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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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Updated on: 19 Dec 2013, 11:41
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Difficulty:

35% (medium)

Question Stats:

73% (01:56) correct 27% (02:13) wrong based on 113 sessions

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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, 11:05.
Last edited by Bunuel on 19 Dec 2013, 11:41, 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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19 Dec 2013, 21:51
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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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Re: The radius of a cylindrical water tank is reduced by 50%. Ho  [#permalink]

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19 Dec 2013, 23:44
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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04 Dec 2017, 10:51
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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:
$$\frac{New-Old}{Old}*100$$

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

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

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25 Dec 2017, 01:22
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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, 01:22
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# The radius of a cylindrical water tank is reduced by 50%. Ho

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