CBSE Class 8

NCERT Solutions for Class 8 Maths Chapter 13 Direct and Indirect Proportions Ex 13.1 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 13 Direct and Indirect Proportions Ex 13.1.

• Direct and Indirect Proportions Class 8 Ex 13.2

NCERT Solutions for Class 8 Maths Chapter 13 Direct and Indirect Proportions Ex 13.1

Question 1.
Following are the car parking charges near a railway station up to
4 hours ₹ 60
8 hours ₹ 100
12 hours ₹ 140
24 hours ₹ 180
Check if the parking charges are in direct proportion to the parking time.
Solution.
We have
\(\frac { 60 }{ 4 } =\frac { 15 }{ 1 } \)
\(\frac { 100 }{ 8 } =\frac { 25 }{ 2 } \)
\(\frac { 140 }{ 12 } =\frac { 35 }{ 3 } \)
\(\frac { 180 }{ 24 } =\frac { 15 }{ 2 } \)
Since all the values are not the same, therefore, the parking charges are not in direct proportion to the parking time.

Question 2.
A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of the base. In the following table, find the parts othe f base that need to be added.

Solution.
Sol. Let the number of parts of red pigment is x and the number of parts of the base is y.
As the number of parts of red pigment increases, a number of parts of the base also increases in the same ratio. So it is a case of direct proportion.
We make use of the relation of the type

Question 3.
In Question 2 above, if 1 part of a red pigment requires 75 mL of the base, how much red pigment should we mix with 1800 mL of base?
Solution.
Let the number of parts of red pigment is x and the amount of base be y mL.
As the number of parts of red pigment increases, the amount of base also increases in the same ratio. So it is a case of direct proportion. We make use of the relation of the type.

Question 4.
A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?
Solution.
Let the machine fill x bottles in five hours. We put the given information in the form of a table as shown below :
Number of bottles filled 840 x2
Number of hours 6 5
More the number of hours, more the number of bottles would be filled. So, the number of bottles filled and the number of hours are directly proportional to each other.
So, \(\frac { { x }_{ 1 } }{ { x }_{ 2 } } =\frac { { y }_{ 1 } }{ { y }_{ 2 } } \)
⇒ \(\frac { 840 }{ { x }_{ 2 } } =\frac { 6 }{ 5 } \)
⇒ \(6{ x }_{ 2 }=840\times 5\)
⇒ \({ x }_{ 2 }=\frac { 840\times 5 }{ 6 } \)
⇒ \({ x }_{ 2 }=700\)
Hence, 700 bottles will be filled.

Question 5.
A photograph of a bacteria enlarged 50,000 times attains a length of 5 cm as shown in the diagram. What is the actual length of the bacteria ? If the photograph is enlarged 20,000 times only, what would be its enlarged length?

Solution.
Actual length of the bacteria
= \(\frac { 5 }{ 50000 } \)cm
= \(\frac { 1 }{ 10000 } \) = \({ 10 }^{ -4 }\)cm
10000
Let the enlarged length be y2 cm. We put the given information in the form of a table as shown below:
Number of times Length attained
photograph enlarged (in cm)
50.000 5
20.000 y₂
More the number of times a photograph of a bacteria is enlarged, more the length attained. So, the number of times a photograph of a bacteria is enlarged and the length attained are directly proportional to each other.
So,\(\frac { { x }_{ 2 } }{ { y }_{ 2 } } =\frac { { x }_{ 2 } }{ { y }_{ 2 } } \)
⇒ \(\frac { 50000 }{ 5 } =\frac { 20000 }{ { y }_{ 2 } } \)
⇒ \(50000{ y }_{ 2 }=5\times 20000\)
⇒ \({ y }_{ 2 }=\frac { 5\times 20000 }{ 50000 } \)
⇒ \({ y }_{ 2 }=2\)
Hence, its enlarged length would be
2 cm.

Question 6.
In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m high. If the length of the ship is 28 m, how long is the model ship ?

Solution.

Question 7.
Suppose 2 kg of sugar contains 9 x 106 crystals. How many sugar crystals are there in
(i) 5 kg of sugar?
(ii) 1.2 kg of sugar?
Solution.
Suppose the amount of sugar is x kg and the number of crystals is y.
We put the given information in the form of a table as shown below:

As the amount of sugar increases, the number of crystals also increases in the same ratio. So it is a case of direct proportion. We make use of the relation of the type \(\frac { { x }_{ 1 } }{ { y }_{ 1 } } =\frac { { x }_{ 2 } }{ { y }_{ 2 } } \)

Question 8.
Rashmi has a roadmap with a scale of 1 cm representing 18 km. She drives on a T’oad for 72 km. What would be her distance covered in the map?
Solution.
Let the distance covered in the map be x cm. Then,
1 : 18 = x : 72
⇒ \(\frac { 1 }{ 18 } =\frac { x }{ 72 }\)
⇒ \(x=\frac { 72 }{ 18 } \)
⇒ x = 4
Hence, the distance covered in the map would be 4 cm.

Question 9.
A5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time
(i) the length of the shadow cast by another pole 10 m 50 cm high
(ii) the height of a pole which casts a shadow 5 m long.
Solution.
Let the height of the vertical pole be x m and the length of the shadow be y m.
We put the given information in the form of a table as shown below:

Question 10.
A loaded truck travels 14 km. in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?
Solution.
Two quantities x and y which vary in direct proportion have the relation

We hope the NCERT Solutions for Class 8 Maths Chapter 13 Direct and Indirect Proportions Ex 13.1 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 13 Direct and Indirect Proportions Ex 13.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers Ex 12.1 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers Ex 12.1.

• Exponents and Powers Class 8 Ex 12.2

NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers Ex 12.1

Question 1.
Evaluate :
(i) \({ 3 }^{ -2 }\)
(ii) \({ -4 }^{ -2 }\)
(iii) \(({ \frac { 1 }{ 2 } ) }^{ -5 }\)
Solution.

Question 2.
Simplify and express the result in power notation with positive exponent.

Solution.

Question 3.
Fmd the value of:

Solution.

Question 4.
Evaluate
(i) \(\frac { { 8 }^{ -1 }\times { 5 }^{ 3 } }{ { 2 }^{ -4 } } \)
(ii) \(({ 5 }^{ -1 }\times { 2 }^{ -1 })\times { 6 }^{ -1 }\)
Solution.

Question 5.
Find the value of m for which \({ 5 }^{ m }+{ 5 }^{ -3 }={ 5 }^{ 5 }\)
Solution.

Question 6.
Evaluate :

Solution.

Question 7.
Simplify:
(i) \(\frac { 25\times { t }^{ -4 } }{ { 5 }^{ -3 }\times 10\times { t }^{ -8 } } \) (t ≠ 0)
(ii) \(\frac { { 3 }^{ -5 }\times { 10 }^{ -5 }\times 125 }{ { 5 }^{ -7 }\times { 6 }^{ -5 } } \)
Solution.

We hope the NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers Ex 12.1 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers Ex 12.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1.

• Mensuration Class 8 Ex 11.2
• Mensuration Class 8 Ex 11.3
• Mensuration Class 8 Ex 11.4

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1

Question 1.
A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?


Solution.
Area of the square field = a x a
= 60 m x 60 m = 3,600 \({ m }^{ 2 }\)
Perimeter of the square field = 4a
= 4 x 60 m = 240 m
∴ Perimeter of rectangular field = 240 m
⇒ 2(l + b) = 240
⇒ 2(80 + b) = 240
where b m is the breadth of the rectangular field
⇒ 80 + b = \(\frac { 240 }{ 2 } \) ⇒ 80 + bx = 120
⇒ b = 120 – 80 = 40
∴ Breadth = 40 m
∴ Area of rectangular field
= l x b = 80 m x 40 m = 3,200 \({ m }^{ 2 }\)
So, the square field (a) has a larger area

Question 2.
Mrs. Kaushik has a square plot ‘ with the measurement as shown in the figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a: garden around the house at the rate of ₹ 55 per \({ m }^{ 2 }\).
Solution.
Area of the square plot = a x a
= 25 x 25 \({ m }^{ 2 }\) = 625 \({ m }^{ 2 }\)
Area of the house = a x b
= 20 x 15 \({ m }^{ 2 }\) = 300 \({ m }^{ 2 }\)
∴ Area of the garden
= Area of the square plot – Area of the house
= 625 \({ m }^{ 2 }\) – 300 \({ m }^{ 2 }\)
= 325 \({ m }^{ 2 }\)
∵ The cost of developing the garden per square metre = ₹ 55.
∴ Total cost of developing the garden
= ₹ 325 x 55
= ₹ 17,875.

Question 3.
The shape of a garden is rectangular in the middle and semicircular at the ends as shown in the diagram. Find the area and the perimeter of this garden.

Solution.

Question 4.
A flooring tile has the shape of a parallelogram whose base is 24 cm and the cor-responding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m2? (If required you can split the tiles in whatever way you want to fill up the corners).
Solution.
Area of a flooring tile = bh
= 24 x 10 \({ cm }^{ 2 }\)
= 240 \({ cm }^{ 2 }\)
Area of the floor
= 1080 \({ m }^{ 2 }\)
= 1080 x 100 x 100 \({ cm }^{ 2 }\)
∵ m2 = 100 x 100 \({ cm }^{ 2 }\)

∴ Number of tiles required to cover the floor
=\(\frac { Area\quad of\quad the\quad floor\quad }{ Area\quad of\quad a\quad flooring\quad tile } \)
= \(\frac { 1080\times 100\times 100\quad }{ 240 } \)
= 45000.

Question 5.
An ant is moving around a few food pieces of different shapes scattered on the floor. For which food-piece would the ant have to take a longer round? Remember, the circumference of a circle can be obtained by using the expression c = 2πr, where r is the radius of the circle.


Solution.

We hope the NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1, drop a comment below and we will get back to you at the earliest.

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.