Two water taps together can fill a tank in 9 ⅜ hours. The tap 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.

By BYJU'S Exam Prep

Updated on: September 25th, 2023

Two water taps together can fill a tank in 9 ⅜ hours. The tap of a larger diameter takes 10 hours less than the smaller one to fill the tank separately. The time in which each tap can separately fill the tank is 25 hours for smaller diameters and 15 hours for larger diameters. Let us consider a tap with a smaller diameter fills the tank alone in x hours

Now let the tap with a larger diameter fills the tank alone in (x – 10) hours.
The tap with the smaller diameter can fill 1/x of the tank in an hour.
The tap with a wider diameter can fill 1/x-10 of the tank in one hour.
The tank is filled up in = 9 ⅜ = 75/8 hours
Therefore, in 1 hour the taps fill 8/75 part of the tank

Table of content

1. Two water taps together can fill a tank in 9 ⅜ hours. The tap 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.

Now, according to the question, we can write as

1/x + 1/x – 10 = 75/8
(x – 10 + x)/ x (x – 10) = 8/75
75 (2x – 10) = 8x (x – 10)
150x – 750 = 8×2 – 80x

On rearranging we get

8×2 – 80x – 150x + 750 = 0
8×2 – 230x + 750 = 0
4×2 – 115x + 375 = 0
4×2 – 100x – 15x + 375 = 0
4x (x – 25) – 15 (x – 25) = 0
(4x – 15) (x – 25) = 0

On simplification we get

4x – 15 = 0 x – 25 = 0
4x = 15x = 25
x = 15/4

There are currently two x values.

Consequently, there will be two cases.

Case (a) – When x = 15/4 hours, then
(x – 10) = 15/4 – 10 = -25/4 hours

Time cannot be negative, therefore. Consequently, this scenario is not feasible.

Case (b) – When x = 25 hours
(x – 10) = 25 – 10 = 15 hours

As a result, the smaller tap can independently fill the tank in 25 hours, while the larger tap needs 15 hours to do so.

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