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

NCERT Solutions Class 8 Mathematics 
Chapter – 13 (Direct and Inverse Proportions) 

The NCERT Solutions in English Language for Class 8 Mathematics Chapter – 13 Direct and Inverse Proportions Exercise 13.1 has been provided here to help the students in solving the questions from this exercise. 

Chapter 13: Direct and Indirect Proportions

Exercise – 13.1

1. Following are the car parking charges near a railway station upto:
4 hours        Rs.60
8 hours        Rs.100
12 hours      Rs.140
24 hours     Rs.180
NCERT Maths Solutions Class 8
Check if the parking charges are in direct proportion to the parking time.

Solution – Let’s find the parking charge for 1 hour in all the four cases shown below:
60/4 = 15/1 = ₹15
100/8 = 25/2 = ₹12.5
140/12 = 35/3 = ₹11.67
180/24 = 15/2 = ₹7.50
We see that the 1 hour charge in all the four cases shown below is not equal.
Thus, the parking charges are not in direct proportion.

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

Solution – Let the ratio of parts of red pigment and parts of base be \frac{a}{b}.
Case 1: Here, a1 = 1, b1 = 8
\frac{a_1}{b_1} = \frac{1}{8} =  k (say)

Case 2: When  a2 = 4 , b2 = ?
k = \frac{a_2}{b_2}  

b2 = \frac{a_2}{k}  = \frac{4}{\frac{1}{8}} = 4 × 8 = 32

Case 3: When  a3 = 7 , b3 = ?
k = \frac{a_3}{b_3}  

b3 = \frac{a_3}{k} = \frac{7}{\frac{1}{8}} = 7 × 8 = 56

Case 4: When a4 = 12 , b4 =?
k = \frac{a_4}{b_4}

b4 = \frac{a_4}{k} = \frac{12}{\frac{1}{8}} = 12 × 8 = 96

Case 5: When  a5 = 20 , b5 = ?
k = \frac{a_5}{b_5}

b5 = \frac{a_5}{k} = \frac{20}{\frac{1}{8}} = 20 × 8 = 160
Combine results for all the cases, we have

Parts of red pigment 1 4 7 12 20
Parts of Base 8 32 56 96 160

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?

Solution –  Let the parts of red pigment mix with 1800 mL base be  x.

Parts of red pigment 1 x
Parts of Base 75 1800

Since it is in direct proportion.
\frac{1}{75} = \frac{x}{1800}

Ncert solutions class 8 chapter 13-10 75 × x = 1 × 1800
x = \frac{1\times 1800}{75} = 24
Hence with base 1800 mL, 24 parts red pigment should be mixed.

4. A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?
NCERT Maths Solutions Class 8
Solution – Let the number of bottles filled in five hours be x.
Here ratio of hours and bottles are in direct proportion.
\frac{6}{840}=\frac{5}{x}

6x = 5 × 840

x = \frac{5\times 840}{6} = 700
Hence machine will fill 700 bottles in five hours.

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 – Let enlarged length of bacteria be  x .
NCERT Maths Solutions Class 8
Here length and enlarged length of bacteria are in direct proportion.
\frac{5}{50000} =\frac{x}{20000}  

Ncert solutions class 8 chapter 13-17 x × 50000 = 5 × 20000

x = \frac{5\times 20000}{50000}
x = 2cm
Hence the enlarged length of bacteria is 2 cm.

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 if the ship is 28 m, how long is the model ship?
NCERT Maths Solutions Class 8
Solution – Let the length of model ship be  x .
NCERT Maths Solutions Class 8
Here length of mast and actual length of ship are in direct proportion.
\frac{12}{9} = \frac{28}{x}

Ncert solutions class 8 chapter 13-23 x × 12 = 28 × 9

x = \frac{28\times 9}{12}
x = 21 cm
Hence length of the model ship is 21 cm.

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?

Solution –
(i) Let sugar crystals be  x.
NCERT Maths Solutions Class 8
Here, weight of sugar and number of crystals are in direct proportion.
\frac{2}{9\times 10^6} = \frac{5}{x}

Ncert solutions class 8 chapter 13-28 x × 2 = 5 × 9 × 106Ncert solutions class 8 chapter 13-29

= x = \frac{2\times 9 \times 10^6}{2}  

= 22.5 × 106
= 2.25 × 107
Hence the number of sugar crystals is 2.25 × 107 

(ii)  Let sugar crystals be  x.
Here weight of sugar and number of crystals are in direct proportion.
NCERT Maths Solutions Class 8
\frac{2}{9\times 10^6} = \frac{1.2}{x} 

Ncert solutions class 8 chapter 13-35 x × 2 = 1.2 × 9 × 106

x = \frac{1.2\times 9\times 10^6}{2}

= 0.6 × 9 × 106

= 5.4 × 106
Hence the number of sugar crystals is  5.4×106

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?

Solution – Let distance covered in the map be  x.

Actual Distance (Km) 18 72
Distance covered in map (cm) 1 x

Here actual distance and distance covered in the map are in direct proportion.

\frac{18}{1} = \frac{72}{x}

Ncert solutions class 8 chapter 13-41 x × 18 = 72 × 1

x = \frac{72}{18} = 4 cm
Hence distance covered in the map is 4 cm.

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 5 m long.

Solution – Here height of the pole and length of the shadow are in direct proportion.
1 m = 100 cm
5 m 60 cm = 5×100+60 = 560 cm
3 m 20 cm = 3×100+20 = 320 cm
10 m 50 cm = 10×100+50 = 1050 cm
5 m = 5×100 = 500 cm

(i) Let the length of the shadow of another pole be  x.

Height of pole (cm) 560 1050
Length of shadow (cm) 320  x

\frac{560}{320} = \frac{1050}{x}

Ncert solutions class 8 chapter 13-46 x × 560 = 1050 × 320

x = \frac{1050\times 320}{560}
x= 600 cm = 6m
Hence length of the shadow of another pole is 6 m.

(ii) Let the height of the pole be  x.

Height of pole (cm) 560 x
Length of shadow (cm) 320 500

\frac{560}{320} = \frac{x}{500}  

Ncert solutions class 8 chapter 13-51 x × 320 = 560 × 500

x =\frac{560\times 500}{320}

= 875 cm = 8 m 75 cm
Hence height of the pole is 8 m 75 cm.

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 – Let distance covered in 5 hours be x km.
1 hour = 60 minutes
Therefore, 5 hours = 5×60 = 300 minutes
NCERT Maths Solutions Class 8
Here distance covered and time in direct proportion.
\frac{14}{25}=\frac{x}{300}

Ncert solutions class 8 chapter 13-56 25 × x = 300 × 14

x = \frac{14\times 300}{25}
x = 168
Therefore, a truck can travel 168 km in 5 hours.

 

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