Calculating percentages when options given make calculation

[nikhil.baveja]

Hi, I saw this Question in MGMAT Practice question pool

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 100% increase in R and a 50% decrease in H

a 30% decrease in R and a 300% increase in H

a 10% decrease in R and a 150% increase in H

a 40% increase in R and no change in H

a 50% increase in R and a 20% decrease in H

Now I know answer is E, and the way i calculated was selecting a random number like r=2 and h=4 and then checking each option.
In Mgmat explanation the method was something like this

(E) a 50% increase to R and a 20% decrease to H:
The new volume = (1.5R)2(.8H) = (2.25)(.8)R2H = 1.8R2H. This is the farthest away from twice the original volume.

I wanted to know what other method can one choose to do this because quite frankly, it took me some time in solving this question by checking r=2 and h=4 against each option and MG method also sounded hard to me, as I had to multiply everything in decimals which is again time consuming.
wanted to know if there can be any other smart way of working this out.

Br
Nikhil

[jlucero]

This is (unfortunately) one of those questions where you'll probably have to spend 2 minutes no matter what way you solve this. Picking numbers works well, but you'll still have some computations no matter what numbers you pick. My strategy is just a bit different but still labor intensive:

pi*r^2*h

Since pi is constant, what changes is r^2 * h. So I start with the values of r = 1, h = 1, and deal with increases/decreases in terms of fractions:

a) 4 * 1/2 = 2
b) (7/10)^2 * 4 = 49/100 * 4 = 2ish
c) (9/10)^2 *2.5 = 81/100 * 5/2 = 2ish
d) (14/10)^2 * 1 = 196/100 * 1 = 2ish
e) (15/10)^2 * 4/5 = 225/100 * 4/5 = 900 / 500 = 9/5 (smallest #)