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C

revenue 2008 = 100
profir = 10

revenue 2009 =80
profit=12

ans=(12/10)*100=120%
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Bunuel
In 2008, the profits of Company N were 10 percent of revenues. In 2009, the revenues of Company N fell by 20 percent, but profits were 15 percent of revenues. The profits in 2009 were what percent of the profits in 2008?

A. 80%
B. 105%
C. 120%
D. 124.2%
E. 138%


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In 2008, Let the revenue be 100 => Profit = 10 ---- 1

In 2009, Revenue fell by 20% and Profits were 15%=> R = 80 and P = 15% of 80 =12 ----- 2

From 1 and 2, 12 is 120% of 10.

Hence Answer is C
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In 2008, the profits of Company N were 10 percent of revenues. In 2009, the revenues of Company N fell by 20 percent, but profits were 15 percent of revenues. The profits in 2009 were what percent of the profits in 2008?

A. 80%
B. 105%
C. 120%
D. 124.2%
E. 138%




We can form 3 equations:
p1 = 0.1r1 ; p2 = 0.15r2 ; r2 = 0.8r1

therefore, from the 2nd eq.,
p2 = 0.15 * 0.8 * r1
= 0.15 * 0.8 * p1/0.1
= 1.2 * p1

hence [D]
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We can directly use numbers for simple question to save time.
2008
Revenue = 100
Profit = 10

2009
Revenue = 80
Profit = 0.15*80=12

Required percentage:
(12/10)*100=120%

Answer: C
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Bunuel
In 2008, the profits of Company N were 10 percent of revenues. In 2009, the revenues of Company N fell by 20 percent, but profits were 15 percent of revenues. The profits in 2009 were what percent of the profits in 2008?

A. 80%
B. 105%
C. 120%
D. 124.2%
E. 138%


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KAPLAN OFFICIAL SOLUTION:

In business school you’ll be talking about the process by which a firm would experience reduced revenue but still manage to increase profits (hint: think off-shoring). However, the qualitative side is irrelevant to this problem.

We jump into the problem to organize our data points. In this case, it is best to pick a number to represent both revenues and profits at Company N. Any good ideas on a good number to pick? Let’s go with: Revenues = 100 and profits = 10 in 2008. On test day, you’ll write these numbers down on your noteboard. Then, the next step is to apply the 2009 changes to these figures. In 2009, the revenues and profits were, 80 and 12, respectively.

Ok. We have our figures. Here is where we need to make sure we slow down a bit on test day. Let’s look at our answer choices:

80%
105%
120%
124.2%
138%

Within these answer choices, we see several attractive answers: 80% is the revenue in 2009 compared to 2008 – not the question we are trying to answer. 105% is simply taking the 20 percent revenue decline and adding back the 15 percent profit increase. Totally wrong approach and totally wrong answer (side note: When dealing with % in Problem Solving questions, don’t fall for the answer choice that simply adds or subtracts the percentages – those are almost never right). Finally, answer choice 124.2% is just different – if you had to guess on this question, it is so different from the rest of the choices that it does appear viable; however, those ‘different’ answer choices are not often the correct answer.

It is important to review the actual question in this problem: “The profits in 2009 were what percent of the profits in 2008?”. Make sure you plug the correct numbers into this situation: 2008 profits = 10, 2009 profits = 12. Thus, 120%.

On test day, it is easy to be rushing along and have an answer that is correct for a different question. Kaplan’s time tested method requires you to double check your selected answer against the problem – just to make sure you are answering the right question. In our new flagship GMAT course, we spend time practicing and drilling your ability to quickly outline the relevant material in the question, organize it against several of our strategies, and double check that it is the right answer to the right question.

Good luck as you study!

Brian Fruchey
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If P(x), P(y), R(x), and R(y) represent the profits for 2008 and 2009/revenues for 2008 and 2009, respectively, the goal is to find
[P(y)/P(x)] x 100.

We know
P(x) = 0.1R(x)
R(y)=R(x) - .2R(x) = .8R(x)
P(y) = .15R(y)

Substituting for P(y) and P(x) in the original equation gives
[.15R(y)]/[.1R(x)]
Substituting once again, this time for R(y) gives
[.15(.8)R(x)]/[.1R(x)]

Finally, [(.15 x .8)/.1] x 100
=120%
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