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radius of a cylindrical water tank

by spd143 » Thu Dec 19, 2013 1:03 pm
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
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by theCodeToGMAT » Thu Dec 19, 2013 7:41 pm
Let initial Height be 10 & radius be 10

Initial Volume = pi * 100 * 10 = 1000*pi

Initial speed = 1 * pi

Time Taken = 1000*pi/1*pi = 1000 minutes

Volume after changed radius = pi * 25 * 10 = 250 * pi

Changed Speed = pi*/2

Time Taken = 250* pi / pi*/2 = 500 minutes

Percentage = 500/1000 * 100 = 50% less time

[spoiler]{A}[/spoiler]
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by ceilidh.erickson » Fri Dec 20, 2013 12:25 pm
If we wanted to solve this algebraically, we could say that the original volume of the cylinder was: V = pi*r^2*h

The rate at which the cylinder is filled with water could be called k (for some constant rate). Thus, the original time that it would take to fill the cylinder would be given as: time = volume/rate
(pi*r^2*h)/k

If we reduce both r and k by half, we get:
(pi*((1/2)r)^2*h)/((1/2)k)

In other words, 1/2 times (pi*r^2*h)/k, which was the original time -> half the original time. A.

Perhaps a better way to think about this, though, is conceptually. When we halve the radius, it doesn't merely halve the volume. Because we square the radius to find volume, 1/2 of the radius would yield 1/4 of the volume. If the rate of water didn't change, 1/4 the volume would mean filling 4 times as fast. Since we're slowing the rate by half, though, half of 4x is 2x. Twice as fast = 50% of the original time.
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by Patrick_GMATFix » Fri Dec 20, 2013 11:34 pm
My approach:

Multiplying the radius by 1/2 (50% reduction) means the area of the base will be multiplied by 1/4. In other words, the volume of the new cylinder is 1/4 of the old volume. So if speed remained constant, the new time required to fill the cylinder would be 1/4 of the old time. Thus times would be t and t/4.

Since the speed was caut in half, the new time is doubled (half speed -> takes 2x as long). So the new time is actually 2* (t/4) = t/2.

So the old time was t, and the new time is t/2. This is a 50% reduction in time.

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by Scott@TargetTestPrep » Thu Jan 11, 2018 1:56 pm
spd143 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
We can let the radius of the cylindrical tank be 2 ft and the height be 4 ft and the rate at which the water is filled into the tank be 2π ft^3/min. Thus, the volume of the tank is V = π x 2^2 x 4 = 16π ft^3 and the time it takes to fill it is 16π/2π = 8 min.

Now, let's reduce both the radius and the rate at which water is filled into the tank by 50%. The new radius = 1 ft and the new rate = π ft^3/min. The volume of the new tank is V = π x 1^2 x 4 = 4π ft^3 and the time it takes to fill it is 4π/π = 4 min.

We see that the new time to fill the tank is half as long as the old time. Thus, it takes 50% less time to fill the tank.

Answer: A

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