# A rectangular tank can be filled with water by two pipes in 100/9 minutes. If the larger pipe alone takes 5 minutes less to fill the tank than the...smaller pipe, find the time each pipe takes to...

A rectangular tank can be filled with water by two pipes in 100/9 minutes. If the larger pipe alone takes 5 minutes less to fill the tank than the...

smaller pipe, find the time each pipe takes to fill the tank.

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### 2 Answers

In accordance with the data, we assume the larger of the pipe alone takes x+5 minutes to fill up the water tank and the smaller alone in x minutes.

So, the fraction of the tank filled by the larger pipe in 1 minute is 1/(x+5), and the fraction of tank filled by the smaller pipe in one munute = 1/x.

So the fraction of the tank filled by the two pipes together in one minute =1/(x+5)+1/x =(2x+5)/[x(x+5)]

Therefore, the number of minutes required by them to fill algebraically = 1/ {(2x+5)/[x(x+5)]}= x(x+5)/(2x+5) but this is said to be equal to 100/9 minutes by the data.

Therefore, x(x+5)/(2x+5)=100/9 .

Multiply by the LCM of the denominators both sides:

9x(x+5)=100(2x+5)

9x^2+45x=200x+500

9x^2+45x-200x-500=0

9x^2-180x+25x-500=0

9x(x-20)+25(x-20)=0

(9x+25)(x-20)=0

x=20 and x+5=25

Therefore the time taken by the larger and smaller pipes individully to fill up the water tank : 20 and 25 minutes respectively.

Hope this helps.

The solution posted above contains some errors and leaves some points explained although the final answer given is correct. For example, it is stated:

... we assume the larger of the pipe alone takes x+5 minutes to fill up the water tank and the smaller alone in x minutes.

This statement cannot be correct as the larger pipe should take less time to fill the tank as compared to the smaller pipe.

I suggest the following solution.

Let the time taken by the larger pipe to fill the tank be x minutes.

Then capacity of larger pipe to fill the tank in terms of tanks per minute = 1/x.

Now as the smaller pipe takes five minutes longer than the larger pipe the time taken by it to fill the tank = x + 5 minutes.

Then capacity of smaller pipe to fill the tank in terms of tanks per minute = 1/(x +5).

The total capacity of large plus small pipe = 1/x + 1/(x + 5)

= ([x + 5) + x]/[x*(x + 5)] = (2x + 5)/(x^2 + 5x)

The time taken by large plus small pipe to fill the tank =

1/(total capacity of large plus small pipe) = 1/[(2x + 5)/(x^2 + 5x)]

= (x^2 + 5x)/(2x + 5)

This time is given as 100/9 minutes

Therefore: 100/9 = (x^2 + 5x)/(2x + 5)

Therefore: 100*(2x + 5) = 9*(x^2 + 5x)

Therefore: 200x + 500 = 9*x^2 + 45x

Rearranging this equation with all the terms on left hand side we get:

9*x^2 - 155x + 500 = 0

Therefore: 9*x^2 - 180x + 25x + 500 = 0

Therefore: 9x(x - 20) + 25*(x + 20) = 0

Therefore: (x - 20)*( 9x + 25) = 0

Therefore x = 20, and -25/9

However the time taken to fill the tank cannot be negative therefore we accept only the first value i.e. x = 20

Therefore large pipe will take 20 minutes to fill the tank.

And small pipe will take 5 minutes more than that i.e. 20 + 5 = 25 minutes.