pushpitkc wrote:

Bunuel wrote:

An automobile tire has two punctures. The first puncture by itself would make the tire flat in 9 minutes. The second puncture by itself would make the tire flat in 6 minutes. How long will it take for both punctures together to make the tire flat? (Assume the air leaks out at a constant rate.)

(A) 18/5 minutes

(B) 4 minutes

(C) 21/4 minutes

(D) 15/2 minutes

(E) 15 minutes

If we assume the quantity of air that the tire contains to be 54 units,

the first puncture leaks 6 units/minute and the second puncture leaks 9 units/minute.

Together, they would leak 15 units/minute

Hence, we would need \(\frac{54}{15}\) = \(\frac{18}{5}\)minutes to leak the air if both the punctures leak air(Option A)

Hi,

THAT may not be the correct way..

If you had assumed it to be 15 units, ans would be 15/15=1

One releases air in 9 min ; so in 1 min, it will release 1/9

other releases in 6 minutes ; so in 1 min it will release 1/6

so in 1 minute combined they will release \(\frac{1}{9}+\frac{1}{6} = \frac{15}{54}\)...

so entire air will be released in \(\frac{54}{15} = \frac{18}{5}\)

A

otherwise, if one releases in 9 min and second in less than that, ans has to be <9/2

all but A and B remain

the first puncture would have leaked \(\frac{15}{9}\) or \(\frac{5}{3}\) units in a minute

and the second puncture would have leaked \(\frac{15}{6}\) or \(\frac{5}{2}\) units in a minute

Combined they would have released \(\frac{5}{3}+\frac{5}{2} = \frac{10+15}{6} = \frac{25}{6}\) units in a minute.

The time taken to release the air by both the punctures will now be \(\frac{15 * 6}{25} = \frac{90}{25} = \frac{18}{5}\)(Option A)