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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]  17 Jan 2010, 10:10
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63% (03:23) correct 37% (02:23) wrong based on 586 sessions
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) %
[Reveal] Spoiler: OA
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Re: Last year the price per share of Stock X increased by k [#permalink]  15 Jul 2012, 22:44
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vaivish1723 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) %

Put in some values to figure it out.

Say original Price per share (PPS) = $100 and original Earnings per share (EPS) =$100
Ratio PPS/EPS = 100/100 = 1

Say PPS increased by 20% (k) and became 120
Say EPS increased by 10% (m) and became 110
Ratio PPS/EPS = 12/11

This is an increase of 1/11 = 100/11 %

Only option (D) gives you 100/11 when k = 20 and m = 10
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Re: simple but couldnt do on test [#permalink]  17 Jan 2010, 12:53
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vaivish1723 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
[Reveal] Spoiler:
d

Original price - $$x$$
Original earnings - $$y$$
Original ratio price per earnings - $$\frac{x}{y}$$

Increased price - $$x(1+\frac{k}{100})=\frac{x(100+k)}{100}$$
Increased earnings - $$y(1+\frac{m}{100})=\frac{y(100+m)}{100}$$
New ratio price per earnings- $$\frac{x(100+k)}{y(100+m)}$$

General formula for percent increase or decrease, (percent change):

$$Percent=\frac{Change}{Original}*100$$

$$\frac{\frac{x(100+k)}{y(100+m)}-\frac{x}{y}}{\frac{x}{y}}100=\frac{100(k-m)}{100+m}$$

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Re: simple but couldnt do on test [#permalink]  15 Jul 2012, 05:00
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superpus07 wrote:
Bunuel wrote:

$$\frac{\frac{x(100+k)}{y(100+m)}-\frac{x}{y}}{\frac{x}{y}}100=\frac{100(k-m)}{100+m}$$

Could you explain step-wise how you derived his one? I keep missing one step.

Thanks.

Sure: $$\frac{\frac{x(100+k)}{y(100+m)}-\frac{x}{y}}{\frac{x}{y}}100$$ --> reduce by $$\frac{x}{y}$$:

$$\frac{\frac{x(100+k)}{y(100+m)}-\frac{x}{y}}{\frac{x}{y}}100=\frac{\frac{100+k}{100+m}-1}{1}100=(\frac{100+k}{100+m}-1)*100=(\frac{100+k-100-m}{100+m})*100=\frac{100(k-m)}{100+m}$$.

Hope it's clear.
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Re: simple but couldnt do on test [#permalink]  15 Jul 2012, 03:06
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Bunuel wrote:

$$\frac{\frac{x(100+k)}{y(100+m)}-\frac{x}{y}}{\frac{x}{y}}100=\frac{100(k-m)}{100+m}$$

Could you explain step-wise how you derived his one? I keep missing one step.

Thanks.
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Re: Last year the price per share of Stock X increased by k [#permalink]  06 May 2013, 05:53
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Expert's post
vaivish1723 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) %

Responding to a pm: (super detailed steps)
When you have variables in the question and variables in the options, one method of getting to the answer is by assuming values for the variables. If a question says: x is greater than y/3 but less than y/2. Which of the following relations must hold?
(A) y < x < 2y
(B) 2y < 6x < 3y
...
...

One way of doing this is assuming some value for y. Say, y = 60. So x is greater than 20 but less than 30. Say x could be 25.
Not put y = 60 and x = 25 in the options:
(A) y < x < 2y -----> 60 < 25 < 2*120 ( does not hold)
(B) 2y < 6x < 3y ------> 2*60 < 6*25 < 3*60 (holds so could be the answer)

Mind you, you need to check every option to ensure that the relation holds for only one option. If it holds for more than one options, you will need to check for some other values too.

The point is that we assume values and then plug them in the options to see which relation works.

We can do this question using the same concept.
Last Year:
Price per share (PPS) increased by k%. Let's say k = 20. So if the price was 100, it increased by 20% to become 120 (New PPS).
Earnings per share increased by m%. Let's say m = 10 (m must be less than k). So if earnings per share was also 100, it increased by 10% to become 110 (New EPS).

By what percent did the ratio of price per share to earnings per share increase?

Ratio of PPS/EPS was 100/100 = 1
New ratio = New PPS/ New EPS = 120/110 = 1.0909

%Increase = [(1.0909 - 1)/1] * 100 = 9.09%

Now, in the options, put k = 20, m = 10
Only (D) will give you 9.09%. Let me show you calculations of some of the options.

B. (k – m) %
(20 - 10)% = 10% (Not the answer)

C. [100(k – m)] / (100 + k) %
[100(20 - 10)]/(100 + 20) % = 1000/120 % = 8.3% (Not the answer)

D. [100(k – m)] / (100 + m) %
[100(20 – 10)] / (100 + 10) % = 1000/11 % = 9.09% (Answer)
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Re: Last year the price per share of Stock X increased by k [#permalink]  29 Jul 2014, 07:03
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I think combination of number picking and algebraic approach gives power results. Here is how I solved it:

(P) share price= 100
(E) earnings=100
P/E = 1

New P/E = {(100+K)/100} / {(100+m)/100} = (100+k) / (100+m)

Percentage increase = [(100+k) / (100+m) - 1]* 100 = [(k-m)*100] / ( 100+m) --> Ans D
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Re: simple but couldnt do on test [#permalink]  29 Mar 2013, 09:26
Bunuel wrote:
superpus07 wrote:
Bunuel wrote:

$$\frac{\frac{x(100+k)}{y(100+m)}-\frac{x}{y}}{\frac{x}{y}}100=\frac{100(k-m)}{100+m}$$

Could you explain step-wise how you derived his one? I keep missing one step.

Thanks.

Sure: $$\frac{\frac{x(100+k)}{y(100+m)}-\frac{x}{y}}{\frac{x}{y}}100$$ --> reduce by $$\frac{x}{y}$$:

$$\frac{\frac{x(100+k)}{y(100+m)}-\frac{x}{y}}{\frac{x}{y}}100=\frac{\frac{100+k}{100+m}-1}{1}100=(\frac{100+k}{100+m}-1)*100=(\frac{100+k-100-m}{100+m})*100=\frac{100(k-m)}{100+m}$$.

Hope it's clear.

Hi could you explain please why do you convert x/y into 1?
thanks
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Re: simple but couldnt do on test [#permalink]  29 Mar 2013, 10:18
Bunuel wrote:
superpus07 wrote:
Bunuel wrote:

Could you explain step-wise how you derived his one? I keep missing one step.

Thanks.

Sure: $$\frac{\frac{x(100+k)}{y(100+m)}-\frac{x}{y}}{\frac{x}{y}}100$$ --> reduce by $$\frac{x}{y}$$:

$$\frac{\frac{x(100+k)}{y(100+m)}-\frac{x}{y}}{\frac{x}{y}}100=\frac{\frac{100+k}{100+m}-1}{1}100=(\frac{100+k}{100+m}-1)*100=(\frac{100+k-100-m}{100+m})*100=\frac{100(k-m)}{100+m}$$.

Hope it's clear.

Hi could you explain please why do you convert x/y into 1?
thanks

$$\frac{\frac{x}{y}}{\frac{x}{y}} = \frac{x}{y} * \frac{y}{x}=1$$
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Re: Last year the price per share of Stock X increased by k [#permalink]  25 Jan 2014, 12:52
but this number plugin should only work when you assume that price = earnings ?

VeritasPrepKarishma wrote:
vaivish1723 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) %

Put in some values to figure it out.

Say original Price per share (PPS) = $100 and original Earnings per share (EPS) =$100
Ratio PPS/EPS = 100/100 = 1

Say PPS increased by 20% (k) and became 120
Say EPS increased by 10% (m) and became 110
Ratio PPS/EPS = 12/11

This is an increase of 1/11 = 100/11 %

Only option (D) gives you 100/11 when k = 20 and m = 10
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Last year the price per share of Stock X increased by k [#permalink]  16 Jul 2014, 04:24
Bunuel wrote:
vaivish1723 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
[Reveal] Spoiler:
d

Original price - $$x$$
Original earnings - $$y$$
Original ratio price per earnings - $$\frac{x}{y}$$

Increased price - $$x(1+\frac{k}{100})=\frac{x(100+k)}{100}$$
Increased earnings - $$y(1+\frac{m}{100})=\frac{y(100+m)}{100}$$
New ratio price per earnings- $$\frac{x(100+k)}{y(100+m)}$$

General formula for percent increase or decrease, (percent change):

$$Percent=\frac{Change}{Original}*100$$

$$\frac{\frac{x(100+k)}{y(100+m)}-\frac{x}{y}}{\frac{x}{y}}100=\frac{100(k-m)}{100+m}$$

HI Bunuel,

Similar kind of question is mentioed on following link

currently-y-percent-of-the-members-on-the-finance-committee-33127.html

I just want to know what is the best way. I tried the above question is doable without picking numbers. but the question mentioned on above link i stuck there. Please provide your comments on the question mentioed in above link.

Thanks
Math Expert
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Posts: 27238
Followers: 4237

Kudos [?]: 41157 [0], given: 5672

Re: Last year the price per share of Stock X increased by k [#permalink]  17 Jul 2014, 09:05
Expert's post
PathFinder007 wrote:
Bunuel wrote:
vaivish1723 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
[Reveal] Spoiler:
d

Original price - $$x$$
Original earnings - $$y$$
Original ratio price per earnings - $$\frac{x}{y}$$

Increased price - $$x(1+\frac{k}{100})=\frac{x(100+k)}{100}$$
Increased earnings - $$y(1+\frac{m}{100})=\frac{y(100+m)}{100}$$
New ratio price per earnings- $$\frac{x(100+k)}{y(100+m)}$$

General formula for percent increase or decrease, (percent change):

$$Percent=\frac{Change}{Original}*100$$

$$\frac{\frac{x(100+k)}{y(100+m)}-\frac{x}{y}}{\frac{x}{y}}100=\frac{100(k-m)}{100+m}$$

HI Bunuel,

Similar kind of question is mentioed on following link

currently-y-percent-of-the-members-on-the-finance-committee-33127.html

I just want to know what is the best way. I tried the above question is doable without picking numbers. but the question mentioned on above link i stuck there. Please provide your comments on the question mentioed in above link.

Thanks

Number plugging approach is in Karishma's post there: currently-y-percent-of-the-members-on-the-finance-committee-33127.html#p1318632
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Re: Last year the price per share of Stock X increased by k [#permalink]  10 Jan 2015, 06:32
Earlier year share price = P
Earlier earnings = E
Ratio = P/E

Assume:
k = 50%
m = 25%

Therefore:
New share price = 1.5P
New earnings = 1.25E
New Ratio: 1.5P/1.25E = 150P/125E = 6P/5E = [1 + (1/5)]P/E
i.e. P/E ratio increased by 1/5 or 20%

Now, plug k = 50 and m = 25 in solutions. Only D yields 20%

Re: Last year the price per share of Stock X increased by k   [#permalink] 10 Jan 2015, 06:32
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