CBSE Class 8

NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.1 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.1.

• Comparing Quantities Class 8 Ex 8.2
• Comparing Quantities Class 8 Ex 8.3

NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.1

Question 1.
Find ratio of the following:
(a) Speed of a cycle 15 km per hour to the speed of scooter 30 km per hour.
(b) 5 m to 10 km
(c) 50 paise ₹ 5
Solution.
(a)
Speed of a cycle 15 km per hour to the speed of scooter 30 km per hour
Ratio of the speed of cycle which is 15 km per hour to the speed of scooter which is 30 km per hour
= 15 km per hour : 30 km per hour = 15 : 30
= \(\frac { 15 }{ 30 } =\frac { 1 }{ 2 } \) or 1 : 2

(b)
5 m to 10 km
10 km = 10 x 1000 m = 10000 m
∴ Ratio of 5 m to 10 km
= 5 m : 10 km
= 5 m : 10000 m
= 5 : 10000
= \(\frac { 5 }{ 10000 } =\frac { 1 }{ 2000 } \) or 1 : 2000

(c)
50 paise to ₹ 5
₹ 5 = 5 x 100 = 500 paise
∴ Ratio of 50 paise to ₹ 5
= 50 paise : ₹ 5
= 50 paise : 500 paise
= 50 : 500
= \(\frac { 50 }{ 500 } =\frac { 1 }{ 10 } \) or 1 : 10

Question 2.
Convert the following ratios to percentages:
(a) 3: 4
(b) 2 : 3
Solution.
(a) 3: 4

(b) 2 : 3

Question 3.
72% of 25 students are good in mathematics. How many are not good in Mathematics?
Solution.
Total number of students = 25
Students good in mathematics = 72%
∴ Students who are not good in mathematics
= (100 – 72)% = 28%
∴ Number of those students who are not good in mathematics
= 28% of 25
= \(\frac { 28 }{ 100 } \times 25=7\)
Hence, 7 students are not good in mathematics.

Question 4.
A football team won 10 matches out of the total number of matches they played. If their win percentage was 40, then how many matches did they play in all?
Solution.
∵ If 40 matches were won, then the total number of matches played = 100
∴ If 1 match was won, then the total number of matches played = \(\frac { 100 }{ 40 } \)
∴ If 10 matches were won, then the total number of matches played = \(\frac { 100 }{ 40 } \times 10=25\)
Hence, they played 25 matches in all.
Aliter: According to the question, 40% of (total number of matches) = 10
⇒ \(\frac { 40 }{ 100 } \) x (total number of matches) = 10
⇒ Total number of matches = \(\frac { 10\times 100 }{ 40 } \) = 25
Hence, they played 25 matches in all.

Question 5.
If Chameli had ₹ 600 left after spending 75% of her money, how much did she have in the beginning?
Solution.
Percentage of money left = (100 – 75)% = 25%
∵ If Chameli had ₹ 25 left, then the money she had in the beginning = 100
∴ If Chameli had ₹ 1 left, then the money she had in the beginning = \(\frac { 100 }{ 25 } \)
∴ If Chameli has ₹ 600 left, then the money she had in the beginning = \(\frac { 100 }{ 25 } \times 600=2400\)
Hence, the money she had in the beginning was ₹ 2400.
Aliter:
According to the question, 25% of total money = ₹ 600
⇒ \(\frac { 25 }{ 100 } \) x total money = ₹ 600
⇒ Total money = ₹ \(\frac { 600\times 100 }{ 25 } \) = ₹ 2400
Hence, the money she had in the beginning was ₹ 2400

Question 6.
If 60% of people in a city like a cricket, 30% like football and the remaining like other games then what percent of the people like other games? If the total number of people are 50 lakh, find the exact number who like each type of game.
Solution.
People who like other games = [100 – (60 + 30)]%
= (100 – 90)% = 10%.
Total number of people = 50 lakh
= 5000000
∴ Number of people who like cricket = 60% of 5000000
= 5000000 x \(\frac { 60 }{ 100 } \)
= 3000000 = 30 lakh
Number of people who like football = 30% of 5000000
= 5000000 x \(\frac { 30 }{ 100 } \)
= 1500000 = 15 lakh
Number of people who like the other games
= 10% of 5000000
= 5000000 x \(\frac { 10 }{ 100 } \)
= 5,00,000 = 5 lakh

We hope the NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.1 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Ex 7.1 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Ex 7.1.

• Cubes and Cube Roots Class 8 Ex 7.2

NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Ex 7.1

Question 1.
Which of the following numbers are not perfect cubes?
(i) 216
(ii) 128
(iii) 1000
(iv) 100
(v) 46656
Solution.
(i) 216

(ii) 128

(iii) 1000

(iv) 100

(v) 46656

Question 2.
Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.
(i) 243
(ii) 256
(iii) 72
(iv) 675
(v) 100
Solution.
(i) 243

(ii) 256

(iii) 72

(iv) 675

(v) 100

Question 3.
Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.
(i) 81
(ii) 128
(iii) 135
(iv) 192
(v) 704
Solution.
(i) 81

(ii) 128

(iii) 135

(iv) 192

(v) 704

Question 4.
Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?
Solution.
Volume of a cuboid = 5 x 2 x 5 \({ cm }^{ 3 }\).
Since there is only one 2 and only two 5’s in the prime factorization, so, we need 2 x 2 x 5, i.e., 20 to make a perfect cube. Therefore, we need 20 such cuboids to make a cube.

We hope the NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Ex 7.1 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Ex 7.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.1 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.1.

• Squares and Square Roots Class 8 Ex 6.2
• Squares and Square Roots Class 8 Ex 6.3
• Squares and Square Roots Class 8 Ex 6.4

NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.1

Question 1.
What will be the unit digit of the squares of the following numbers ?
(i) 81
(ii) 272
(iii) 799
(iv) 3853
(v) 1234
(vi) 26387
(vii) 52698
(viii) 99880
(ix) 12796
(x) 55555
Solution.
(i) 81
∵ \(1\times 1\)
∴ The unit digit of the square of the number 81 will be 1.

(ii) 272
∵ \(2\times 2\)
∴ The unit digit of the square of the number 272 will be 4.

(iii) 799
∵ \(9\times 9\)
∴ The unit digit of the square of the number 799 will be 1.

(iv) 3853
∵ \(3\times 3\)
∴ The unit digit of the square of the number 3853 will be 9

(v) 1234
∵ \(4\times 4\)
∴ The unit digit of the square of the number 1234 will be 6.

(vi) 26387
∵ \(7\times 7\)
∴ The unit digit of the square of the number 26387 will be 9.

(vii) 52698
∵ \(8\times 8\)
∴ The unit digit of the square of the number 52698 will be 4.

(viii) 99880
∵ \(0\times 0\)
∴ The unit digit of the square of the number 99880 will be 0.

(ix) 12796
∵ \(6\times 6\)
∴ The unit digit of the square of the number 12796 will be 6.

(x) 55555
∵ \(5\times 5\)
∴ The unit digit of the square of the number 55555 will be 5.

Question 2.
The following numbers are obviously not perfect squares. Give reason.
(i) 1057
(ii) 23453
(iii) 7928
(iv) 222222
(v) 64000
(vi) 89722
(vii) 222000
(viii) 505050.
Solution.
(i) 1057
The number 1057 is not a perfect square because it ends with 7 at unit’s place whereas the square numbers end with 0, 1, 4, 5, 6 or 9 at unit’s place.

(ii) 23453
The number 23453 is not a perfect square because it ends with 3 at unit’s place whereas the square numbers end with 0, 1, 4, 5, 6 or 9 at unit’s place.

(iii) 7928
The number 7928 is not a perfect square because it ends with 8 at unit’s place whereas the square numbers end with 0, 1, 4, 5, 6 or 9 at unit’s place.

(iv) 222222
The number 222222 is not a perfect square because it ends with 2 at unit’s place whereas the Square numbers end with 0, 1, 4, 5, 6 or 9 at unit’s place.

(v) 64000
The number 64000 is not a square number because it has 3 (an odd number of) zeros at the end whereas the number of zeros at the end of a square number ending with zeros is always even.

(vi) 89722
The number 89722 is not a square number because it ends in 2 at unit’s place whereas the square numbers end with 0, 1, 4, 5, 6 or 9 at unit’s place.

(vii) 222000
The number 222000 is not a square number because it has 3 (an odd number of) zeros at the end whereas the number of zeros at the end of a square number ending with zeros is always even.

(viii) 505050
The number 505050 is not a square number because it has 1 (an odd number of) zeros at the end whereas the number of zeros at the end of a square number ending with zeros is always even.

Question 3.
The squares of which of the following would be odd numbers ?
(i) 431
(ii) 2826
(iii) 7779
(v) 82004.
Solution.
(i) 431
∵ 431 is an odd number
∴ Its square will also be an odd number.

(ii) 2826
∵ 2826 is an even number
∴ Its square will not be an odd number.

(iii) 7779
∵ 7779 is an odd number
∴ Its square will be an odd number.

(iv) 82004
∵ 82004 is an even number
∴ Its square will not be an odd number.

Question 4.
Observe the following pattern and find the missing digits:

Solution.

Question 5.
Observe the following pattern and supply the missing numbers:

Solution.

Question 6.
Using the given pattern, find the missing numbers:

Solution.

Question 7.
Without adding, find the sum:
(i) 1 + 3 + 5+7 + 9
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 +17 + 19
(iii) 1+3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 +23.
Solution.
(i)
l + 3 + 5 + 7 + 9 = sum of first five odd natural numbers = \({ 5 }^{ 2 }\) = 25

(ii)
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = sum of first ten odd natural numbers = \({ 10 }^{ 2 }\) = 100

(iii)
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = sum of first twelve odd natural numbers = \({ 12 }^{ 2 }\)= 144.

Question 8.
(i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
Solution.
(i)
49 (= \({ 7 }^{ 2 }\))
= 1 + 3 + 5 + 7 + 9 + 11 + 13

(ii)
121 (= \({ 11 }^{ 2 }\))
= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21.

Question 9.
How many numbers lie between squares of the following numbers ?
(i) 12 and 13
(ii) 25 and 26
(iii) 99 and 100.
Solution.
(i) 12 and 13
Here, n = 12
∴ 2n = 2 x 12 = 24
So, 24 numbers lie between squares of i the numbers 12 and 13.

(ii) 25 and 26
Here, n = 25
∴ 2n = 2 x 25 = 50
So, 50 numbers lie between squares of the numbers 25 and 26.

(iii) 99 and 100
Here, n = 99
∴ 2n = 2 x 99 = 198
So, 198 numbers lie between squares of of the numbers 99 and 100.

We hope the NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.1 are part help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.1 are part, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.1 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.1.

• Data Handling Class 8 Ex 5.2
• Data Handling Class 8 Ex 5.3

NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.1

Question 1.
For which of these would you use a histogram to show the data?
(a) The number of letters for different areas in a postman’s bag.
(b) The height of competitors in an athletics meet.
(c) The number of cassettes produced by 5 companies.
(d) The number of passengers boarding trains from 7:00 a.m. to 7:00 p.m at a station.
Give reasons for each.
Solution.
For (b) and (d) because in these two cases data can be grouped into class intervals.

Question 2.
The shoppers who come to a departmental store are marked as man (M), woman (W), boy (B) or girl (G). The following list gives the shoppers who came during the first hour in the morning.
WWWGBWWMGGMMWWWWG
BMWBGGMWWMMWW
WMWBWGMWWWWGWMMWW
MWGWMGWMMBGGW
Make a frequency distribution table using tally marks. Draw a bar graph to illustrate it.
Solution.

Question 3.
The weekly wages (in of 30 workers in a factory are:
830, 835, 890, 810, 835, 836, 869, 845, 898, 890, 820, 860,
832, 833, 855, 845, 804, 808, 812, 840, 885, 835, 835, 836,
878, 840, 868, 890, 806, 840.
Solution.

Question 4.
Draw a histogram for the frequency table made for the data in Question 3, and answer the following questions:
(i) Which group has the maximum number of workers?
(ii) How many workers earnt 850 and more?
(iii) How many workers earn less than? 850?
Solution.

(i) Group 830—840, has the maximum number of workers.
(ii) 1 + 3 + 1 + 1 + 4 = 10, workers earn ? 850 and more.
(iii) 3 + 2 + l + 9 + 5 = 20, workers earn less than ? 850.

Question 5.
The number of hours for which students of a particular class watched television during holidays is shown through the given graph:

Answer the following:
(i) For how many hours did the maximum number of students watch TV?
(ii) How many students watched TV for less than 4 hours?
(iii) How many students spent more than 5 hours watching TV?
Solution.
(i) The maximum number of students watch TV for 4 to 5 hours.
(ii) 4 + 8 + 22 = 34 students watch TV for less than 4 hours.
(iii) 8 + 6 = 14 students spend more than 5 hours in watching TV.

We hope the NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.1 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.1 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.1.

• Practical Geometry Class 8 Ex 4.2
• Practical Geometry Class 8 Ex 4.3
• Practical Geometry Class 8 Ex 4.4
• Practical Geometry Class 8 Ex 4.5

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.1

Question 1.
Construct the following quadrilaterals:
(i) Quadrilateral ABCD
AB = 4.5 cm
BC = 5.5 cm
CD = 4 cm
AD = 6 cm
AC = 7 cm

(ii) Quadrilateral JUMP
JU = 3.5 cm
UM = 4 cm
MP = 5 cm
PJ = 4.5 cm
PU 6.5 cm

(iii) Parallelogram MORE
OR = 6 cm
RE = 4.5 cm
EO = 7.5 cm

(iv) Rhombus BEST
BE = 4.5 cm
ET = 6 cm
Solution.
(i) Steps of Construction

1. Draw AB 4.5 cm
2. With A as centre and radius AC = 7 cm, draw an arc.
   
3. With B as center and radius BC = 5.5 cm, draw another arc to intersect the arc drawn in step (2) at C.
4. With A as center and radius AD = 6 cm, draw an arc on the side of AC, opposite to that of B.
5. With C as center and radius CD = 4 cm, draw another arc to intersect the arc drawn in step (4) at D.
6. Join BC, CD, DA, and AC.

Then, ABCD is the required quadrilateral.

(ii) Steps of Construction

1. Draw JU = 3.5 cm
2. With J as center and radius JP = 4.5 cm, draw an arc.
3. With U as center and radius UP = 6.5 cm, draw another arc to intersect the arc drawn in step 2 at P.
   
4. With U as center and radius UM = 4 cm, draw an arc on the side of PU opposite to that of J.
5. With P as center and radius PM = 5 cm, draw another arc to intersect the arc drawn in step 4 at M.
6. Join UM, MP, PJ, and UP.

Then, JUMP is the required quadrilateral.

(iii) Steps of Construction
[We know that in a parallelogram, opposite sides are equal in length.
∴ MO = ER = 4.5 cm and ME – OR = 6 cm]

1. Draw MO = 4.5 cm
2. With M as center and radius ME = 6 cm, draw an arc.
   
3. With O as center and radius OE = 7.5 cm, draw an arc to intersect the arc drawn in step 2 at E.
4. With O as center and radius OR = 6 cm, draw an arc on the side of OE opposite to that of M.
5. With E as center and radius ER = 4.5 cm, draw another arc to intersect the arc drawn in step 4 at R.
6. Join OR, RE, EM, and EO.

Then, MORE is the required parallelogram.

(iv) Steps of Construction
[We know that in a rhombus, all the sides are equal in length.
∴ BE = ES = ST = TB = 4.5 cm]

1. Draw BE = 4.5 cm
2. With B as centre and radius BT = 4.5 cm, draw an arc.
   
3. With E as center and radius
   ET = 6 cm, draw another arc to intersect the arc drawn in step 2 at T.
4. With E as center and radius
   ES = 4.5 cm, draw an arc on the side of ET opposite to that of B.
5. With T as center and radius
   TS = 4.5 cm, draw another arc to
   intersect the arc drawn in step 4 at S.
6. Join ES, ST, TB, and TE.

Then, BEST is the required rhombus.

We hope the NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.1 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.1, drop a comment below and we will get back to you at the earliest.