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.

 

Asked on by lucky092569

2 Answers | Add Yours

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.

 

krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

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.

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