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by rkhicha » Tue Aug 17, 2010 11:21 pm
Last year the price per share of Stock X increased by k percent and the earnings per share of Stock X increased by m percent, where k is greater than m. By what percent did the ratio of price per share to earnings per share increase, in terms of k and m?

A. k/m %
B. (k - m) %
C. [100(k - m)] / (100 + k) %
D. [100(k - m)] / (100 + m) %
E. [100(k - m)] / (100 + k + m) %

OA later
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Source: — Problem Solving |

by kvcpk » Tue Aug 17, 2010 11:57 pm
rkhicha wrote:Last year the price per share of Stock X increased by k percent and the earnings per share of Stock X increased by m percent, where k is greater than m. By what percent did the ratio of price per share to earnings per share increase, in terms of k and m?

A. k/m %
B. (k - m) %
C. [100(k - m)] / (100 + k) %
D. [100(k - m)] / (100 + m) %
E. [100(k - m)] / (100 + k + m) %

OA later
Let price per share increase by 100%. whicmeans price per share is doubled. k=100
Let earnings per share did not increase at all. m=0

Earlier ratio was p/e
now, 2p/e
perc increase in ratio is 100%

On substitution, only B and D remain.

B is obvous answer and most probably wrong. Under time pressure, we can choose D from here and move on.

Else put , K=100, m=50
B says ratio increases by 50%
which means new ratio should be 3P/2e

But actually new ratio is 2p/3e/2 = 4p/3e

Hence pick D

Hope this helps!!
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by gmatmachoman » Wed Aug 18, 2010 3:16 am
Praveen was spot on... wording seems to be damn tricky...

But here goes algebraic approach:

P= stock price

the price per share of Stock X increased by k percent Pnew= P+( k/100 )* X
= P ( 1 + {K/100})

earnings per share = E
the earnings per share of Stock X increased by m percent ; Enew= E ( 1 + {m/100})

[(Pnew/Enew) - ( P/E) ]/ {P/E} --- Equation 1

Pnew/Enew = P ( 1 + {K/100}) / {E ( 1 + {m/100})}
P/E ( (100+K)/(100 +m)

Substutute the values of Pnew/Enew

you will get = [100(k - m)] / (100 + m) %

Pick D
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by rkhicha » Wed Aug 18, 2010 4:50 am
Much clearer now. OA is D
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