Question

a swimming pool is filled in 10 hours by three tankers A,B,C The tank c is Twice as fast as B and B is twice as fast A. How much time will tanker A alone take to fill the swimming pool ?

Answer 1

Answer:

70 hrs

Step-by-step explanation:

Given a swimming pool is filled in 10 hours by three tankers A,B,C The tank c is Twice as fast as B and B is twice as fast A. How much time will tanker A alone take to fill the swimming pool

We have C = 2B and B = 2A

Let the tank A fill in x hrs, so tank B will be x/2 hrs and tank C will be x/4 hrs.

In x hrs tank A will be filled.

So in 1 hr it will fill 1 / x tank will be filled

In x/2 hrs tank B will be filled

So in 1 hr it will fill x/2 / 1 = 2/x

In x/4 hrs tank C will be filled

So in 1 hr it will fill x/4 /1 = 4/x

So A + B + C = 1/x + 2/x + 4/x = 7/x in 1 hr

Therefore tank A, B, C will fill in x/7 hrs

Now it is given it is filled in 10 hrs, so x/7 = 10

x = 10 x 7 = 70 hrs.

So tanker A will fill the swimming pool in 70 hrs

Answer 2

➡️Answer:

Tanker A alone fills the swimming pool in 70 hour

➡️Solution:

Let tanker A fills the swimming pool in 'x' hours

ATQ

than tanker B fills the Swimming pool in $\frac{x}{2} \: \: hour$

Since it is twice fast as A,so fill the pool in half time

and tanker C fills the swimming pool in $\frac{x}{4} \: hours$
Since it is twice fast as B,so it fills the pool in half time of B

Now, in one hour

Tanker A fills = $\frac{1}{x} \: part \: of \: pool$

Tanker B fills $= \frac{2}{x} \: part \: of \: pool$

Tanker C fills$= \frac{4}{x} \: part \: of \: pool$

In one hour pool fills 1/10 th part

$\frac{1}{x} + \frac{2}{x} + \frac{4}{x} = \frac{1}{10} \\ \\ \frac{7}{x} = \frac{1}{10} \\ \\ x = 70$

Tanker A fills the swimming pool in 70 hour

Tanker B fills the swimming pool in 35 hour

Tanker C fills the swimming pool in 17.5 hour

Hope it helps you.