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505-555 (Easy)|   Percent and Interest Problems|                     
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Bunuel
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Sorry All but I was not able to view the options given. I am using Firefox browser.
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Sorry All but I was not able to view the options given. I am using Firefox browser.

This is a data sufficiency question. Options for DS questions are always the same.

The data sufficiency problem consists of a question and two statements, labeled (1) and (2), in which certain data are given. You have to decide whether the data given in the statements are sufficient for answering the question. Using the data given in the statements, plus your knowledge of mathematics and everyday facts (such as the number of days in July or the meaning of the word counterclockwise), you must indicate whether—

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
D. EACH statement ALONE is sufficient to answer the question asked.
E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

Hope this helps.
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Bunuel
The price per share of Stock X increased by 10 percent over the same time period that the price per share of Stock Y decreased by 10 percent. The reduced price per share of Stock Y was what percent of the original price per share of Stock X ?

Let the initial price per share of stock X be \(x\), so after increase by 10% it would become \(1.1x\);
Let the initial price per share of stock Y be \(y\), so after decrease by 10% it would become \(0.9y\).

Question: \(\frac{0.9y}{x}=\frac{9y}{10x}=?\)

(1) The increased price per share of Stock X was equal to the original price per share of Stock Y --> \(1.1x=y\) --> \(\frac{9y}{10x}=\frac{9*1.1x}{10x}=0.99\) or 99%. Sufficient.

(2) The increase in the price per share of Stock X was 10/11 the decrease in the price per share of Stock Y --> \(1.1x-x=\frac{10}{11}*(y-0.9y)\) --> \(0.1x=\frac{10}{11}*0.1y\) --> \(1.1x=y\) the same info as in (1). Sufficient

Answer: D.


Shouldn't the highlighted part be [ y- .1y] as its 10/11 of the decease and it is to be decreased by 10 %, correct me if I'm wrong.
Regards
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Bunuel
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Bunuel
The price per share of Stock X increased by 10 percent over the same time period that the price per share of Stock Y decreased by 10 percent. The reduced price per share of Stock Y was what percent of the original price per share of Stock X ?

Let the initial price per share of stock X be \(x\), so after increase by 10% it would become \(1.1x\);
Let the initial price per share of stock Y be \(y\), so after decrease by 10% it would become \(0.9y\).

Question: \(\frac{0.9y}{x}=\frac{9y}{10x}=?\)

(1) The increased price per share of Stock X was equal to the original price per share of Stock Y --> \(1.1x=y\) --> \(\frac{9y}{10x}=\frac{9*1.1x}{10x}=0.99\) or 99%. Sufficient.

(2) The increase in the price per share of Stock X was 10/11 the decrease in the price per share of Stock Y --> \(1.1x-x=\frac{10}{11}*(y-0.9y)\) --> \(0.1x=\frac{10}{11}*0.1y\) --> \(1.1x=y\) the same info as in (1). Sufficient

Answer: D.


Shouldn't the highlighted part be [ y- .1y] as its 10/11 of the decease and it is to be decreased by 10 %, correct me if I'm wrong.
Regards

It's should be the way it is written.

The decrease in the price per share of Stock Y is \(y - 0.9y\). 10/11 times that is \(\frac{10}{11}*(y-0.9y)\). I think you misunderstood the statement.
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Bunuel
The price per share of Stock X increased by 10 percent over the same time period that the price per share of Stock Y decreased by 10 percent. The reduced price per share of Stock Y was what percent of the original price per share of Stock X ?

Let the initial price per share of stock X be \(x\), so after increase by 10% it would become \(1.1x\);
Let the initial price per share of stock Y be \(y\), so after decrease by 10% it would become \(0.9y\).

Question: \(\frac{0.9y}{x}=\frac{9y}{10x}=?\)

(1) The increased price per share of Stock X was equal to the original price per share of Stock Y --> \(1.1x=y\) --> \(\frac{9y}{10x}=\frac{9*1.1x}{10x}=0.99\) or 99%. Sufficient.

(2) The increase in the price per share of Stock X was 10/11 the decrease in the price per share of Stock Y --> \(1.1x-x=\frac{10}{11}*(y-0.9y)\) --> \(0.1x=\frac{10}{11}*0.1y\) --> \(1.1x=y\) the same info as in (1). Sufficient

Answer: D.




Shouldn't the highlighted part be [ y- .1y] as its 10/11 of the decease and it is to be decreased by 10 %, correct me if I'm wrong.
Regards

It's should be the way it is written.

The decrease in the price per share of Stock Y is \(y - 0.9y\). 10/11 times that is \(\frac{10}{11}*(y-0.9y)\). I think you misunderstood the statement.




But the stem states that decrease in y is 10 %, Would we not write this question 1 -.1 =.9 . Sorry, i guess I am overlooking something. I am not understanding why are you subtracting .9y when question states that decrease is 10 % not 90 %.
Kindly explain
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But the stem states that decrease in y is 10 %, Would we not write this question 1 -.1 =.9 . Sorry, i guess I am overlooking something. I am not understanding why are you subtracting .9y when question states that decrease is 10 % not 90 %.
Kindly explain


The price per share of Stock Y decreased by 10 percent:
Let the initial price per share of stock Y be \(y\), so after decrease by 10% it would become \(0.9y\).
Amount by which it decreased is initial price - new price = y-0.9y.
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Walkabout
The price per share of Stock X increased by 10 percent over the same time period that the price per share of Stock Y decreased by 10 percent. The reduced price per share of Stock Y was what percent of the original price per share of Stock X ?

(1) The increased price per share of Stock X was equal to the original price per share of Stock Y.
(2) The increase in the price per share of Stock X was 10/11 the decrease in the price per share of Stock Y.

Important: Bunuel´s explanations (above) ARE FUNDAMENTAL to students who are beginning their "GMAT Quant development".
The solution we present below is for the students who already know "the basics".

(That´s why we postpone Data Sufficiency learning, in our course, till the student´s maturity has changed considerably!)

In our method we make a clear distinction between Problem Solving and Data Sufficiency.

In Data Sufficiency problems, we are FOCUSED in the UNIQUENESS OR NOT of a (numerical) value, not in calculations related to the possible value(s)...

Please read our solution, that starts below, with that in mind!

\(\left( {{\text{DATA}}} \right)\,\,\,\,X\,\,\, \to \,\,\,\frac{{11}}{{10}}X\,\,\,\,\,\,\,\,;\,\,\,\,\,\,\,\,Y\,\,\, \to \,\,\,\frac{9}{{10}}Y\)

\(\left( {{\text{FOCUS}}} \right)\,\,\,\,? = \frac{{\,\frac{9}{{10}}Y\,}}{X}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\boxed{? = \frac{Y}{X}}\,\,\,\,\,\,\,\,\,\,\,\left( {X \ne 0\,\,\,\,{\text{implicitly}}} \right)\)

\(\left( 1 \right)\,\,\,\frac{{11}}{{10}}X = Y\,\,\,\, \Rightarrow \,\,\,\,\,\frac{Y}{X}\,\,\,{\text{unique}}\,\,\,\, \Rightarrow \,\,\,\,\,{\text{SUFF}}.\)

\(\left( 2 \right)\,\,\,\frac{1}{{10}}X = \frac{{10}}{{11}}\left( {\frac{1}{{10}}Y} \right)\,\,\,\, \Rightarrow \,\,\,\,\,\frac{Y}{X}\,\,\,{\text{unique}}\,\,\,\, \Rightarrow \,\,\,\,\,{\text{SUFF}}.\)

(If you realize this is BRUTAL, SAFE and POWERFUL, you understood correctly. )


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Here are my two cents for this question:

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We are asked .9b is what % of a ----- (I)

From statement 1 we have:

1.1 a= b
or
a=\(\frac{10}{11} b\)

Since we have a relation between a and b we can definitely solve for the equation (I)
.9b=\(\frac{x}{100} \frac{10}{11} b\)

Statement 2 we have:
Increase in per share price of stock X is .1 a , and\( \frac{10}{11}\) decrease in price per share of stock y is \(\frac{1}{11} b \)
and we are told that .1a = \(\frac{1}{11} b \)
or a=\( \frac{10}{11}\) b which is essentially same as statement 1 .

So both 1 and 2 Statement individually help to answer our question
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Bunuel
The price per share of Stock X increased by 10 percent over the same time period that the price per share of Stock Y decreased by 10 percent. The reduced price per share of Stock Y was what percent of the original price per share of Stock X ?

Let the initial price per share of stock X be \(x\), so after increase by 10% it would become \(1.1x\);
Let the initial price per share of stock Y be \(y\), so after decrease by 10% it would become \(0.9y\).

Question: \(\frac{0.9y}{x}=\frac{9y}{10x}=?\)

(1) The increased price per share of Stock X was equal to the original price per share of Stock Y --> \(1.1x=y\) --> \(\frac{9y}{10x}=\frac{9*1.1x}{10x}=0.99\) or 99%. Sufficient.

(2) The increase in the price per share of Stock X was 10/11 the decrease in the price per share of Stock Y --> \(1.1x-x=\frac{10}{11}*(y-0.9y)\) --> \(0.1x=\frac{10}{11}*0.1y\) --> \(1.1x=y\) the same info as in (1). Sufficient

Answer: D.


But the stem states that decrease in y is 10 %, Would we not write this question 1 -.1 =.9 . Sorry, i guess I am overlooking something. I am not understanding why are you subtracting .9y when question states that decrease is 10 % not 90 %.
Kindly explain

mtk10
the \(decrease\) in the price per share of Stock Y is \(0.1y\). (The decreased price is 0.9y). Both are not the same cases.
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Please correct if i'm wrong here. Assuming original prices are X and Y, we need to find 0.9Y/X*100.

St. 1 -- Sufficient. It tells us 1.1X = Y. We can find out 0.9Y/X %.
St. 2 -- Sufficient. It tells us 0.1X = (10/11)*0.1Y. We can again find out 0.9Y/X % through the given information.

Therefore, D is the correct answer.
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Walkabout
The price per share of Stock X increased by 10 percent over the same time period that the price per share of Stock Y decreased by 10 percent. The reduced price per share of Stock Y was what percent of the original price per share of Stock X ?

(1) The increased price per share of Stock X was equal to the original price per share of Stock Y.
(2) The increase in the price per share of Stock X was 10/11 the decrease in the price per share of Stock Y.





Nick Slavkovich, GMAT/GRE tutor with 20+ years of experience

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