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NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.1 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.1.

• Visualising Solid Shapes Class 8 Ex 10.2
• Visualising Solid Shapes Class 8 Ex 10.3

NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.1

Question 1.
For each of the given solid, the two views are given. Match for each solid the corresponding top and front views. The first one is done for you.

Solution.

Question 2.
For each of the given solid, the three views are given. Identify for each solid the corresponding top, front and side views.

Solution.

Question 3.
For each given solid, identify the top view, front view, and side view.

Solution.

Question 4.
Draw the front view, side view and top view of the given objects,

Solution.

We hope the NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.1 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.1 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.1.

• Algebraic Expressions and Identities Class 8 Ex 9.2
• Algebraic Expressions and Identities Class 8 Ex 9.3
• Algebraic Expressions and Identities Class 8 Ex 9.4
• Algebraic Expressions and Identities Class 8 Ex 9.5

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.1

Question 1.
Identify the terms, their coefficients for each of the following expressions:
(i) \({ 5xyz }^{ 2 }-3zy\)
(ii) \(1+x+{ x }^{ 2 }\)
(iii) \(4{ x }^{ 2 }{ y }^{ 2 }-4{ x }^{ 2 }{ y }^{ 2 }{ z }^{ 2 }+{ z }^{ 2 }\)
(iv) 3 – pq + qr – rp
(v) \(\frac { x }{ 2 } +\frac { y }{ 2 } -xy\)
(vi) 0.3a – 0.6ab + 0.5b.
Solution.
(i) \({ 5xyz }^{ 2 }-3zy\)

(ii) \(1+x+{ x }^{ 2 }\)

(iii) \(4{ x }^{ 2 }{ y }^{ 2 }-4{ x }^{ 2 }{ y }^{ 2 }{ z }^{ 2 }+{ z }^{ 2 }\)

(iv) 3 – pq + qr – rp

(v) \(\frac { x }{ 2 } +\frac { y }{ 2 } -xy\)

(vi)0.3a – 0.6ab + 0.5b.

Question 2.
Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

Solution.

Question 3.
Add the following.
(i) ab – be, be – ca, ca – ab
(ii) a -b + ab, b – c + be, c – a + ac
(iii) \(2{ p }^{ 2 }{ q }^{ 2 }-3pq+4,\quad 5+7pq-3{ p }^{ 2 }{ q }^{ 2 }\)
(iv) \({ l }^{ 2 }+{ m }^{ 2 },\quad { m }^{ 2 }+{ n }^{ 2 },\quad { n }^{ 2 }+{ l }^{ 2 }\), 2lm + 2mn + 2nl.
Solution.
(i) ab – be, be – ca, ca – ab

(ii) a -b + ab, b – c + be, c – a + ac

(iii) \(2{ p }^{ 2 }{ q }^{ 2 }-3pq+4,\quad 5+7pq-3{ p }^{ 2 }{ q }^{ 2 }\)

(iv) \({ l }^{ 2 }+{ m }^{ 2 },\quad { m }^{ 2 }+{ n }^{ 2 },\quad { n }^{ 2 }+{ l }^{ 2 }\), 2lm + 2mn + 2nl.

Question 4.
(a) Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 56 – 3
(b) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz
(c) Subtract \(4{ p }^{ 2 }q-3pq+5p{ q }^{ 2 }-8p+7q-10\) from \(18-3p-11q+5pq-2p{ q }^{ 2 }+5{ p }^{ 2 }q\)
Solution.

We hope the NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.1 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.1, drop a comment below and we will get back to you at the earliest.

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 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.