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An automobile tire has two punctures. The first puncture by itself wou

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An automobile tire has two punctures. The first puncture by itself wou [#permalink]

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New post 26 Oct 2017, 02:58
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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
[Reveal] Spoiler: OA

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An automobile tire has two punctures. The first puncture by itself wou [#permalink]

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New post 26 Oct 2017, 04:18
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)
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Re: An automobile tire has two punctures. The first puncture by itself wou [#permalink]

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New post 26 Oct 2017, 05:20
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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
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Re: An automobile tire has two punctures. The first puncture by itself wou [#permalink]

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New post 26 Oct 2017, 05:58
chetan2u wrote:
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


If I had assumed 15 units,
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)
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Re: An automobile tire has two punctures. The first puncture by itself wou [#permalink]

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New post 29 Oct 2017, 07:47
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


The combined rate of the two punctures is:

1/9 + 1/6 = 2/18 + 3/18 = 5/18

Thus, the time to make a flat with both punctures is 1/(5/18) = 18/5 minutes.

Answer: A
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Re: An automobile tire has two punctures. The first puncture by itself wou [#permalink]

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New post 25 Nov 2017, 14:51
Hi All,

This is an example of a "Work Formula" question - when you have two entities working together on a task, you can use the following formula to determine how long it will take the two entities to complete the task:

(A)(B)/(A+B) = time to complete the task (where A and B are the two times it takes the individuals to complete the task when working alone)

We're told that the first puncture would create a flat tire in 9 minutes and the second puncture would make a flat tire in 6 minutes. Plugging in those values, we'd have...

(9)(6)/(9+6) = 54/15 = 18/5 = 3 3/5 minutes to make a flat tire

Final Answer:
[Reveal] Spoiler:
A


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Re: An automobile tire has two punctures. The first puncture by itself wou   [#permalink] 25 Nov 2017, 14:51
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