# NCERT Solutions Class 8 Maths Chapter 13 Exercise 13.1

Chapter 13 of Mathematics Class 8 is Direct and Inverse Proportions. Two quantities are said to be in Direct Proportion if, on increasing the value of the first quantity, the value of the second quantity increases such that the ratio of the first quantity to the second quantity remains constant. However, in inverse proportion two quantities may change in such a way that if one quantity increases, the other quantity decreases, and vice versa. Mathematics is the science of quantity, space, and number, either as abstract concepts or as applied to other disciplines such as Physics and Engineering. If students study Mathematics carefully, they can score maximum marks in the subject. The curriculum of Class 8 Mathematics introduces students to a wide range of concepts that can be challenging for them. Therefore, Extramarks provides students with  Class 8 Maths Chapter 13 Exercise 13.1, so that they can properly understand the concepts of the chapter and resolve their doubts. In order to prepare for any examinations, NCERT solutions are a good resource. Once students thoroughly go through these solutions, they are able to solve any complicated problem they may face during the examinations. Along with Class 8 Maths Exercise 13.1, Extramarks also provides students with NCERT Solutions for all the classes and many other credible study materials.

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## NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions (EX 13.1) Exercise 13.1

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### Access NCERT Solutions for Class 8 Chapter 13 – Direct and Inverse Proportions

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### NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions (Ex 13.1) Exercise 13.1

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Q.1 Following are the car parking charges near a railway station upto

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.

Ans

The given information is represented in a table.

 Number of hours 4 8 12 24 Parking charges (in ₹ ) 60 100 140 180

The ratios of the parking charges to the number of hours are as follows:

$\frac{60}{4}=15,\frac{100}{8}=\frac{25}{2},\frac{140}{12}=\frac{35}{3},\frac{180}{24}=\frac{15}{2}$

Since all the ratios are different from each other, the parking charges are not in a direct proportion to the parking time.

Q.2

$\begin{array}{l}\begin{array}{l}\text{A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base.}\\ \text{In the following table, find the parts of base that need to be added.}\end{array}\\ \begin{array}{cccccc}\text{Parts of red pigment}& \text{1}& \text{4}& \text{7}& \text{12}& \text{20}\\ \text{Parts of base}& \text{8}& \dots & \dots & \dots & \dots \end{array}\end{array}$

Ans

The given mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. Since the ratio of red pigment and the base is same every time, therefore, the parts of red pigments and the parts of base are in direct proportion.

Let us take the unknown values as x1, x2, x3 and x4.
The given information in the form of a table is as follows.

$\begin{array}{cccccc}\text{Parts of red pigment}& 1& 4& 7& 12& 20\\ \text{Parts of base}& 8& {x}_{1}& {x}_{2}& {x}_{3}& {x}_{4}\end{array}$ $\begin{array}{l}\text{According to direct proportion},\\ \frac{1}{8}=\frac{4}{{\mathrm{x}}_{1}}⇒{\mathrm{x}}_{1}=8×4⇒{\mathrm{x}}_{1}=32\\ \frac{1}{8}=\frac{7}{{\mathrm{x}}_{2}}⇒{\mathrm{x}}_{2}=8×7⇒{\mathrm{x}}_{2}=56\\ \frac{1}{8}=\frac{12}{{\mathrm{x}}_{3}}⇒{\mathrm{x}}_{3}=8×12⇒{\mathrm{x}}_{3}=96\\ \frac{1}{8}=\frac{20}{{\mathrm{x}}_{4}}⇒{\mathrm{x}}_{4}=8×20⇒{\mathrm{x}}_{4}=160\end{array}$

Therefore, the parts of a base to be added are shown as follows:

$\begin{array}{cccccc}\text{Parts of red pigment}& 1& 4& 7& 12& 20\\ \text{Parts of base}& 8& 32& 56& 96& 160\end{array}$

Q.3 In Question 2 above, if 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base?

Ans

Let the parts of red pigment required to mix with 1800 mL of base be x.

The given information in the form of a table is as follows.

$\begin{array}{ccc}\text{Parts of red pigment}& 1& \mathrm{x}\\ \text{Parts of base(in mL)}& 75& 1800\end{array}$

The parts of red pigment and the parts of base are in direct proportion.
Therefore, we obtain

$\begin{array}{l}\text{According to direct proportion},\\ \frac{1}{75}=\frac{\mathrm{x}}{1800}⇒75\mathrm{x}=1800⇒\mathrm{x}=24\end{array}$

Thus, 24 parts of red pigments should be mixed with 1800 mL of base.

Q.4 A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?

Ans

Let the number of bottles filled by the machine in five hours be x.

The given information in the form of a table is as follows.

$\begin{array}{ccc}\text{Number of bottles}& 840& \mathrm{x}\\ \text{Time taken(in hours)}& 6& 5\end{array}$

The number of bottles and the time taken to fill these bottles are in direct proportion. Therefore, we obtain

$\begin{array}{l}\frac{840}{6}=\frac{\mathrm{x}}{5}⇒6\mathrm{x}=840×5\\ ⇒6\mathrm{x}=4200\\ ⇒\mathrm{x}=700\end{array}$ 

Thus, 700 bottles will be filled in 5 hours.

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

Ans

Let the actual length of the bacterium be x cm.
The given information in the form of a table is as follows.

$\begin{array}{ccc}\text{Length of the bacterium (in cm)}& 5& x\\ \begin{array}{l}\text{Number of times photograph of}\\ \text{the bacterium was enlarged}\end{array}& 50000& 1\end{array}$

The number of times the photograph of the bacterium was enlarged and the lengths of the bacterium are in direct proportion.
Therefore, we obtain

$\begin{array}{l}\frac{5}{50000}=\frac{x}{1}\\ ⇒50000x=5\\ ⇒x=\frac{1}{10000}\\ ⇒x={10}^{-4}\\ \\ \text{Hence},\text{the actual length of the bacterium is 1}{0}^{-\text{4}}\text{cm}.\end{array}$

Now, we have to find its enlarged length if the photograph is enlarged 20,000 times only.

Let the enlarged length of the bacterium be y cm.
The given information in the form of a table is as follows.

$\begin{array}{ccc}\text{Length of the bacterium (in cm)}& 5& \mathrm{y}\\ \begin{array}{l}\text{Number of times photograph}\\ \text{of the bacterium was enlarged}\end{array}& 50000& 20000\end{array}$

The number of times the photograph of the bacterium was enlarged and the lengths of bacterium are in direct proportion.

Therefore, we obtain

$\begin{array}{l}\frac{5}{50000}=\frac{\mathrm{y}}{20000}\\ ⇒100000=50000\mathrm{y}\\ ⇒\mathrm{y}=\frac{100000}{50000}\\ ⇒\mathrm{y}=2\\ \text{Hence},\text{the enlarged length of the bacterium is 2 cm}.\end{array}$

Q.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?

Ans

Let the length of the mast of the model ship be x cm.
The given information in the form of a table is as follows:

$\begin{array}{ccc}& \text{Height of mast}& \text{Length of ship}\\ \text{Model of ship}& 9& \mathrm{x}\\ \text{Actual ship}& 12& 28\end{array}$

The dimensions of the actual ship and the model ship are directly proportional to each other.

Therefore, we obtain:

$\begin{array}{l}\frac{9}{12}=\frac{\mathrm{x}}{28}\\ ⇒9×28=12\mathrm{x}\\ ⇒\mathrm{x}=\frac{252}{12}\\ ⇒\mathrm{x}=21\\ \end{array}$

Thus, the length of the model ship is 21 cm.

Q.7 Suppose 2 kg of sugar contains 9 × 106 crystals. How many sugar crystals are there in
(i) 5 kg of sugar? (ii) 1.2 kg of sugar?

Ans

(i) Let the number of sugar crystals in 5 kg of sugar be x.
The given information in the form of a table is as follows:

$\begin{array}{ccc}\text{Amount of sugar(in kg)}& 2& 5\\ \text{Number of crystals}& 9×{10}^{6}& \mathrm{x}\end{array}$

The amount of sugar and the number of crystals it contains are directly proportional to each other.
Therefore, we obtain

$\begin{array}{l}\frac{2}{9×{10}^{6}}=\frac{5}{\mathrm{x}}\\ ⇒2\mathrm{x}=5×9×{10}^{6}\\ ⇒\mathrm{x}=\frac{5×9×{10}^{6}}{2}\\ ⇒\mathrm{x}=22.5×{10}^{6}\\ ⇒\mathrm{x}=2.25×{10}^{7}\\ \end{array}$

Hence, the number of sugar crystals is 2.25 × 107.

(ii) Let the number of sugar crystals in 1.2 kg of sugar be y.

The given information in the form of a table is as follows:

$\begin{array}{ccc}\text{Amount of sugar(in kg)}& 2& 1.2\\ \text{Number of crystals}& 9×{10}^{6}& \mathrm{y}\end{array}$

The amount of sugar and the number of crystals it contains are directly proportional to each other.

$\begin{array}{l}\frac{2}{9×{10}^{6}}=\frac{1.2}{\mathrm{y}}\\ ⇒2\mathrm{y}=1.2×9×{10}^{6}\\ ⇒\mathrm{x}=\frac{1.2×9×{10}^{6}}{2}\\ ⇒\mathrm{x}=5.4×{10}^{6}\\ \\ \text{Hence},\text{the number of sugar crystals is}5.4×{10}^{6}.\end{array}$

Therefore, we obtain

Q.8 Rashmi has a road map with a scale of 1 cm representing 18 km. She drives on a road for 72 km. What would be her distance covered in the map?

Ans

Let the distance represented on the map be x cm.
The given information in the form of a table is as follows.

$\begin{array}{ccc}\text{Distance covered on map(in cm)}& 1& \mathrm{x}\\ \text{Distance covered on road(in km)}& 18& 72\end{array}$

The distances covered on road and represented on map are directly proportional to each other.
Therefore, we obtain

$\begin{array}{l}\frac{1}{18}=\frac{\mathrm{x}}{72}\\ ⇒72=18\mathrm{x}\\ ⇒\mathrm{x}=\frac{72}{18}\\ ⇒\mathrm{x}=4\end{array}$

Hence, the distance represented on the map is 4 cm.

Q.9 A 5 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 5m long.

Ans

Let the length of the shadow of the other pole be x m.
The given information in the form of a table is as follows.

$\begin{array}{ccc}\text{Height of the pole(in m)}& 5.60& 10.50\\ \text{Length of the shadow(in m)}& 3.20& \mathrm{x}\end{array}$

The height of an object and length of its shadow are directly proportional to each other.
Therefore, we obtain

$\begin{array}{l}\frac{5.60}{3.20}=\frac{10.50}{\mathrm{x}}\\ ⇒5.60\mathrm{x}=10.50×3.20\\ ⇒\mathrm{x}=\frac{33.6}{5.60}\\ ⇒\mathrm{x}=6\end{array}$

Hence, the length of the shadow will be 6 m.

(ii) Let the height of the pole be y m.
The given information in the form of a table is as follows.

$\begin{array}{ccc}\text{Height of the pole(in m)}& 5.60& \mathrm{y}\\ \text{Length of the shadow(in m)}& 3.20& 5\end{array}$

The height of an object and length of its shadow are directly proportional to each other.
Therefore, we obtain

$\begin{array}{l}\frac{5.60}{3.20}=\frac{\mathrm{y}}{5}\\ ⇒5.60×5=3.20\mathrm{y}\\ ⇒\mathrm{y}=\frac{5.60×5}{3.20}\\ ⇒\mathrm{y}=8.75\end{array}$

Thus, the height of the pole is 8.75 m or 8 m 75 cm.

Q.10 A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?

Ans

Let the distance travelled by the truck in 5 hours be x km.
We know, 1 hour = 60 minutes
∴ 5 hours = (5 × 60) minutes = 300 minutes
The given information in the form of a table is as follows.

$\begin{array}{ccc}\text{Distance travelled(in km)}& 14& \mathrm{x}\\ \text{Time(in min)}& 25& 300\end{array}$

The distance travelled by the truck and the time taken by the truck are directly proportional to each other.

Therefore, we obtain

$\begin{array}{l}\frac{14}{25}=\frac{\mathrm{x}}{300}\\ ⇒300×14=25\mathrm{x}\\ ⇒\mathrm{x}=\frac{4200}{25}\\ ⇒\mathrm{x}=168\end{array}$

Hence, the distance travelled by the truck is 168 km.

## FAQs (Frequently Asked Questions)

### 1. Where can students find the NCERT Solutions For Class 8 Maths Chapter 13 Exercise 13.1?

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### 2. Are the concepts of the NCERT Solutions For Class 8 Maths Chapter 13 Exercise 13.1 challenging for students?

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### 3. Is it essential to practice all the NCERT Solutions For Class 8 Maths Chapter 13 Exercise 13.1?

Yes, it is essential for the students to go through all the NCERT Solutions For Class 8 Maths Chapter 13 Exercise 13.1. Each question covers a different concept of the chapter. As well as, the question paper for the Class 8 Mathematics examination is based on the NCERT curriculum. As a result, any question from the NCERT textbook can be a part of the examinations. The NCERT Solutions For Class 8 Maths Chapter 13 Exercise 13.1 also help students strengthen their fundamentals so that they can solve any complicated questions in the examination.