In this question, we will have to test each answer choice. The final volume is approximately doubled.
Imagine
initialvolume=pi∗R2∗H. Hence
finalvolume=2∗pi∗R2∗H. We need to find which answer choice is farthest from the final volume.
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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∗22∗R2∗(0.5)H=2∗pi∗R2∗H. Exactly the final volume. Eliminate.
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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.
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C. A 10% decrease in R and a 150% increase in H
Again, new radius = (0.9)*R, new height = (2.5)*H.
Newmultiple=(0.9)2∗2.5, coming slightly greater than 2. Eliminate.
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D. A 40% increase in R and no change in H
Similarly, new radius = (1.4)*R, new height = H.
Newmultiple=(1.4)2=1.96, coming not very close to 2. Hold to compare.
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E. A 50% increase in R and a 20% decrease in H
New radius = (1.5)*R, new height = (0.8)*H.
Newmultiple=(1.5)2∗(0.8)=1.8, coming farthest from 2 so far.
Hence, E is the correct answer.