NCERT Solutions (Ex - 13.1) - Direct and Inverse Proportions

# NCERT Solutions (Ex - 13.1) - Direct and Inverse Proportions Notes - Class 8

## Document Description: NCERT Solutions (Ex - 13.1) - Direct and Inverse Proportions for Class 8 2022 is part of Class 8 preparation. The notes and questions for NCERT Solutions (Ex - 13.1) - Direct and Inverse Proportions have been prepared according to the Class 8 exam syllabus. Information about NCERT Solutions (Ex - 13.1) - Direct and Inverse Proportions covers topics like and NCERT Solutions (Ex - 13.1) - Direct and Inverse Proportions Example, for Class 8 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for NCERT Solutions (Ex - 13.1) - Direct and Inverse Proportions.

Introduction of NCERT Solutions (Ex - 13.1) - Direct and Inverse Proportions in English is available as part of our Class 8 preparation & NCERT Solutions (Ex - 13.1) - Direct and Inverse Proportions in Hindi for Class 8 courses. Download more important topics, notes, lectures and mock test series for Class 8 Exam by signing up for free. Class 8: NCERT Solutions (Ex - 13.1) - Direct and Inverse Proportions Notes - Class 8
 1 Crore+ students have signed up on EduRev. Have you?

Exercise 13.1

Question 1:

Following are the car parking charges near a railway station up to:

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

Answer 1:

Here, the charges per hour are not same, i.e., C1 ≠ C≠ C3 ≠C4

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 base. In the following table, find the parts of base that need to be added.

 Parts of red pigment 1 4 7 12 20 Parts of base 8 — — — —

Answer 2:

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

Question 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?

Answer 3:

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

 Parts of red pigment 1 X Parts of base 75 1800

Hence, with base 1800 mL, 24 parts red pigment should be mixed.

Question 4:

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

Answer 4:

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

 Hours 1 X Bottles 75 1800

Hence, the number of bottles filled in five hours be 700.

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?

Answer 5:

 Length 5 X Enlarged length 50,000 20,000

Here length and enlarged length of bacteria are in direct proportion

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 if the ship is 28 m, how long is the model ship?

Answer 6:

Let the length of model ship be x

 Length of actual ship (in m) 12 28 Length of model ship (in cm) 9 X

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

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

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?

Answer 7:

Let sugar crystals be x.

 Weight of sugar (in kg) 2 5 No. of crystals 9x106 X

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

(ii)

Let sugar crystals be x.

 Weight of sugar (in kg) 2 1.2 No. of crystals 9x106 X

Hence, the number of sugar crystals is 5.4 x 106.

Question 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?

Answer 8:

Let distance covered in the map be x.

 Actual distance (in km) 18 72 Distance covered in map (in cm) 1 X

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

Hence, distance covered in the map is 4 cm

Question 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.

Answer 9:

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

And 1 m = 100 cm

5 m 60 cm = 5 x 100 + 60 = 560 cm

3 m 20 cm = 3 x 100 + 20 = 320 cm

10 m 50 cm = 10 x 100 + 50 = 1050 cm

5 m = 5 x 100 - 500 cm

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

 Height of pole (in cm) 560 1050 Length of shadow (in cm) 320 X

Hence, the length of the shadow of another pole be 6m.

Let the hight of the pole be x.

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

Hence, the hight of the pole be 8m 75 cm

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?

Answer 10:

Let distance covered in 5 hours be .r km.

1 hour = 60 minutes

5 hours = 5 x 60 = 300 minutes

 Distance (in km] 14 X Time (in minutes) 25 300

Here distance covered and time in direct proportion.

The document NCERT Solutions (Ex - 13.1) - Direct and Inverse Proportions Notes - Class 8 is a part of Class 8 category.
All you need of Class 8 at this link: Class 8
 Use Code STAYHOME200 and get INR 200 additional OFF

### Download free EduRev App

Track your progress, build streaks, highlight & save important lessons and more!

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;