Problem Solving

sanju09 wrote:Cylindrical vessel X can leak from bottom into another cylindrical vessel Y at the rate of 30 cubic inches per minute. The radii of bases of the vessels X and Y are 12 inches and 8 inches respectively and their heights are 40 inches and 36 inches respectively. The cylindrical vessel X, when full of water, starts leaking into cylindrical vessel Y. Which of the following is the best approximation for the ratio in the depths of water, in the vessels X and Y respectively, 30 minutes after the start of leakage?
A. 10:9
B. 3:2
C. 5:3
D. 5:2
E. 8:1

Made Up!

GMATSUCKER wrote:

sanju09 wrote:Cylindrical vessel X can leak from bottom into another cylindrical vessel Y at the rate of 30 cubic inches per minute. The radii of bases of the vessels X and Y are 12 inches and 8 inches respectively and their heights are 40 inches and 36 inches respectively. The cylindrical vessel X, when full of water, starts leaking into the empty cylindrical vessel Y. Which of the following is the best approximation for the ratio in the depths of water, in the vessels X and Y respectively, 30 minutes after the start of leakage?
A. 10:9
B. 3:2
C. 5:3
D. 5:2
E. 8:1

Made Up!

Is this a GMAT Question ? To Quote Stuart Kowinsky :-" What's the Source ? "

Assuming that vessel Y was empty initially the approximate ratio of depth of water comes out to be 33.75 : 14.06, so the approximate depth is IMO[spoiler]D.5:2[/spoiler]

Cheers
GMATSUCKER

sanju09 wrote:
Hi GMATSUCKER,

I understand why you had to make the assumption that the vessel Y was initially vacant, it should have been mentioned in the problem. But still, D doesn't seem to be the best approximation either, even after the right assumption is made or the repair is done.

In 30 minutes, the leaked volume is 30 times 30 equal to 900 cubic inches of water inside the empty vessel Y.

Hence, 900 = π (8)^2 (depth in Y), where 3 is the best approximation for π, hence depth in y is approximately 75/16. Whereas vessel X would lose 900 (or 300 π) cubic inches of water from its π(12)^2 (40) cubic inches of water, or

5460 π = π(12)^2 (depth in X). Hence, depth in X = 455/12.

The approximated required ratio is therefore (455/12):(75/16) which is very close to [spoiler]8:1.

Mine's E[/spoiler]

GMATSUCKER wrote:

sanju09 wrote:
Hi GMATSUCKER,

I understand why you had to make the assumption that the vessel Y was initially vacant, it should have been mentioned in the problem. But still, D doesn't seem to be the best approximation either, even after the right assumption is made or the repair is done.

In 30 minutes, the leaked volume is 30 times 30 equal to 900 cubic inches of water inside the empty vessel Y.

Hence, 900 = π (8)^2 (depth in Y), where 3 is the best approximation for π, hence depth in y is approximately 75/16. Whereas vessel X would lose 900 (or 300 π) cubic inches of water from its π(12)^2 (40) cubic inches of water, or

5460 π = π(12)^2 (depth in X). Hence, depth in X = 455/12.

The approximated required ratio is therefore (455/12):(75/16) which is very close to [spoiler]8:1.

Mine's E[/spoiler]

Let's ignore Pi for the moment cause it will eventually cancel out anyway. Vessel Y: 900= 8^2*Height. or Height = 14.06. Vessel X 5760-900= 4860 = 12^2*Height or Height=33.75. So Ratio 33.75:14.06 or 5:2.
Maybe I'm missing something !