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

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17 Jan 2010, 11:10

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63% (03:29) correct
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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. k/m % B. (k – m) % C. [100(k – m)] / (100 + k) % D. [100(k – m)] / (100 + m) % E. [100(k – m)] / (100 + k + m) %

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

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17 Jan 2010, 13:53

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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) %

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):

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

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15 Jul 2012, 23:44

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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 _________________

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

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06 May 2013, 06:53

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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) %

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.

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

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25 Jan 2014, 13:52

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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

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

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16 Jul 2014, 05:24

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) %

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):

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.

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

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17 Jul 2014, 10:05

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) %

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):

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.

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

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10 Jan 2015, 07:32

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]

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10 Dec 2015, 04:17

Dear Bunuel, Would you be so kind to share with us similar questions of hard level?

Thanks in advance

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) %

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):

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

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13 Mar 2016, 08:11

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

Karishma, in what cases would you advise to put in values rather than solve algebraically?

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

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13 Mar 2016, 21:21

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) %

Let Original price =100 Let original earnings per share =100 Price/ earnings =1

New price = 100+k New earnings per share= 100+m price /earnings = (100+k)/(100+m)

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

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13 Mar 2016, 21:37

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) %

I just plugged in values for k = 20 and m = 10. The answer you get is 100/11 Again, I plugged in values for k and m in the options. Only option D is true.

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

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14 Mar 2016, 01:19

Expert's post

Viktoriaa wrote:

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

Karishma, in what cases would you advise to put in values rather than solve algebraically?

When the question doesn't give any concrete numbers and talks in terms of variables (the answer options would also be in terms of variables in that case), you can plug in values to get to the answer. Percentage questions often provide an opportunity to plug in values too, say 100 for a base value. _________________

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