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 cylinderLet 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 cylinderRadius 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 fillHow 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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