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A cylindrical tank of radius R and height H must be redesigned to hold
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29 Mar 2021, 06:00
29 Mar 2021, 06:00
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A cylindrical tank of radius R and height H must be redesigned to hold approximately twice as much liquid. Which of the following changes would be farthest from the new design requirements?
A. A 100% increase in R and a 50% decrease in H
B. A 30% decrease in R and a 300% increase in H
C. A 10% decrease in R and a 150% increase in H
D. A 40% increase in R and no change in H
E. A 50% increase in R and a 20% decrease in H
A. A 100% increase in R and a 50% decrease in H
B. A 30% decrease in R and a 300% increase in H
C. A 10% decrease in R and a 150% increase in H
D. A 40% increase in R and no change in H
E. A 50% increase in R and a 20% decrease in H
A cylindrical tank of radius R and height H must be redesigned to hold
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29 Mar 2021, 23:32
29 Mar 2021, 23:32
3
In this question, we will have to test each answer choice. The final volume is approximately doubled.
Imagine \(initial volume = pi * R^2 * H\). Hence \(final volume = 2 * pi * R^2 * H\). We need to find which answer choice is farthest from the final volume.
R will become 2R, H will become (0.5)H. Plugging in these values, we find that \(pi * 2^2*R^2 * (0.5)H = 2 * pi * R^2 * H\). Exactly the final volume. Eliminate.
Similarly, new radius = (0.7)*R, new height = 4*H. New volume is a \((0.7)^2 * 4 = 0.49 * 4\) multiple of initial volume. Close to double the initial volume. Eliminate.
Again, new radius = (0.9)*R, new height = (2.5)*H.\( New multiple = (0.9)^2 * 2.5\), coming slightly greater than 2. Eliminate.
Similarly, new radius = (1.4)*R, new height = H. \(New multiple = (1.4)^2 = 1.96\), coming not very close to 2. Hold to compare.
New radius = (1.5)*R, new height = (0.8)*H. \(New multiple = (1.5)^2 * (0.8) = 1.8\), coming farthest from 2 so far.
Hence, E is the correct answer.
Imagine \(initial volume = pi * R^2 * H\). Hence \(final volume = 2 * pi * R^2 * H\). We need to find which answer choice is farthest from the final volume.
Quote:
A. A 100% increase in R and a 50% decrease in H
R will become 2R, H will become (0.5)H. Plugging in these values, we find that \(pi * 2^2*R^2 * (0.5)H = 2 * pi * R^2 * H\). Exactly the final volume. Eliminate.
Quote:
B. A 30% decrease in R and a 300% increase in H
Similarly, new radius = (0.7)*R, new height = 4*H. New volume is a \((0.7)^2 * 4 = 0.49 * 4\) multiple of initial volume. Close to double the initial volume. Eliminate.
Quote:
C. A 10% decrease in R and a 150% increase in H
Again, new radius = (0.9)*R, new height = (2.5)*H.\( New multiple = (0.9)^2 * 2.5\), coming slightly greater than 2. Eliminate.
Quote:
D. A 40% increase in R and no change in H
Similarly, new radius = (1.4)*R, new height = H. \(New multiple = (1.4)^2 = 1.96\), coming not very close to 2. Hold to compare.
Quote:
E. A 50% increase in R and a 20% decrease in H
New radius = (1.5)*R, new height = (0.8)*H. \(New multiple = (1.5)^2 * (0.8) = 1.8\), coming farthest from 2 so far.
Hence, E is the correct answer.
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Re: A cylindrical tank of radius R and height H must be redesigned to hold
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29 Mar 2021, 23:41
29 Mar 2021, 23:41
2
GeminiHeat wrote:
A cylindrical tank of radius R and height H must be redesigned to hold approximately twice as much liquid. Which of the following changes would be farthest from the new design requirements?
A. A 100% increase in R and a 50% decrease in H
B. A 30% decrease in R and a 300% increase in H
C. A 10% decrease in R and a 150% increase in H
D. A 40% increase in R and no change in H
E. A 50% increase in R and a 20% decrease in H
A. A 100% increase in R and a 50% decrease in H
B. A 30% decrease in R and a 300% increase in H
C. A 10% decrease in R and a 150% increase in H
D. A 40% increase in R and no change in H
E. A 50% increase in R and a 20% decrease in H
Volume of the cylinder \(= πr^2H\)
Twice as much liquid means twice the Volume
A. A 100% increase in R and a 50% decrease in H
\(V = π(2r)^2(0.5H) = 2πr^2H\)
B. A 30% decrease in R and a 300% increase in H
\(V = π(0.7r)^2(4H) = 1.96πr^2H\)
C. A 10% decrease in R and a 150% increase in H
\(V = π(0.9r)^2(2.5H) = 2.025πr^2H\)
D. A 40% increase in R and no change in H
\(V = π(1.4r)^2(H) = 1.96πr^2H\)
E. A 50% increase in R and a 20% decrease in H
\(V = π(1.5r)^2(0.8H) = 1.8πr^2H\)
Hence, option E
