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# Not a tough one, but tricky enough to share it... There are

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Not a tough one, but tricky enough to share it... There are [#permalink]  19 Mar 2008, 10:36
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Not a tough one, but tricky enough to share it...

There are many taps in the pool. One tap can fill the pool in 50 hours. Someone opens the first tap and further opens a new tap each following hour. In approximately how many hours the pool will be filled?

(A) 6
(B) 8
(C) 9
(D) 10
(E) 12
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Re: PS: Taps [#permalink]  19 Mar 2008, 11:19
Expert's post
D

1 hour: rate=1/50
2 hour: rate=2*1/50
n hour: rate=n*1/50

1=1/50+2/50+...+n/50=n*(n+1)/(2*50)

n^2+n=100

if n=9 then n^2+n=90<100: the pool is not filled
if n=10 then n^2+n=110>100: the pool is filled

Therefore, we need 10 hours

P.S. we can find precise time that will be equal 9.5 hours
but if we had got 9.45 hours, what option would we have preferred?
I would have chosen 10 hours because at 9 hours we have got unfilled pool. Any thoughts?
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Re: PS: Taps [#permalink]  19 Mar 2008, 12:56
I get slightly below 10 also..

here is how

after 1st hr=1/50
after 2nd hr=2/50
after 3rd=3/50

so the equation becomes

time to fill= 1/50+2/50+3/50+-....1/50(1+2+3..)=1

plug in u will see that it take about 9.5 hrs..
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Re: PS: Taps [#permalink]  20 Mar 2008, 06:56
D is the correct answer indeed.
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Re: PS: Taps [#permalink]  27 Mar 2008, 20:11
anyone else with explanations?

I tried doing this: (i'm wondering why this was wrong)

Hour1: 1/50t = 1
Hour2: 2/50t = 49/50
Hour3: 3/50t = 48/50
...
Hour7: 7/50t = 44/50

let t =7 49/50 is greater then 44/50, so pool is filled

Can anyone help me out?
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Re: PS: Taps [#permalink]  27 Mar 2008, 21:52
sreehari wrote:
Not a tough one, but tricky enough to share it...

There are many taps in the pool. One tap can fill the pool in 50 hours. Someone opens the first tap and further opens a new tap each following hour. In approximately how many hours the pool will be filled?

(A) 6
(B) 8
(C) 9
(D) 10
(E) 12

in fact, we need 50 hours of work by all taps.

since each hour a tap is added and each tap's performance can be counted in 1 hour interval, 50 = 1 + 2 + 3 + ......... + n

if n = 9, then sum is 45. if n = 10, then sum is 55.
so it should be between 9 and 10 hours but it cannot be 9. sould be approx 10.
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Re: PS: Taps [#permalink]  27 Mar 2008, 22:29
I have a different way of thinking (trying not to use any math formulas)

End of Hour 1: 1 tap for 1 hour -> 1/50
End of Hour 2: Hour 1 + 2 taps for 1 hour -> 1/50 + 2/50 = 3/50
End of Hour 3: Hour 2 + 3 taps for 1 hour -> 3/50 + 3/50 = 6/50
End of Hour 4: Hour 3 + 4 taps for 1 hour -> 6/50 + 4/50 = 10/50
End of Hour 5: Hour 4 + 5 taps for 1 hour -> 10/50 + 5/50 = 15/50
End of Hour 6: Hour 5 + 6 taps for 1 hour -> 15/50 + 6/50 = 21/50
Hour 7 -> 28/50
Hour 8 -> 36/50
Hour 9 -> 45/50
Hour 10 -> 55/50 overflow ..

Hence hour 10.
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Re: PS: Taps [#permalink]  28 Mar 2008, 15:47
thanks guys...all clear...stupid mistake
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Re: PS: Taps [#permalink]  28 Mar 2008, 17:46
GMAT TIGER wrote:
sreehari wrote:
Not a tough one, but tricky enough to share it...

There are many taps in the pool. One tap can fill the pool in 50 hours. Someone opens the first tap and further opens a new tap each following hour. In approximately how many hours the pool will be filled?

(A) 6
(B) 8
(C) 9
(D) 10
(E) 12

in fact, we need 50 hours of work by all taps.

since each hour a tap is added and each tap's performance can be counted in 1 hour interval, 50 = 1 + 2 + 3 + ......... + n

if n = 9, then sum is 45. if n = 10, then sum is 55.
so it should be between 9 and 10 hours but it cannot be 9. sould be approx 10.

Small edit: hours should be tapes. since 9 and a fraction tapes are not possible. so it must be 10 tapes even if the no. of tapes comes as 9.01.
so it is 10 taps.
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Re: PS: Taps   [#permalink] 28 Mar 2008, 17:46
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