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 * hTo 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.

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Karishma

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