Question

9. The water taps together can fill a tank in
$9 \times \frac{3}{8}$
hours. The top of larger diameter takes 10
hours less than the smaller one to fill the tank separately. Find the time in which each tap
can separately fill the tank.​

Answer 1

Answer:

Let the tap with smaller diameter fills the tank alone in x hours

Let the tap with larger diameter fills the tank alone in (x – 10) hours.

In 1 hour, the tap with a smaller diameter can fill 1/x part of the tank

In 1 hour, the tap with larger diameter can fill 1/(x – 10) part of the tank.

The tank is filled up in 75/8 hours.

Thus, in 1 hour the taps fill 8/75 part of the tank.

1/x + 1/(x-10) = 8/75  

(x-10) + x / x(x-10) = 8/75  

2x – 10/x(x-10) = 8/75  

75 (2x-10) = 8(x2-10x) by cross multiplication  

150x – 750 = 8x2 – 80x  

8x2 − 230x + 750 = 0  

4x2−115x + 375 = 0  

4x2 − 100x −15x + 375 = 0  

4x(x−25)−15(x−25) = 0  

(4x−15)(x−25) = 0  

4x−15 = 0 or x – 25 = 0  

x = 15/4 or x = 25

Case 1: When x = 15/4  

Then x – 10 = 15/4 – 10  

⇒ 15-40/4  

⇒ -25/4  

Time can never be negative so x = 15/4 is not possible.  

Case 2: When x = 25 then  

x – 10 = 25 – 10 = 15

∴ The tap of smaller diameter can separately fill the tank in 25 hours and the time taken by the larger tap to fill the tank = ( 25 – 10 ) = 15 hours.