2flY wrote:

GMAT TIGER wrote:

{[100 + k - 100 - m] / (100 + m)}

100 (k - m) / (100 + m)%

Why doesn't the last step reduce to:

k - m / (100 + m)

as the 100 drops out?

\frac{New P}{New E} = \frac {P*(1 + k/100)}{E*(1 + m/100)}\frac{New P}{New E} = \frac{P}{E} * \frac{(k+100)}{(m + 100)}So P/E increases by a factor of

\frac{(k+100)}{(m + 100)} = (1 +\frac{p}{100}) where p is the percentage increase in P/E

p = [\frac{(k+100)}{(m + 100)} - 1]* 100 = (k - m)/(m + 100)*100So answer is (D)

You don't drop the 100 because you need to get the value of p in % terms (the answer options have % at the end) i.e. if the value of p comes out to be 20%, you need to get the number 20 so you retain the 100.

Take numbers to understand this. (Anyway, I would use numbers to solve this question)

Say k = 20%, m = 10%

NewP/NewE = 1.2P/1.1E = P/E * 12/11 = P/E * (1 + 1/11)

Now, we know that 1/11 = 9.09/100 i.e. 9.09%. The value given in the options are of 9.09.

Put k = 20 and m = 10 in the option (D).

You get [100(k – m)] / (100 + m) % = [100(20 – 10)] / (100 + 10) % = 100/11 % = 9.09%

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Karishma

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