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
OA [spoiler] E. But the suggested solution provided is to just check every single one of them and compare...Anyone got any tricks or shortcuts to work through this faster? [/spoiler].
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are there really problems this tedious?
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First the problem is not tedious at all. Very simple. This is a problem that tests your ability to manipulate percentages.abcdefg 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 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
OA [spoiler] E. But the suggested solution provided is to just check every single one of them and compare...Anyone got any tricks or shortcuts to work through this faster? [/spoiler].
You are looking for the value that is furthest away from 2pir^2 h
A. pi 4r^2 h/2 is the same so eliminate
B .49r^2 x 4h =1.96
C. .81r^2 x 2.5=2.025
D 1.96 r^2 = 1.96
E. 2.25 r^2 x .8 h= 1.8
The suggested solution is the easiest. You can solve this under 1 min. Once u execute A, you don't have to write r^2 or h on your paper. U are just squaring one number and multiply by another. The form you put these numbers in will make it easier or difficult. How fast you can process percent increase or decrease is another factor. Knowing the squares of all numbers from 1 to 20 is still another factor. So overall not a tedious problem at all.
