# Variation – Exercise 3.6.3 – Class IX

1. If 7 workers can build a trench in 25 hours, how many workers will be required to be the same work in 35 hours?

Solution:

workers                  hours

7                                 25

x                                  35

Then,

x = 7×25/35  = 5 workers

1. A farmer has a stock of food enough to feed 28 animals for nine days, he buys 8 more animals which take same quantity of food. How long would the food last now?

Solution:

No. of animals                      days

28                                              9

36                                              x

Inverse proportion as number of animals increase the same food required number of days decrease.

x = 9×28/36 = 7 days

1. Suppose the rate of melting, M grams per second, of a sphere of ice is inversely proportional to the square of the radius, r cm. when r = 20, M = 0.6.

Solution:

Mα1

r = 20

M = 0.6

(a) find the constant of proportionality

M = k/(R2)

0.6 = k/(20×20)

k = 0.6 x 20 x 20 = 240.0 = 240

(b) Find the rate of melting when r = 40 cm

M = k/(R2)  = 240/40×40 = 6/40 = 3/20 gm

(c)Find the radius when the rate of melting is 1 gram per second; give your answer correct to 2 decimal places.

R = ? M = 1gram/sec

M = k/(R2)

1 = 240/(R2)

R = √240

R = 15.49cm

1. When a ball is thrown upwards, the time, t seconds , during which the ball remains in the air is directly proportional to the square root of the height, h meters, reached. We know that T = 4.45 sec when h = 25m

(a)Find a formula for T in terms of h

(b)Find the value of T when  h = 50; give your answer correct to three decimal places

(c)If the ball is thrown upwards and remains in the air for 5 seconds find the height reached(correct to two decimal places)

Solution:

Tα√h  ; T= 4.47sec when h = 25m

T = k√h

(a)

T = k√h

4.47 = k √25

k = 5/4.47 = 0.894

T=0.894√h

(b)

h = 50, T=?

T = k√h

=0.894x√(20) = 0.894x√(25×2)=0.894×5√2=4.47√2 sec = 6.32sec

(c)

T = 5sec ; h =? ; T = 0.894√h

T = k√h

5 = 0.894√h

√h = 5/0.894

h = 31.27m

1. Given that the relation is either y α x or y α 1/x, find which law is followed in the adjacent table. Find x when y= 2. Find y when x = 5.
 X 8 10 12 20 Y 15 12 1 0.6

Solution:

Y α 1/X

Y = k. 1/X

k = XY

k = 8×1.5 = 12

(i) when Y=2

Y = k/X

2 = 12/X

X = 12/2

X = 6

(ii)When x = 5

Y = k/X

Y = 12/5

Y = 2.4

1. Each pair of values form an inverse variation. Find the missing value:

(i) (3, 7), (8, y)

(ii) (2, 5),(4, y)

(iii)(4, 6), (x, 3)

(iv) (2.6, 4.5),(x, 6.3)

Solution:

(i) (3, 7), (8, y)

3/7 = y/8

y = 3×8/7 = 24/7 = 3.42

(ii) (2, 5),(4, y)

3/7 = 4/y

y = 2×4/5 = 8/5 = 1.6

(iii)(4, 6), (x, 3)

4/6 = x/3

x = 6×3/4 = 18/4 = 4. 5

(iv) (2.6, 4.5),(x, 6.3)

2.6/4.5 = x/6.3

x = 6.3×4.5/2.6 = 10.903

1. Suppose that y varies inversely with the square of x and y = 50 when x =4. Find y when x = 5.

Solution:

1/x2  ;

y = 50              x = 4

y = ?                 x = 5

1/x2

Y= k. 1/x2

50 = k. 1/42 = k . 1/16  = k/16

k = 50×16 = 800

y = k. 1/X2 = 800 . 1/52 = 800/25 = 32