arjunbt 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
The solution I am going to give has already been provided but I am still writing it down to highlight the approximations you could use to save time.
Volume must be doubled.
We know volume of a cylinder \(= \pi * r^2 * h\)
To double the volume, either you double the height 'h' OR you multiply r by \(\sqrt{2}\) i.e. by 1.41 i.e. increase radius by a little over 40%.
OR you use a combination.
A. A 100% increase in R and a 50% decrease in H
Volume \(= \pi * 4r^2 * \frac{h}{2} = 2* \pi * r^2 * h\) (matches the design requirement)
B. A 30% decrease in R and a 300% increase in H
Volume \(= \pi * 0.49 r^2 * 4h = \pi * \frac{r^2}{2} * 4h = 2* \pi * r^2 * h\) (matches the design requirement)
Note that .49 is approximately 0.5 i.e. 1/2
C. A 10% decrease in R and a 150% increase in H
Volume \(= \pi * .81 r^2 * \frac{5}{2}h = 2* \pi * \frac{4}{5}r^2 * \frac{5}{2} h = 2* \pi * r^2 * h\) (matches the design requirement)
Notice that .81 is approximately .80 i.e. 80% i.e. 4/5
D. A 40% increase in R and no change in H
We discussed it above. A 40% increase in R matches our design requirement.
E. A 50% increase in R and a 20% decrease in H
Volume \(= \pi * 2.25 r^2 * 0.8 h\)
Compare with (C). 2.5*0.8 gives you 2 so 2.25*0.8 will give you less than 2. Hence doesn't match the requirement.
_________________
Karishma
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