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# Last year the price per share of Stock X increased by k percent and

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Joined: 16 Oct 2010
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Location: Pune, India
Re: Last year the price per share of Stock X increased by k percent and [#permalink]

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19 Jun 2016, 21:24
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

How did you go from 12/11 to an increase of 100/11 percent?

According to original prices, PPS/EPS = 1
After the increase, PPS/EPS = 12/11

Percentage Increase in PPS/EPS $$=\frac{ (\frac{12}{11} - 1)}{1} * 100 = \frac{100}{11} %$$
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Re: Last year the price per share of Stock X increased by k percent and [#permalink]

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08 Jul 2016, 08:59
1
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.

how are you getting the -m in this step?

$$(\frac{100+k}{100+m}-1)*100=(\frac{(100+k-100-m}{100+m})*100$$
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Re: Last year the price per share of Stock X increased by k percent and [#permalink]

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08 Jul 2016, 09:16
nycgirl212 wrote:

how are you getting the -m in this step?

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

$$\frac{100+k}{100+m}-1=\frac{100+k}{100+m}-\frac{100+m}{100+m}=\frac{100+k-100-m}{100+m}$$
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Last year the price per share of Stock X increased by k percent and [#permalink]

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Updated on: 13 Jan 2018, 07:22
Top Contributor
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's use the INPUT-OUTPUT approach.

Another approach is to plug in values.

Let $100 be the original price per share of Stock X Choose a "nice" value for k. How about k = 200 So, after a 200% increase, the new price per share =$300

Let $100 be the original earnings per share of Stock X Choose a "nice" value for m. How about m = 100 So, after a 100% increase, the new earnings per share =$200

Original ratio of price/earnings = $100/$100 = 1
New ratio of price/earnings = $300/$200 = 1.5

By what percent did the ratio of price per share to earnings per share increase?
So, the percent increase (from 1 to 1.5) is 50%.
In other words, when k = 200 and m = 100, the ratio increases 50%

Now, plug in 200 for k, and 100 for m, and look for the answer choice that also yields 50%.

A. k/m = 200/100 = 2 (nope)

B. (k - m) = 200 - 100 = 100 (nope)

C. [100(k - m)] / (100 + k) = 10,000/300 = 33.333 (nope)

D. [100(k - m)] / (100 + m) = 10,000/200 = 50 GREAT!

E. [100(k - m)] / (100 + k + m) = 10,000/400 = 25 (nope)

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Originally posted by GMATPrepNow on 26 Jul 2016, 11:21.
Last edited by GMATPrepNow on 13 Jan 2018, 07:22, edited 1 time in total.
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Re: Last year the price per share of Stock X increased by k percent and [#permalink]

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15 Oct 2016, 09:37
Top Contributor
SW4 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 %

One approach is to plug in values.

Let $100 be the original price per share of Stock X Choose a "nice" value for k. How about k = 200 So, after a 200% increase, the new price per share =$300

Let $100 be the original earnings per share of Stock X Choose a "nice" value for m. How about m = 100 So, after a 100% increase, the new earnings per share =$200

Original ratio of price/earnings = $100/$100 = 1
New ratio of price/earnings = $300/$200 = 1.5

By what percent did the ratio of price per share to earnings per share increase?
So, the percent increase (from 1 to 1.5) is 50%.
In other words, when k = 200 and m = 100, the ratio increases 50%

Now, plug in 200 for k, and 100 for m, and look for the answer choice that also yields 50%.

A. k/m = 200/100 = 2 (nope)

B. (k - m) = 200 - 100 = 100 (nope)

C. [100(k - m)] / (100 + k) = 10,000/300 = 33.333 (nope)

D. [100(k - m)] / (100 + m) = 10,000/200 = 50 GREAT!

E. [100(k - m)] / (100 + k + m) = 10,000/400 = 25 (nope)

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

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12 May 2017, 17:56
I solved it as below:

Old Price = x, Old Earnings = y, Old Ratio = x/y

New Price = x + k% x = x (1+k/100) = x ( k + 100)/100
New Earnings = y + m% y = y ( 1+m/100) = y ( m+100)/100

New Ration = x(k+100) / y(m+100)
We need to find New Ratio is how much % greater than old ratio. This is the key point to understand and represent.

Expressing in Mathematical Form:
New Ratio = p% more than Old Ratio
New Ratio = Old Ratio + p% Old Ratio
x(100+k)/y(100+m) = x/y ( 1 + p/100)
Simplifying
p = (k-m)*100/m+100.
Option D
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Re: Last year the price per share of Stock X increased by k percent and [#permalink]

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04 Jul 2017, 12:31
Because this percent question doesn't ask for the actual numbers anywhere, we can plug in numbers

Let us say the original price = 100, and the original earnings = 200 (I chose different numbers just to make sure they didn't accidentally cancel each other out somewhere)

Word problems are hard for me, and especially when it's asking for multiple things things like this one (a ratio, a difference of ratios, and a percent) so I wrote down in a formula what the last sentence was asking for:

The percent increase of the two ratios = 100* [(ratio 2 - ratio 1)/(ratio 1)

Ratio 1 = original price/original earnings = 100/200 = 1/2

Ratio 2 = new price/new earnings
new price = 100*(1+(k/100) = 100 + (100k/100) = 100 + k
New earning = 200*(1+m/100) = 200 + 200m/100 = 2(100+m
Ratio 2 = (100+k)/2(100+m)

Solve for the percent inscrease of the two ratios
((100+k)/2(100+m)) - 1/2)/1/2
((100+k)/2(100+m) - (100 + m)/2(100+m))/12
(100+k-100-m)/2(100+m) / (1/2)
(k-m)/2(100+m) * 2
(k-m)/(100+m) * 100 (we multiply by 100 to get it in % form)
100(k-m)/(100+m)
D!
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Re: Last year the price per share of Stock X increased by k percent and [#permalink]

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09 Sep 2017, 21:51
SravnaTestPrep 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) %

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)

Percent increase =((100+k)/(100+m) - 1) *100
= (k-m)*100/(100+m)

hi

Would you please speak few more words about the expression "((100+k)/(100+m) - 1)" ....? especially about 1...? Is this something as such that "1", when subtracted from a fraction reveals the change, increase or decrease ....?

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

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09 Sep 2017, 22:03
gmatcracker2017 wrote:
SravnaTestPrep 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) %

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)

Percent increase =((100+k)/(100+m) - 1) *100
= (k-m)*100/(100+m)

hi

Would you please speak few more words about the expression "((100+k)/(100+m) - 1)" ....? especially about 1...? Is this something as such that "1", when subtracted from a fraction reveals the change, increase or decrease ....?

Hi,

Percent increase =( (New value - original value)/ original value ) * 100
New value = (100+k)/(100+m)

Original value=1

Substituting the above values you get the answer.
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Re: Last year the price per share of Stock X increased by k percent and [#permalink]

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01 Dec 2017, 14:27
Experts,

Could you kindly suggest other similar problems to practice on?

Aside from the link below that is:
https://gmatclub.com/forum/currently-y- ... l#p1318632

Thanks a lot!
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Re: Last year the price per share of Stock X increased by k percent and [#permalink]

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14 Dec 2017, 01:36
I find it impossible to be solved in <2'
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Joined: 03 Jul 2015
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Re: Last year the price per share of Stock X increased by k percent and [#permalink]

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19 Dec 2017, 01:33
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. $$\frac{k}{m} %$$

B. $$(k – m) %$$

C. $$\frac{100(k – m)}{(100 + k)} %$$

D. $$\frac{100(k – m)}{(100 + m)} %$$

E. $$\frac{100(k – m)}{(100 + k + m)} %$$

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):

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

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

Hi Bunuel,

Why cant we use the following formula to get the answer:

numerator: the delta increase, calculated as follows:

x . k% divided by y . m% -> this will give us the delta increase of the ratio

denominator: x divided by y (the original ratio)

both numerator and denominator then multipled by 100 to get the percent. I did this calculation and got 100k/m% as a result - did i miss anything ?
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Re: Last year the price per share of Stock X increased by k percent and [#permalink]

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27 Dec 2017, 00:46
Fernandocma wrote:
I find it impossible to be solved in <2'

I agree, it is time consuming one... Substitution would work if you pick up convenient numbers like 100/100 and increase by 200 and 100, otherwise, substitution is time consuming...
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Last year the price per share of Stock X increased by k percent and [#permalink]

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18 Apr 2018, 11:52
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. $$\frac{k}{m} %$$

B. $$(k – m) %$$

C. $$\frac{100(k – m)}{(100 + k)} %$$

D. $$\frac{100(k – m)}{(100 + m)} %$$

E. $$\frac{100(k – m)}{(100 + k + m)} %$$

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):

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

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

hello generis, so what book did you read today ? please tell me in two words

the more i tag you the more new english words i learn from you that will help me during RC and of course working on math

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

why do we deduct price from percentage or is it increased price denoted as percentage ?

And why are we dividing by fraction $$\frac{x}{y}$$, ?? we already have fraction $$\frac{x}{y}$$ in numerator ...

have a lovely jovely day / night
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Joined: 22 May 2016
Posts: 1825
Last year the price per share of Stock X increased by k percent and [#permalink]

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18 Apr 2018, 16:01
1
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. $$\frac{k}{m} %$$

B. $$(k – m) %$$

C. $$\frac{100(k – m)}{(100 + k)} %$$

D. $$\frac{100(k – m)}{(100 + m)} %$$

E. $$\frac{100(k – m)}{(100 + k + m)} %$$

dave13 wrote:
Bunuel wrote:

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):

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

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

hello generis, so what book did you read today ? please tell me in two words

the more i tag you the more new english words i learn from you that will help me during RC and of course working on math

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

why do we deduct price from percentage or is it increased price denoted as percentage ?

And why are we dividing by fraction $$\frac{x}{y}$$, ?? we already have fraction $$\frac{x}{y}$$ in numerator ...

have a lovely jovely day / night

Hi dave13 ,

Are you on a mission lately to find demonic
algebra?
I am teasing you. I'm always happy to help if I can.

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

why do we deduct price from percentage . . .
If you refer to $$\frac{x}{y}$$:

1) $$\frac{x}{y} = \frac{P}{E}$$ ratio

$$x$$ = original price per share
$$y$$ = original earnings per share

2) Bunuel is not deducting price from percentage

$$\frac{x}{y}$$ is the ORIGINAL or OLD $$\frac{P}{E}$$ ratio

He is deducting old $$\frac{P}{E}$$ ratio from new $$\frac{P}{E}$$ ratio
which must be done
in order to calculate percent change
See below

or is it increased price denoted as percentage ?
Not sure what "it" refers to.
I think it is $$\frac{x}{y}$$

The numerator, $$x$$, is indeed
The new price expressed in terms of percent increase (good catch):

$$x$$ = original price per share
That price increases by $$k$$ percent

$$k$$ percent = $$\frac{k}{100}$$

So $$x$$ increases by $$k$$ percent OF $$x$$:
$$x + (\frac{k}{100}*x)$$ = new price

Simplify. Factor out the $$x$$
New Price = $$x(1 + \frac{k}{100})$$

Simplify inside the parentheses. $$(1 +\frac{k}{100})$$
$$1 = \frac{100}{100}$$, so $$\frac{100}{100} + \frac{k}{100} = \frac{100 +k}{100}$$
The $$x$$ outside the parentheses is still there. Result: $$\frac{x(100+k)}{100}$$

I can't decide whether you're confused about
subtracting $$\frac{x}{y}$$ or $$m$$ from $$k$$

Either way, both subtractions happen because . . .
This expression is needed to calculate percent change.

Percent change can be defined as

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

To make the concept of "change" clearer,
percent change can be defined equivalently as

$$\frac{New - Old}{Old} * 100$$

New: $$\frac{x(100+k)}{y(100+m)}$$

Old: $$\frac{x}{y}$$

Try plugging those values into the
second variation of the percent change formula.

And why are we dividing by fraction $$\frac{x}{y}$$? we already have fraction $$\frac{x}{y}$$ in numerator ...

We are dividing by $$\frac{x}{y}$$ because we are
calculating percent change.

$$\frac{x}{y}$$ is . . . the old / original $$\frac{P}{E}$$ ratio

See the footnote, which contains an example of fraction to fraction percent change.

This problem is hard.
It contains a strange sort of ratio
(price-per-share to earnings-to-share is
hugely influential but not, IMO, intuitive);
the difference of ratios
(new and old, in which price growth k percent > earnings growth m percent);
AND a percent increase.

You are not a fan
of substituting values, it appears.

try going through both of Bunuel's posts
with pencil and paper in hand.
By "both posts" I mean the one you quoted and the one HERE.

I hope that helps.

Example of percent change with a fraction
(please note that I must divide by the original $$\frac{1}{4}$$):

Last year at the zoo, 1 animal in 4 was a newborn
This year at the zoo, 2 animals in 5 were newborns

By what percent did the ratio of newborns to all animals increase?
Percent change:
$$(\frac{New - Old}{Old} * 100)$$
$$\frac{\frac{2}{5}-\frac{1}{4}}{\frac{1}{4}}*100$$
$$\frac{\frac{3}{20}}{\frac{1}{4}}*100 = \frac{3}{20}*\frac{4}{1}*100=$$
$$\frac{12}{20}*100=.6 * 100 = 60$$ percent increase

P.S. The book I [am] read[ing] today? In two words?
What kind of restriction is THAT?

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

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18 Apr 2018, 17:42
Hi
This is one of the basic question and can very well done with in one minute. All you require is practice.Practice a large number of quality questions from gmatclub.

Fernandocma wrote:
I find it impossible to be solved in <2'

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

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19 Apr 2018, 06:39
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.[/quote]

how are you getting the -m in this step?

$$(\frac{100+k}{100+m}-1)*100=(\frac{(100+k-100-m}{100+m})*100$$[/quote]

generis, many thanks for taking time to explain i simply would love to improve algebra never was good at it also I like to learn to solve questions in different ways, its not only about GMAT just in life you know

can you please explain in the above mentioned explanation by Bunuel when he reduces by $$\frac{x}{y}$$:

before we reduce by $$\frac{x}{y}$$ in the numerator there are TWO fractions of $$\frac{x}{y}$$: and in the denominator there is ONLY one fraction $$\frac{x}{y}$$

so when we reduce by $$\frac{x}{y}$$: BOTH fractions of $$\frac{x}{y}$$: dissapear in numerator why ? normally one fraction $$\frac{x}{y}$$ should be left after reduction ... by which one ... there is one fraction in fron of brackets and another one with minus sign any idea why this happens /... both fraction in numerator dissapear ?

p.s
books with titles containg more than two words could be some fantasy book
you can tell me two words and i will guess haha like first word and last one:)
Re: Last year the price per share of Stock X increased by k percent and   [#permalink] 19 Apr 2018, 06:39

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