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.

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