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Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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by sukh » Thu Jul 28, 2011 8:52 am
Machine A can complete a certain job in x hours. Machine B can complete the same job in y hours. If A and B work together at their respective rates to complete the job, which of the following represents the fraction of the job that B will not have to complete? job that B will not have to complete? a) x - y / x + y b ) x/ y - x c )x + y / xy d)y /x - y e)y / x + y


When a cylindrical tank is filled with water at a rate of 22 cubic meters per hour, the level of water in the tank rises at a rate of 0.7 meters per hour. Which of the following best approximates the radius of the tank in meters ? a) under-root 10 / 2 b) under-root 10 c )4 d ) 5 e ) 10
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by bblast » Thu Jul 28, 2011 9:10 am
1st question :

rate a = 1/x
rate b = 1/y

total rate = x+y/xy, thus time = xy/x+y

thus a and b both work for a total of xy/x+y hours

fraction of work done by a = rate of a * time

= 1/x * xy/x+y = y/x+y

choose E
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by bblast » Thu Jul 28, 2011 9:14 am
2nd question :

0.7 m ^3 of the tank corresponds to 22 m^3 of water

thus 1 m of the tank corresponds to 220/7 m^3 of water


volume of cylinder = pi * r^2 * h

assume height = 1. You can assume anything u want- solution will be same

pi * r^2 * h = 220/7

r^2 = 10

r = root 10.

pick B
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by gmatboost » Tue Aug 02, 2011 10:22 pm
bblast provided great solutions, I just want to add a couple of comments

Question 1:
bblast used the formula rate * time = work and set work = 1 to get the rates.
This is a good thing to do, though as long as you set work equal to the same thing in each part of the problem, the answer will not change.

For example, if you had decided to set work = W, then, adjusting bblast's work:
rate a = W/x
rate b = W/y
total rate = (Wx+Wy)/(xy), thus time = (xy)/(Wx+Wy)
thus a and b both work for a total of (xy)/(Wx+Wy) hours
fraction of work done by a = rate of a * time
= W/x * (xy)/(Wx+Wy) = (Wxy)/[xW(x+y)] = y/(x+y)
Question 2:

In case bblast's "assume height" point wasn't clear, you can think about the situation exactly 1 hour after the start. The volume will be 22 and the height will be 0.7

Using the formula: volume of cylinder = pi * r^2 * h
22 = pi * r^2 * 0.7

We are trying to find r, so divide both sides by 0.7:
22/0.7 = pi * r^2

You multiply 22/0.7 by 10/10 to get the equivalent 220/7. Since 210/7 = 30, we can figure out that 217/7 = 31 and 220/7 = 31 3/7, or around 31.4.

So, 31.4 = pi * r^2
You should recognize that 31.4 is about 10*pi, so 10*pi = pi * r^2
Divide both sides by pi to get 10 = r^2
So, r^2 is around 10 and r is around root(10).
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by MrR » Thu Aug 04, 2011 11:12 pm
For the second one:
Since for a cylinder:

V=pi*r^2*h
its also true that
delta V= pi*r^2*delta h

Deltas are used for "the change in" in Physics and other mathematical sciences.
We know the change in the volume corresponds to the change in the height because all other aspects of the problem are constant.

Now using the formula and the fact that pi can be approximated with the ratio 22/7 will get you to the same answer with less simplifications.

dV=pi*r^2*dh
22=22/7*r^2*.7
22=2.2r^2
10=r^2
sqrt(10)=r
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