AP State Syllabus AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions Ex 10.3 Textbook Questions and Answers.

## AP State Syllabus 8th Class Maths Solutions 10th Lesson Direct and Inverse Proportions Exercise 10.3

Question 1.
Siri has enough money to buy 5 kg of potatoes at the price of ₹ 8 per kg. How much can she buy for the same amount if the price is increased to ₹ 10 per kg?
Solution:
Number of kgs of potatoes to their price are in inverse proportion.
∴ x1y1 = x2 y2
⇒ 8 × 5 = 10 × x
⇒ x = $$\frac{8 \times 5}{10}$$ = 4 kgs
∴ 4 kgs of potatoes will be purchased at the rate of ₹ 10 per kg.

Question 2.
A camp has food stock for 500 people for 70 days. ¡f200 more people join the camp, how long will the stock last?
Solution:
Number of persons and their food stock are in inverse proportion.
⇒ x1y1 = x2 y2 (Let y2 = x say)
⇒ 500 × 70 = (500 + 200) × x
⇒  x = $$\frac{500 \times 70}{700}$$ = 5 × 10
∴ x = 50
∴ The food will be stock for (200 + 500) 700 men = 50 days

Question 3.
36 men can do a piece of work in 12 days. ¡n how many days 9 men can do the same work?
Solution:
Number of workers and number of days are in inverse proportion
∴ x1y1 = x2 y2 let y2 = x (say)
= 36 × 12 = 9 × x
x = $$\frac{36 \times 12}{9}$$ = 48
∴ x = 48 days

Question 4.
A cyclist covers a distance of28 km in 2 hours. Find the time taken by him to cover a distance of 56 km with the same speed.
Solution:
Time and distance are in direct proportion.
∴ $$\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}$$ , Let y2 = x (say)
⇒ $$\frac{28}{2}$$ = $$\frac{56}{x}$$
⇒ x = $$\frac{56}{14}$$
∴ x = 4 hours

Question 5.
A ship can cover a certain distance in 10 hours at a speed of 16 nautical miles per hour. By how much should its speed be increased so that it takes only 8 hours to cover the same distance? (A nautical mile in a unit of measurement used at sea distance or sea water i.e. 1852 metres).
Solution:
Speed and distance are in inverse proportion.
⇒ x1y1 = x2 y2 , Let x2 = x (say)
⇒ 16 × 10 = x × 8
⇒ x = $$\frac{16 \times 10}{8}$$= 20
∴ x = 20
∴ The speed to be increased
= 20 – 16 = 4 nautical miles

Question 6.
5 pumps are required to fill a tank in 1$$\frac { 1 }{ 2 }$$ hours. How many pumps of the same type are used to fill the tank in half an hour.
Solution:
Number of pumps and time to fill the tanks are in inverse proportion.
⇒ x1y1 = x2 y2
⇒ 5 × 1$$\frac { 1 }{ 2 }$$ = x x 1$$\frac { 1 }{ 2 }$$
⇒ 5 × $$\frac { 3 }{ 2 }$$ = x x $$\frac { 1 }{ 2 }$$
⇒ x = 5 × 3 = 15
∴ Number of pumps required = 15

Question 7.
If 15 workers can build a wall in 48 hours, how many workers will be required to do the same work in 30 hours?
Solution:
Number of workers and time are in inverse proportion.
⇒ x1y1 = x2 y2
⇒ 15 × 48 = x × 30
⇒ x = $$\frac{15 \times 48}{30}$$ = 24
∴ Number of workers required = 24

Question 8.
A School has 8 periods a day each of45 minutes duration. How long would each period become ,if the school has 6 periods a day? ( assuming the number of school hours to be the same)
Solution:
Time and number of periods are in inverse proportion.
⇒ x1y1 = x2 y2
⇒ 45 × 8 = x × 6
⇒ $$\frac{45 \times 8}{6}$$
⇒ 60 minutes

Question 9.
If z varies directly as xand inversely as y. Find the percentage increase in z due to an increase of 12% in x and a decrease of 20% in y.
Solution:
Given that
z varies directly as x and inversely as y So, z ∝ x (1); z ∝ 1/y ……………… (2)
From (1) & (2), z ∝ $$\frac{\mathrm{x}}{\mathrm{y}}$$

Let x1 = 100x, x2 = 112x
(∵ It increases 12%)
y1 = 100y, y2 = 80y
(∵ It decreases 20%)
From (3),

∴ z is increased in 40%

Question 10.
If x + 1 men will do the work in x + 1 days, find the number of days that (x + 2) men can finish the same work.
Solution:
Number of workers and number of days are in inverse proportion.
⇒ x1y1 = x2 y2
⇒ (x + 1) (x + 1) = (x + 2) x k
⇒ k = $$\frac{(x+1)(x+1)}{(x+2)}$$
∴ k = $$\frac{(x+1)^{2}}{(x+2)}$$

Question 11.
Given a rectangle with a fixed perimeter of 24 meters, if we increase the length by 1 m the width and area will vary accordingly. Use the following table of values to look at how the width and area vary as the length varies.