the best approximation for the ratio

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!

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

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

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]

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 !

You cannot ignore pi initially at all, as you are going to use to decrease the total volume of water leaked from the vessel X from the total volume of the vessel X which is going to be = {pi* (12*12*40)-900} so it seems wise to round of pi to approximately ~3. I dont know how much should it count I am going to do the calculation both ways to find out the difference in the answer. But this approach sounds much more logical from the GMAT exam viewpoint as it lets you solve it in a much lesser time. The answer comes out finally to be something like 364/45 which is .....

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 !

900 doesn't have π along, better take 300 there in bold.