**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

**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

**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}/(R^{2})

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}/(R^{2}) = ^{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}/(R^{2})

1 = ^{240}/(R^{2})

R = √240

R = 15.49cm

**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

**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

**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

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

Solution:

Yα^{1}/x^{2} ;

y = 50 x = 4

y = ? x = 5

Yα^{1}/x^{2}

Y= k. ^{1}/x^{2}

50 = k. ^{1}/_{4}^{2} = k . ^{1}/_{16 } = ^{k}/_{16}

k = 50×16 = 800

y = k. ^{1}/_{X}^{2} = 800 . ^{1}/_{5}^{2} = ^{800}/_{25} = 32

^{ }

## 1 thought on “Variation – Exercise 3.6.3 – Class IX”

Comments are closed.