cm = 3 x 12 cm

To put a border around this geometric model, we need 36 cm of colored paper.

Solution:

⇒ Perimeter of the 1 st figure = 12 cm + 12 cm + 12 cm = 36 cm

Question 2. Let me find the perimeter of the second geometric figure. The perimeter of the second geometric figure = _____ cm + ________ cm + _________ cm + _________ cm = _______ cm. I shall need ______ cm of colored paper to put a border on it.

Solutions:

⇒ Perimeter of the 2nd figure

= 20 cm + 22 cm + 22 cm + 18 cm = 82 cm

“WBBSE Class 6 Maths Chapter 15 determination of area and perimeter solutions”

Read and Learn More WBBSE Solutions For Class 6 Maths

Question 3. The perimeter of the third geometric shape is = ________ cm + _______ cm + _____ cm + ______ cm + ________ cm + _____ cm = ________ cm

Solution:

⇒ Perimeter of the 3rd figure = 6 cm + 6 cm + 6 cm + 6 cm + 6 cm + 6 cm = 36 cm

Class 6 History	Class 6 Social Science
Class 6 Geography	Class 6 Science
Class 6 Maths	Class 6 Science MCQs
Class 6 General Science	Class 6 Maths Solutions
Class 6 Geography	Class 6 Hindi

Class 6 Math Solutions WBBSE

Question 4. Perimeter of the fourth geometric figure. = _____ cm + _____ cm + ______ cm + ______ cm + ______ cm + _______ cm + _______ cm + ______ cm + = _____ cm

Solution:

⇒ Perimeter of the 4th figure

= 5 cm + 13 cm + 30 cm + 13 cm + 5 cm + 7 cm + 20 cm + 7 cm

= 100 cm

“WBBSE solutions for Class 6 Maths area and perimeter”

Question 5. Perimeter of fifth geometric figure = _____ cm. + _______ cm + _____ cm + ________ cm + ______ cm + _________ cm + _______ cm + _______ cm = _____ cm

Solution:

⇒ Perimeter of the 5th figure

= 16 cm + 4 cm + 6 cm + 10 cm + 4 cm + 10 cm + 6 cm + 4 cm = 60 cm

“Class 6 Maths WBBSE Chapter 15 area and perimeter exercises”

Question 6. Let’s find the perimeter of the sixth geometric figure.

Solution:

⇒ Perimeter of the 6th figure

= 4 cm + 9 cm + 11 cm + 9 cm + 4 cm + 22 cm + 4 cm + 9 cm + 11 cm + 9 cm + 4 cm + 22 cm = 118 cm.

“WBBSE Class 6 Maths Chapter 15 important questions and answers”

Hands-on trial: Now, we shall join points on graph paper to form different geometric figures and try to find the space occupied by these figures.

⇒ We pasted the graph paper on a cardboard. We formed different geometric figures on it by joining points and let’s try to find the area enclosed by them.

“Step-by-step solutions for area and perimeter Class 6 WBBSE”

Example: Let’s fill up the data and let’s measure the area:
Solution:

 

Class 6 Math Solutions WBBSE Chapter 14 Concept Of Line Segment Ray And Point

Let’s draw, observe, and write what I know

My friends and I draw pictures, observe and identify the things we know, let’s note down

Example 1. Figure (1) is the picture of a ____.
Answer:

⇒ Figure (1) is a picture of a cottage or house.

Example 2. In the picture of the house we can find many ___ [Straight lines or Angular points]
Answer:

⇒ In the picture of the house we can find many Angular points.

“WBBSE Class 6 Maths Chapter 14 concept of line segment ray and point”

Read and Learn More WBBSE Solutions For Class 6 Maths

Example 3. Let’s mention the four angular points — ______, ____, _____, ______ and mark them with ‘O’ in the picture.
Answer:

⇒ The four angular points are A, B, C, and D.

Example 4. Again, we find several ____[Ray/Line segment] in the picture of the house. From Figure (1) let us find and name 4 of the line segments ____, ___, _____, and _____
Answer:

⇒ Again, we find several line segments in the picture of the house. From Figure (1) 4 names of the line segments are AB, BC, CD, and GD.

Example 5. We find the line segments DE and DC are intersecting at ______.
Answer:

⇒ We find the line segments DE and DC intersect at D.

Class 6 Math Solutions WBBSE

Example 6. Again, in Figure (1) ______ and ____ line segments have intersected at E.
Answer:

⇒ Again, in Figure (1) DE and EF line segments have intersected at E

Class 6 History	Class 6 Social Science
Class 6 Geography	Class 6 Science
Class 6 Maths	Class 6 Science MCQs
Class 6 General Science	Class 6 Maths Solutions
Class 6 Geography	Class 6 Hindi

Example 7. But, we see the 2 line segments DE, CF, and the line segments DC and _____ are on the same plane but they will never intersect each other, no matter how far they are produced on either side.
Answer:

⇒ But, we see the 2 line segments DE, CF, and the line segments DC and EF are on the same plane but they will never intersect each other, no matter how far they are produced on either side.

Example 8. The line segments DE and DC are on the same plane and intersect each other at ____, hence these lines are called _____(Intersecting/Parallel).
Answer:

⇒ The line segments DE and DC are on the same plane and are intersecting each other at D, hence these lines are called Intersecting.

Example 9. The line segments DE and CF are on the same plane but will never meet, no matter how far they are produced. Hence, these line segments are (Intersecting/Parallel).
Answer:

The line segments DE and CF are on the same plane but will never meet, no matter how far they are produced. Flence these line segments are parallel.

Class 6 WBBSE Math Solutions Chapter 14 Concept Of Line Segment Ray And Point Exercise 14

Question 1. From the picture (2), let’s identify 4 angular points and name them _______, _________, ________, ______
Solution:

⇒ From the picture (2), let’s identify 4 angular points and name them I, J, K, L.

Question 2. Let’s identify and write the four line segments which are named in the picture (2) _______, ______ and ______
Solution:

Let’s identify and write the four line segments that are named in the picture (2) SP, PQ, QR, and SR.

“WBBSE solutions for Class 6 Maths line segment ray and point”

Question 3. From the picture (2), let’s identify two pairs of intersecting line segments on a plane and find the points at which they intersect.
Solution:

1. SP and PQ are two pairs of intersecting line segments at P as intersecting points.
2. SP and SR are two pairs of intersecting line segments at S as an intersecting point.
3. SR and QR are two pairs of intersecting line segments at R as an intersecting point.
4. QR, and PQ are two pairs of intersecting line segments at Q as an intersecting point.

Question 4. In the picture (2), let’s find two pairs of parallel line segments on a plane that are already named in the figure.
Solution:

⇒ PQ || SR and SP || QR are two pairs of parallel line segments on a plane.

Question 5. From picture (3) — let’s identify and write 3 intersecting points and 4 line segments already named in the picture.
Solution:

3 intersecting points: M, N, O.

4 line segments: MN; NO; OP and PM.

Class 6 Maths Solutions WBBSE Chapter 14 Concept Of Line Segment Ray And Point 14.1

Question 1. Let’s draw 6 concurrent straight lines and name them.
Solution:

⇒ The 6 concurrent straight lines are \(\overrightarrow{A B}, \overrightarrow{C D}, \overrightarrow{E F}, \overrightarrow{G H}, \overrightarrow{I J}, \ \overrightarrow{K L}\)

“Class 6 Maths WBBSE Chapter 14 line segment ray and point exercises”

Question 2. 5 non-collinear points are to be drawn and find how many straight lines can be drawn through them.
Solution:

A, B, D, and E, are 5 non-collinear points

⇒ Through the point A, we can draw 4 straight lines.

⇒ Through the point B, we can draw 3 straight lines.

⇒ Through the point C, we can draw 2 straight lines.

⇒ Through the point D, we can draw 1 straight line

Question 3. Let’s draw 5 points that are collinear.
Solution:

The points A, B, C, D, and E are 5 collinear points.

WBBSE Math Solutions Class 6 Chapter 14 Concept Of Line Segment Ray And Point 14.2

Question 1. Let’s identify the points of intersection in the following figures and note them down.

Solution:

1. D, E, and F are the points of intersection.
2. P, Q, R, S, and T are the points of intersection.
3. A, B, C, and D are the points of intersection.
4. K, L, M, N, and O are the points of intersection.

Question 2. In the figures given below, let’s identify line segments and rays and note them down.

Solution:

1. Line segments = \(\overline{X Y}, \overline{X Z}\)

⇒ Rays = \(\overrightarrow{X Y}, \overrightarrow{X Z}\)

2. Line segments = \(\overline{\mathrm{AD}}, \overline{\mathrm{AB}}, \overline{\mathrm{BC}}\)

⇒ Rays = \(\overrightarrow{A D}, \overrightarrow{B C}\)

“Step-by-step solutions for line segment ray and point Class 6 WBBSE”

3. Line segments = \(\overline{\mathrm{AB}}, \overline{\mathrm{BC}}, \overline{\mathrm{CD}}, \overline{\mathrm{DE}}\)

⇒ Rays = \(\overrightarrow{B A}, \overrightarrow{D E}\)

4. Line segments = \(\overline{\mathrm{AB}}, \overline{\mathrm{CD}}\)

⇒ Rays = \(\overrightarrow{O B}, \overrightarrow{O C}\)

Question 3.

1. What are these three points X, Y, and Z called?
2. How many line segments are formed with these three points?

Solution:

⇒ The points X, Y, and Z are called collinear points.

⇒ Three line segments are formed with these points.

Question 4. From the figure:

1. Let’s find and write three pairs of parallel lines.
2. Let us find and write three pairs of intersecting lines.
3. Let us find and write six-line segments.

Solution:

1. \((\stackrel{\leftrightarrow}{A B}, \overleftrightarrow{C D}),(\overleftrightarrow{C D}, \overleftrightarrow{E F}),(\stackrel{\leftrightarrow}{A B}, \overleftrightarrow{E F})\)

2. \((\stackrel{\leftrightarrow}{\mathrm{PQ}}, \stackrel{\leftrightarrow}{\mathrm{MN}}) ;(\stackrel{\leftrightarrow}{\mathrm{PQ}}, \overleftrightarrow{\mathrm{G}-1}) \&(\stackrel{\leftrightarrow}{\mathrm{MN}}, \stackrel{\leftrightarrow}{\mathrm{RS}})\)

3. \((\overline{\mathrm{PB}}, \overline{\mathrm{AC}}, \overline{\mathrm{CE}}, \overline{\mathrm{BD}}, \overline{\mathrm{DF}}, \& \overline{\mathrm{AB}})\)

Question 5. From the figure:

1. Let’s find the points of intersection.
2. Let’s write all sets of collinear points separately.
3. Let’s write the line segments separately.
4. Let’s write the concurrent line segments.

Solution:

1. A, F, B, D, C, E and 0.
2. (A, F, B); (A, O, D), (A, E, C); (B, D, C); (C, 0, F), (B, O, E).
3. AB, BC, CA, AF, FB, AB, AD, OD, BO, BE, OE, CO, CF, OF, AE EC BD DC
4. AD, BE and CF.

Question 6. Let us put (√) for correct and (x) for wrong answers

1. On the line segment \(\overrightarrow{\mathrm{YW}}\), Z and W are coliinear.

2. \(\overrightarrow{\mathrm{ZV}}\) and \(\overrightarrow{\mathrm{WV}}\) are the same rays.

3. \(\overrightarrow{\mathrm{ZV}}\) and \(\overrightarrow{\mathrm{ZX}}\) are same rays.

4. On the ray \(\overrightarrow{\mathrm{YX}}\), Z is a point.

“WBBSE Class 6 Maths Chapter 14 important questions and answers”

Solution:

1. On the line segment \(\overrightarrow{\mathrm{YW}}\), Y, Z and W are coliinear. (√)

2. \(\overrightarrow{\mathrm{ZV}}\) and \(\overrightarrow{\mathrm{WC}}\), \(\overrightarrow{\mathrm{WV}}\) WV are the same rays- (√)

3. \(\overrightarrow{\mathrm{ZV}}\) and \(\overrightarrow{\mathrm{ZX}}\) are the same raYs- (X)

4. On the ray \(\overrightarrow{\mathrm{YX}}\), Z is a point. (X)

Question 7. Let’s try and find the answers

1. How many straight lines can be drawn through one fixed point?
Solution: Infinite

2. How many straight lines can be drawn through two fixed points?
Solution: One

3. How many line segments can be drawn through three non-collinear points?
Solution: Three

4. How many endpoints are there in a line segment \(\overline{\mathrm{AB}}\)?
Solution: A, B

5. How many endpoints are there in ray \(\overrightarrow{\mathrm{AB}}\)?
Solution: A, B

6. Straight line, line segment, and ray — of these three which one has a fixed length?
Solution: Line segment

7. Are the rays \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{BA}}\) same?
Solution: Not Same

“Solved examples of line segment ray and point WBBSE Class 6”

8. Are the line segments \(\overrightarrow{\mathrm{AB}}\) and \(\overrightarrow{\mathrm{BA}}\) equal? In what way are they equal?
Solution: Yes

9. What is the maximum number of points at which two line segments can meet?
Solution: One

WBBSE Math Solutions Class 6

10. Let’s find the maximum number of points at which three non-concurrent lines can intersect.
Solution: Three

Question 8. Let’s draw ourselves

1. \(\overline{\mathrm{PQ}}\) and \(\overline{\mathrm{RS}}\) are two line segments interseceting at O.

2. Using a scale, let us draw \(\stackrel{\leftrightarrow}{\mathrm{AB}}\) and \(\stackrel{\leftrightarrow}{\mathrm{CD}}\) as two Parallel lines

3. Let’s draw a ray \(\overrightarrow{\mathrm{MN}}\) on which S is a point.

4. Let’s draw two line segments AC and DC meeting at C.

Solution:

1. PQ and RS are two line segments interesting at O.
2. AB and CD are two parallel line segments.
3. MN is a rayon which S is a point.
4. AC and CD are two line segments meeting at C.

Question 9. From the figure given below, let’s measure the lengths with scale and fill in the blank spaces accordingly.

1. \(\overline{\mathrm{PQ}}+\overline{\mathrm{QR}}\)=____ \(\mathrm{cm} \cdot=\overline{\mathrm{PR}}\)

2. \(\overline{\mathrm{QR}}+\overline{\mathrm{RS}}\) = _____ \(\mathrm{cm}\) = _____

3. \(\overline{\mathrm{PS}}\) = ____ + \(\overline{\mathrm{QR}}\) + _____

4. \(\overline{\mathrm{PR}}-\overline{\mathrm{QR}}\) = _______ cm = _______

5. PR – PQ = ______ cm

6. QS – QR = _______ cm.

7. QS – RS = ______ cm.

Solution:

1. \(\overline{\mathrm{PQ}}+\overline{\mathrm{QR}}=(3+3) \mathrm{cm} \cdot=\overline{\mathrm{PR}}=6 \mathrm{~cm}\)

2. \(\overline{\mathrm{QR}}+\overline{\mathrm{RS}}=(3+3) \mathrm{cm} .=\overline{\mathrm{QS}}=6 \mathrm{~cm}\)

3. \(\overline{\mathrm{PS}}=\overline{\mathrm{PQ}}+\overline{\mathrm{QR}}+\overline{\mathrm{RS}}\)

4. \(\overline{\mathrm{PR}}-\overline{\mathrm{QR}}=3 \mathrm{~cm} .=\overline{\mathrm{PQ}}\)

5. \(\mathrm{PR}-\mathrm{PQ}=(6-3) \mathrm{cm}=3 \mathrm{~cm}\)

6. \(\mathrm{QS}-\mathrm{QR}=(6-3) \mathrm{cm}=3 \mathrm{~cm}\)

7. \(\mathrm{QS}-\mathrm{RS}=(6-3) \mathrm{cm}=3 \mathrm{~cm}\)

Class 6 Math Solutions WBBSE Chapter 13 Data Handling And Analysis Exercise 13

Example: Like every year, this year the Physical Education teacher of our school took measurements (in cm) of the height of the students of class 6 who would be taking part in school sports. The data obtained is given below

Given: The height of the students of class 6

⇒ The shortest height = 120 and the tallest height = 127.

Question 1. In a class test of 15 marks, I prepare a list showing marks obtained by 20 students. Let’s write down these marks by giving tally marks and prepare a frequency distribution chart. 9, 8, 6, 10, 2, 1,15, 12, 8, 6, 9, 2, 8, 5, 9, 10, 5, 9, 10, 8
Solution:

Given

In a class test of 15 marks, I prepare a list showing marks obtained by 20 students.

9, 8, 6, 10, 2, 1,15, 12, 8, 6, 9, 2, 8, 5, 9, 10, 5, 9, 10, 8

⇒ Frequency Distribution Table

Read and Learn More WBBSE Solutions For Class 6 Maths

“WBBSE Class 6 Maths Chapter 13 data handling solutions”

Question 2. In Jahanara’s school classes were held for 22 days only this month. The number of students attending classes on these 22 days is given below. 30, 28, 34,29,25,30,28,26,29,30, 22,25,26,29, 30, 31,21,27, 25,13, 32, 28. Using the above raw data, let’s prepare a frequency distribution chart with tally marks.

Class 6 Math Solutions WBBSE
Solution:

Given

In Jahanara’s school classes were held for 22 days only this month. The number of students attending classes on these 22 days are given below. 30, 28, 34,29,25,30,28,26,29,30, 22,25,26,29, 30, 31,21,27, 25,13, 32, 28.

⇒ Frequency distribution table

Question 3. Given below is the data showing what fruits my 20 friends like to have. Let me draw a bar graph based on this data.
Solution:

Class 6 History	Class 6 Social Science
Class 6 Geography	Class 6 Science
Class 6 Maths	Class 6 Science MCQs
Class 6 General Science	Class 6 Maths Solutions
Class 6 Geography	Class 6 Hindi

Question 4. I prepared raw data showing the number of students present in the class for 6 days of this week. From this raw data, I prepare a list using tally marks. Let’s write down from the list the day of the week on which the least number of students are present.
Solution:

⇒ I prepare a raw data showing number of students present in the class for 6 days of the week.

Class 6 Math Solutions WBBSE

⇒ Frequency distribution table

Question 5. I write down the weight (kg) of 30 students of class VI of Soham’s school. 32, 32, 37, 34, 37, 35, 35, 36,37, 39, 40, 36, 37, 36, 33,31,32,36, 37,38, 40, 34, 36, 34, 35, 33, 34, 35, 32, 35 Using the above raw data, let’s prepare a frequency distribution chart.
Solution:

⇒ Weight (kg) of 30 students of class 32, 32, 37, 34, 37, 35, 35, 36, 37, 39, 40, 36, 37, 36, 33, 31,32, 36, 37, 38, 40, 34, 36, 34, 35, 33, 34, 35, 32, 35

Class 6 History	Class 6 Social Science
Class 6 Geography	Class 6 Science
Class 6 Maths	Class 6 Science MCQs
Class 6 General Science	Class 6 Maths Solutions
Class 6 Geography	Class 6 Hindi

⇒ Frequency Distribution Table

Class 6 Math Solutions WBBSE

Question 6. From a survey conducted on 150 students of our school, it is seen how many students like to study a special subject. The raw data from the survey is given below
Solution:

“WBBSE solutions for Class 6 Maths data handling and analysis”

Question 7. Ram da has a shop selling bags in Haidar Para. Ram da himself makes the bags. I prepare a list of bags sold from the shop this week.

Let take a scale of my choice and draw a bar diagram.
Solution:

“Class 6 Maths WBBSE Chapter 13 data handling exercises”

Question 8. I have prepared a bar diagram showing the number of students appearing at the Madhymik Examination each year during the five years.

1. I have prepared a set of questions from the bar diagram and let’s answer the questions.
2. Let us see in which year a maximum number of students have sat for the exam.
3. Let us see in which year the least number of students have sat for the exam.
4. Let’s see how many more students appeared for the examination in 2011 than in 2010.
5. In 2010, let us see how many less students have appeared for the exam than in 2009.
6. Let’s see how many total students have sat for examination between 2008 and 2010.

“Step-by-step solutions for data handling Class 6 WBBSE”

Solution:

1. 2012
2. 2010
3. 120-100 = 20
4. 100-90 = 10
5. 110 + 100 + 90