NCERT Solutions Class 8th Maths Chapter 11 Direct and Indirect proportions Exercise 11.1

May 16, 2022

NCERT Solutions Class 8th Maths Chapter 11 Direct and Indirect proportions

Textbook	NCERT
Class	8^th
Subject	Mathematics
Chapter	11^th
Chapter Name	Direct and Indirect proportions
Category	Class 8th Maths Solutions
Medium	English
Source	Last Doubt

NCERT Solutions Class 8th Maths Chapter 11 Direct and Indirect proportions

Chapter – 11

Direct and Indirect proportions

Exercise 11.1

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

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Check if the parking charges are in direct proportion to the parking time.
Solution:
Charges per hour:
C1 = 60/4 = Rs. 15
C2 = 100/8 = Rs. 12.50
C3 = 140/12 = Rs. 11.67
C4 = 180/24 = Rs.7.50
Here, the charges per hour are not same, i.e., C1 ≠ C2 ≠ C3 ≠ C4
Therefore, the parking charges are not in direct proportion to the parking time.

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.

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Solution:

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Let the ratio of parts of red pigment and parts of base be a/b .
Case 1: Here, a1 = 1, b1 = 8
a1/b1 = 1/8 = k (say)
Case 2: When a2 = 4 , b2 = ?
b2 = a2/k = 4/(1/8) = 4×8 = 32
Case 3: When a3 = 7 , b3 = ?
b3 = a3/k = 7/(1/8) = 7×8 = 56
Case 4: When a4 = 12 , b4 =?
b4 = a4/k = 12/(1/8) = 12×8 = 96
Case 5: When a5 = 20 , b5 = ?
b5 = a5/k = 20/(1/8) = 20×8 = 160
Combine results for all the cases, we have

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.

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Since it is in direct proportion.

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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?

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Solution:
Let the number of bottles filled in five hours be x.
Here ratio of hours and bottles are in direct proportion.

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6x = 5×840
x = 5×840/6 = 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 .
Actual length of bacteria = 5/50000 = 1/10000 cm = 10-4 cm

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Here length and enlarged length of bacteria are in direct proportion.

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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?

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Solution:
Let the length of model ship be x .

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Here length of mast and actual length of ship are in direct proportion.

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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.

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Here, weight of sugar and number of crystals are in direct proportion.

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Hence the number of sugar crystals is

(ii) Let sugar crystals be x.

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Here weight of sugar and number of crystals are in direct proportion.

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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.

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

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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.
And 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.

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