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

Question

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

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

bhatiamanu05 · Answer

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

zerosleep · Answer

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.

Ans- (A)

generis · Answer

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=  
 -50  %

The negative sign means  less  time.

Answer A

NDND · Answer

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

LeoN88 · Answer

t1 is old time. s is old speed
t2 is new time.

t1 * s = PI(r^2)h

t2 * (s/2) = PI{(r/2)^2}h

Equate to find t2 = (1/2) * t1, so  50% less time taken. (A)

bumpbot · Answer

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

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