AP State Syllabus AP Board 7th Class Maths Solutions Chapter 6 Ratio – Applications Ex 2 Textbook Questions and Answers.

## AP State Syllabus 7th Class Maths Solutions 6th Lesson Ratio – Applications Exercise 2

Question 1.

Find the missing in the following proportion in the table given below.

Solution:

Question 2.

Write true or false.

(i) 15 : 30 : : 30 : 40

(ii) 22 : 11 : : 12 : 6

(iii) 90 : 30 : : 36 : 12

(iv) 32 : 64 : : 6 : 12

(v) 25 : 1 : : 40 : 160

Solution:

i) Product of extremes = 15 x 40 600

Prod uct of means 30 x 30 = 900

600 ≠ 900

ii) 22 11:; 12:6 -True

Product oi extremes 22 x 6 = 132

Product of means = 11 x 12 = 132

iii) 90 : 30 :: 36 : 12 -True

Product oi extremes = 90 x 12 = 1080

(5 x 2 = 10 M)

Product of means = 30 x 36 = 1080

1080 = 1080

iv) 32 : 64 :: 6 : 12 -True

Product of extremes = 32 x 12 = 384

Product of means = 64 x 6 = 384

v) 25 : 1 :: 40 : 1.60 – True

Product of extremes = 25 x 1.60 = 40.00

Product of means = 1 x 40 = 40

Question 3.

Madhu buys 5 kg of potatoes at the market. If the cost of 2 kg is ₹ 36, how much will Madhu pay?

Solution:

Cost of 2 kg potatoes = ₹.36

∴ Cost of 1 kg potatoes = \(\frac{36}{2}\) = ₹.18

Cost of 5kg potatoes @ ₹.18 = 5 × ₹.18 = ₹.90

∴ Madhu pays ₹.90

(OR)

Let Madhu pays ₹.x

2 : 36 :: 5 : x

x = \(\frac{36 \times 5}{2}\) = ₹ 90

Question 4.

Physics tells us that weights of an object on the moon is proportional to its weight on Earth.

Suppose a 90 kg man weighs 15 kg on the moon what will a 60 kg woman weigh on the moon?

Solution:

Man’s weight on earth = 90 kg

Man’s weight on moon = 15 kg

Woman’s weight on earth = 60 kg

Let the womans weight on moon = x kg

then 90 : 15 :: 60 : x

Product of means Product of extremes

90x = 15 × 60

x = \(\frac{15 \times 60}{90}\) = 10

∴ Womans weight on the moon = 10 kg

Question 5.

A disaster relief team consists of engineers and doctors in the ratio of 2 : 5.

(i) If there are 18 engineers, find the number of doc tors.

(ii) If there are 65 doctors, find the number of engineers.

Solution:

i) Ratio of engineers and doctors= 2 : 5

The number of engineers = 18

Let the number of doctors = x

Now 2 : 5 :: 18 : x .

By the rule of proportion

2x = 5 × 18

= \(\frac{5 \times 18}{2}\) = 45

∴ The number of doctors 45

ii) Numbers of doctors = 65

Let the number of engineers = x

then 2 : 5 :: x : 65

By the rule of proportion

5x = 2 × 65

x = \(\frac{2 \times 65}{2}\) = 26

∴ The number of engineers = 26

Question 6.

The ratio of two angles is 3: 1. Find the

(i) larger angle if the smaller is 180°

(ii) smaller angle if the larger is 63°.

Solution:

i) Given that the ratio of angles = 3 : 1

Smaller angle is given as = 180°

Let the larger angle be x°

then 3: 1 :: x : 180

∴ By rule of proportion

1 . x = 180° × 3

x = 540°

∴ The required larger angle = 540°

ii) The larger angle is given as = 63°

Let the smaller angle be y°

then 3 : 1 :: 63°: y

∴ By rule of proportion

3 × y = 63 × 1

y = \(\frac{63}{3}\) = 21°

∴ The required smaller angle = 21°