Question

Two water taps together can fill a tank 9 and 3/8 hours. The tap of larger diameter takes 10 hour 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:

25 hrs

Step-by-step explanation:

Let us consider the time taken by the smaller diameter tap = x

The time is taken by the larger diameter tap = x – 10

Totaltimeistakentofillatank=938

Total time is taken to fill a tank = 75/8

In one hour portion filled by smaller diameter tap = 1/x

In one hour portion filled by larger diameter tap = 1/(x – 10)

In one hour portion filled by taps = 8/75

⇒ 1x+1x−10=875 x−10+xx(x−10)=875 2x−10x2−10x=875

75 (2x-10) = 8(x2-10x)

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.