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The radius of a cylindrical water tank is reduced by 50%. Ho

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The radius of a cylindrical water tank is reduced by 50%. Ho  [#permalink]

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New post Updated on: 19 Dec 2013, 12:41
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The radius of a cylindrical water tank is reduced by 50%. However, the speed by which water is filled into the tank is also decreased by 50%. How much more or less time will it take to fill the tank now?

(A) 50% less time
(B) 50% more time
(C) 75% less time
(D) 75% more time
(E) 100% more time

Originally posted by sdpp143 on 19 Dec 2013, 12:05.
Last edited by Bunuel on 19 Dec 2013, 12:41, edited 1 time in total.
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Re: The radius of a cylindrical water tank is reduced by 50%. Ho  [#permalink]

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New post 19 Dec 2013, 22:51
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sdpp143 wrote:
The radius of a cylindrical water tank is reduced by 50%. However, the speed by which water is filled into the tank is also decreased by 50%. How much more or less time will it take to fill the tank now?

(A) 50% less time
(B) 50% more time
(C) 75% less time
(D) 75% more time
(E) 100% more time


It can be solved in very simpler way.

(VC)Volume of the cylinderical vessal is directly proportional to R^2.

So if radius is 50% less volume will be 1/4th of the original volume.(VC/4)

Now if with velocity V tank can be filled in T1 time of volume VC

So now Velocity is 50% less i..e V/2

So time taken to fill the capacity VC/4 by V/2 velocity is T2.

VT1 = VC

V/2*T2 = VC/4

So T1/T2 = 1/2

So Tank will be filled in less time. that is 50 % less.

Thanks,
AB


+1 Kudos if you like and understand.


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Thanks,
AB

+1 Kudos if you like and understand.

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Re: The radius of a cylindrical water tank is reduced by 50%. Ho  [#permalink]

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New post 20 Dec 2013, 00:44
Since radius is reduced by 50%, the new volume becomes 1/4 of the original volume. Also the speed is reduced by 50% i.e. new speed is 1/2 of the original speed.
If the voume would have been same, then with half of the speed the time taken would be double. ALso the volume is reduced by 1/4 of original. Hence time taken effectively will be 2/4 i.e. 1/2 of the original. Therefore 50% less time than normal.

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The radius of a cylindrical water tank is reduced by 50%. Ho  [#permalink]

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New post 04 Dec 2017, 11:51
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sdpp143 wrote:
The radius of a cylindrical water tank is reduced by 50%. However, the speed by which water is filled into the tank is also decreased by 50%. How much more or less time will it take to fill the tank now?

(A) 50% less time
(B) 50% more time
(C) 75% less time
(D) 75% more time
(E) 100% more time

Pick smart numbers, and this question can be answered pretty quickly.

Original cylinder
Let r = 4 ft
Let h = 2 ft
Find volume, then pick a rate.

Volume of original cylinder, in cubic feet:
\(\pi r^2h=(\pi*16*2)= 32\pi\)

Choose a smart fill rate for \(32\pi\).
Let fill rate, in cu. feet per hr = \(\frac{16\pi}{1hr}\)

Time to fill original cylinder:
\(\frac{Volume}{rate}= Time\)

\(\frac{32\pi}{(\frac{16\pi}{1})}= 32\pi* \frac{1}{16\pi}= 2\) hours

New cylinder

Radius decreases by 50 percent:
r = .50(4) = 2 feet
h = 2 feet
Volume of new cylinder, in cu. feet:
\(\pi r^2h=(\pi*4*2)= 8\pi\)

Fill rate decreases by 50 percent:
\(\frac{16\pi}{1hr}*(\frac{1}{2}) =
\frac{8\pi}{1hr}\)


Time to fill new cylinder:
\(\frac{Volume}{rate}=Time\)

\(\frac{8\pi}{(\frac{8\pi}{1})}= 8\pi* \frac{1}{8\pi}= 1\) hour

Percent change in time to fill
How much more or less time will it take to fill the tank now?

Percent change:
\(\frac{New-Old}{Old}*100\)

\(\frac{1-2}{2}*100=-\frac{1}{2}*100=-.50*100=-50\)
%

Answer A
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Re: The radius of a cylindrical water tank is reduced by 50%. Ho  [#permalink]

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New post 25 Dec 2017, 02:22
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Volume = \(πr^2h\)
New Volume= \(π(r/2)^2h\) = \(1/4 (πr^2h)\)

Time =V/R
New Time= 1/4V ÷ R/2= 1/2 (V/R) = 50% less time

A is the correct Answer
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Re: The radius of a cylindrical water tank is reduced by 50%. Ho &nbs [#permalink] 25 Dec 2017, 02:22
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