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Question

Two water taps together can fill a tank in 938hours. The tap of larger diameter takes 10hrs less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.


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Solution

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

Let the tap with larger diameter fills the tank alone in (x10)hours.

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

In 1 hour, the tap with larger diameter can fill 1x-10 part of the tank.

The tank is filled up in =938=758hrs

Thus, in 1 hour the taps fill 875 part of the tank.

Now, according to question,

1x+1x-10=758x-10+xxx-10=875752x-10=8xx-10150x-750=8x2-80x8x2-80x-150x+750=08x2-230x+750=04x2-115x+375=04x2-100x-15x+375=04xx-25-15x-25=04x-15x-25=04x-15=0x-25=0x=154x=25

Now, there are two values of x.

Therefore, we will have two cases.

Case (a)- When x=154hrs, then

x-10=154-10=-254hrs.

Since, time cannot be in negative. therefore this case is not possible.

Case (b)- When x=25hrs

x-10=25-10=15hrs.

∴ Hence, 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 15hours.


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