NCERT Solutions for Class 8 Maths Chapter 13, Direct and Inverse Proportions is a combination of questions and answers for all textbook problems, prepared by BYJUâ€™S subject experts. These NCERT Class 8 exercise questions make students familiar with the various concepts. CBSE Chapter 13 consist of standard questions which comprise of various topics carrying different marks, so students must be well aware of all concepts to score well. Download mathematics NCERT Solutions and sharp your skills. These NCERT solutions is an easy way to practice and learn the concepts covered in class 8 chapter 13.

### Download PDF of NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Exercise 13.1

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Exercise 13.2 Solutions : 11 Questions (Long answers)

### Access Answers of Maths NCERT class 8 Chapter 13 Direct and Inverse Proportions Exercise 13.1 Page number 208

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

**4 hoursÂ Â Rs.60**

**8 hoursÂ Â Rs.100**

**12 hoursÂ Â Rs.140**

**24 hoursÂ Â Rs.180**

**Check if the parking charges are in direct proportion to the parking time.**

**Solution:**

Charges per hour:

C1 = 60/4 = Rs. 15

Â C2 = 100/8 = Rs. 12.50

C3 = Â 140/12 = Rs. 11.67

C4 = 180/24 Â = Rs.7.50

Here, the charges per hour are not same, i.e., C1 â‰ C2 â‰ C3 â‰ 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.

Solution:

Â Let the ratio of parts of red pigment and parts of base be a/b .

Case 1: Here, a_{1 }= 1, b_{1 }= 8

a_{1}/b_{1} = 1/8 =Â k (say)

Case 2: WhenÂ a_{2 }= 4 , b_{2 }= ?

Â Â

b_{2} = a_{2}/k = 4/(1/8) = 4Ã—8 = 32

Case 3: WhenÂ a_{3} = 7 , b_{3 }= ?

Â Â

b_{3 }= a_{3}/k = 7/(1/8) = 7Ã—8 = 56

Case 4: When a_{4 }= 12 , b_{4 }=?Â

Â Â

b_{4 }= a_{4}/k = 12/(1/8) = 12Ã—8 = 96

Case 5: WhenÂ a_{5} = 20 , b_{5 }= ?

Â Â

b_{5} = a_{5}/k = 20/(1/8) = 20Ã—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 beÂ x.

Since it is in direct proportion.

Â

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?**

**Solution**:

Â Let the number of bottles filled in five hours beÂ x.

Here ratio of hours and bottles are in direct proportion.

Â

6x = 5Ã—840

x = 5Ã—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 theÂ *actual*Â 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 beÂ x .

Actual length of bacteria =Â 5/50000 = 1/10000 Â cm = 10^{-4} Â cm

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

Â

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?

**Solution**:

Let the length of model ship beÂ x .

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

Â

x = 21 cm

Hence length of the model ship is 21 cm.

7. Suppose 2 kg of sugar contains 9Ã—10^{6} crystals. How many sugar crystals are there in

(i) 5 kg of sugar? (ii) 1.2 kg of sugar?

**Solution:**

Â (i)Â Let sugar crystals beÂ x.

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

=Â

Hence the number of sugar crystals isÂ

(ii)Â Let sugar crystals beÂ x.

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

Â

=Â

Hence the number of sugar crystals isÂ 5.4Ã—10^{6}

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 beÂ x.

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

Â

Â

Â

Â

x = 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:**

Â Here height of the pole and length of the shadow are in direct proportion.

And 1 m = 100 cm

5 m 60 cm = 5Ã—100+60 = 560 cm

3 m 20 cm = 3Ã—100+20 = 320 cm

10 m 50 cm = 10Ã—100+50 = 1050 cm

5 m = 5Ã—100 = 500 cm

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

Â

x= 600 cm = 6m

Hence length of the shadow of another pole is 6 m.

(ii)Â Let the height of the pole beÂ x.

Â

= 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 = 5Ã—60 = 300 minutes

Here distance covered and time in direct proportion.

Â

25x = 300(14)

Â

x = 168

Therefore, a truck can travel 168 km in 5 hours.

NCERT Class 8 Maths Chapter 13 Direct and Inverse Proportions Exercise 13.1, based on direct proportion. Some of the topics that one can be asked are; To calculate the proportions between one and the other quantity, The proportions between the different parts of a base or comparing the parking charges with the hours and many more.