# Moderate Time and Work Solved QuestionAptitude Discussion

 Q. Two pipes can fill a tank in 20 and 24 minutes respectively and a waste pipe can empty 3 gallons per minute. All the three pipes working together can fill the tank in 15 minutes. The capacity of the tank in gallons is
 ✖ A. 100 ✖ B. 110 ✔ C. 120 ✖ D. 140

Solution:
Option(C) is correct

Work done by the waste pipe in 1 minute

$=\dfrac{1}{15}-\left[\dfrac{1}{20}+\dfrac{1}{24}\right]$

$=-\dfrac{1}{40}$

volume of  $\dfrac{1}{40}$ part = 3 gallons.

Therefore, Volume of whole =$(3×40)$ gallons = 120 gallons.

Edit: For an alternative solution, check comment by Sm Mudabbir.

## (4) Comment(s)

Jisha
()

The LCM of 20 and 24 is 120.

To find gallons per minute for both pipe-

1st: 120/20= 6 gallons/minute

2nd: 120/24=5 gallons/ minute

(We can solve with just one of the above figure only. We will take the first pipe for example)

The first pipe fills 1/20th part(volume) of vessel by a rate of 6 gallons/minute.

Total part (total volume) = 6x20= 120 gallons.

Atharva Joshi
()

i don't know if my method is correct or not but basically if we check the options only 120 gallons will fit the perfect divisibility test and considering that the answer is in whole numbers that should be the answer

Hashmita
()

By LCM method.

LCM of (20, 24, 15) is 120

thus, the rate of pipe1 will be 6

pipe2 will be 5

and P1 + P2 - P3 = 8

so, P3 = 3

thus, P3 empties tank in => total-work/rate i.e, 120/3 => 40min

and also as given in question [a waste pipe can empty 3 gallons per minute] which implies (40min * 3) = 120 gallons

Sm Mudabbir
()

Let, total be $x$ gallons

Tank filled by the waste pipe in 15 min $=15 \times 3=45$ gallons

Now, pipe $a$ will fill $\dfrac{15x}{20}$ in 15 min.

Pipe $b$ will fill $\dfrac{15x}{24}$ in 15 min.

Therefore,

$45=\dfrac{15x}{20}+\dfrac{15x}{24}-x$

$x= \textbf{120 gallons}$