Last year the price per share of Stock X increased by k : GMAT Problem Solving (PS)
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# Last year the price per share of Stock X increased by k

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Last year the price per share of Stock X increased by k [#permalink]

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07 Dec 2007, 17:39
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75% (02:31) correct 25% (02:05) wrong based on 63 sessions

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

OPEN DISCUSSION OF THIS QUESTION IS HERE: last-year-the-price-per-share-of-stock-x-increased-by-k-89307.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 21 Mar 2013, 02:49, edited 1 time in total.
Moved to PS forum.
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07 Dec 2007, 20:32
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Ravshonbek 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. m/k %
B. (k – m) %
C. [100(k – m)] / (100 + k) %
D. [100(k – m)] / (100 + m) %
E. [100(k – m)] / (100 + k + m) %

OA D

Let P = the old price per share; E = the old earning per share. Thus P/E is the price to earning ratio before the increases

After the increase the new price is: P*(1+k/100) and the new earning is: E*(1 + m/100)

The new P/E is: (1+k/100)P/(1+m/100)E

The Percent of P/E increase = (new P/E - P/E)/(P/E). Subsititute new P/E to the equation we have:

[(1+k/100)/(1+m/100)*P/E - P/E]/(P/E)*100%. Simplifly the expression and you should get the answer to be:

D: 100*(k-m)/(100+m) %
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07 Dec 2007, 23:37
Ravshonbek 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. m/k %
B. (k – m) %
C. [100(k – m)] / (100 + k) %
D. [100(k – m)] / (100 + m) %
E. [100(k – m)] / (100 + k + m) %

OA D

D. old P/E ratio = p/e
new P/E ratio = p(1+k)/e(1+m)
% change in P/E ratio = [p(100+k)/e(100+m) - p/e] / [p/e]
[p(100+k) - p (100+m)] / [e (100+m)]} / {p/e}
{[100 + k - 100 - m] / (100 + m)}
100 (k - m) / (100 + m)%
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20 Mar 2013, 09:37
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?
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20 Mar 2013, 21:06
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Expert's post
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
$$p = [\frac{(k+100)}{(m + 100)} - 1]* 100 = (k - m)/(m + 100)*100$$

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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Re: Last year the price per share of Stock X increased by k [#permalink]

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21 Mar 2013, 02:48
Ravshonbek 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. m/k %
B. (k – m) %
C. [100(k – m)] / (100 + k) %
D. [100(k – m)] / (100 + m) %
E. [100(k – m)] / (100 + k + m) %

OA D

OPEN DISCUSSION OF THIS QUESTION IS HERE: last-year-the-price-per-share-of-stock-x-increased-by-k-89307.html
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Re: Last year the price per share of Stock X increased by k   [#permalink] 21 Mar 2013, 02:48
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