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x2y = 10,000y – 100xy [multiplying by the common denominator 10,000] x2y + 100xy – 10,000y = 0 [everything to one side, because it’s quadratic] y(x2 + 100x – 10,000) = 0 [factoring] Therefore, the answer to the prompt question is affirmative if either x2 + 100x – 10,000 = 0 or y = 0.

(1) SUFFICIENT: This statement rearranges to give = 0.

(2) INSUFFICIENT: y cannot be 0, but no information is provided about x, making it impossible to determine whether x2 + 100x – 10,000 = 0.

Re: MGMAT test 4: X Percent of X Percent [#permalink]

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21 Jun 2010, 17:06

Expert's post

4

This post was BOOKMARKED

zisis wrote:

Is x% of x% of y equal to x% less than y ?

(1) x(x + 100) = 10,000 (2) y(y + 1) = 1

OA: A

x2y = 10,000y – 100xy [multiplying by the common denominator 10,000] x2y + 100xy – 10,000y = 0 [everything to one side, because it’s quadratic] y(x2 + 100x – 10,000) = 0 [factoring] Therefore, the answer to the prompt question is affirmative if either x2 + 100x – 10,000 = 0 or y = 0.

(1) SUFFICIENT: This statement rearranges to give = 0.

(2) INSUFFICIENT: y cannot be 0, but no information is provided about x, making it impossible to determine whether x2 + 100x – 10,000 = 0.

But, I believe D: A sufficient B: y=1 or y+1=1 therefore y=0 therefore B sufficient...

what am i missing? please help

Question: is \(\frac{x}{100}*\frac{x}{100}*y=y(1-\frac{x}{100})\)? --> is \(x^2y=100y(100-x)\)? --> is \(x^2y=y(10,000-100x)\) --> is \(y(x^2+100x-10,000)=0\)?

Basically question is does \(y=0\) or/and \(x^2+100x-10,000=0\)?

(2) \(y(y+1)=1\). Here it's clear that \(y\neq{0}\), (substitute \(y=0\) in this equation: \(0(0+1)=0\neq{1}\)). So we know that \(y\neq{0}\), but don't know whether \(x^2+100x-10,000=0\)? Not sufficient.

To elaborate more: the problem with your solution is that you solved incorrectly \(y(y+1)=1\). \(y(y+1)=1\) --> \(y^2+y-1=0\) --> solving for \(y\): \(y=\frac{-1-\sqrt{5}}{2}\) or \(y=\frac{-1+\sqrt{5}}{2}\), so \(y\neq{0}\).

Question: is \(\frac{x}{100}*\frac{x}{100}*y=y(1-\frac{x}{100})\)? --> is \(x^2y=100y(100-x)\)? --> is \(x^2y=y(10,000-100x)\) --> is \(y(x^2+100x-10,000)=0\)?

Basically question is \(y=0\) or/and \(x^2+100x-10,000=0\)?

(2) \(y(y+1)=1\). Here it's clear that \(y\neq{0}\), (substitute \(y=0\) in this equation: \(0(0+1)=0\neq{1}\)). So we know that \(y\neq{0}\), but don't know whether \(x^2+100x-10,000=0\)? Not sufficient.

Re: MGMAT test 4: X Percent of X Percent [#permalink]

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07 Jul 2012, 04:22

Bunuel wrote:

zisis wrote:

Is x% of x% of y equal to x% less than y ?

(1) x(x + 100) = 10,000 (2) y(y + 1) = 1

OA: A

x2y = 10,000y – 100xy [multiplying by the common denominator 10,000] x2y + 100xy – 10,000y = 0 [everything to one side, because it’s quadratic] y(x2 + 100x – 10,000) = 0 [factoring] Therefore, the answer to the prompt question is affirmative if either x2 + 100x – 10,000 = 0 or y = 0.

(1) SUFFICIENT: This statement rearranges to give = 0.

(2) INSUFFICIENT: y cannot be 0, but no information is provided about x, making it impossible to determine whether x2 + 100x – 10,000 = 0.

But, I believe D: A sufficient B: y=1 or y+1=1 therefore y=0 therefore B sufficient...

what am i missing? please help

Question: is \(\frac{x}{100}*\frac{x}{100}*y=y(1-\frac{x}{100})\)? --> is \(x^2y=100y(100-x)\)? --> is \(x^2y=y(10,000-100x)\) --> is \(y(x^2+100x-10,000)=0\)?

Basically question is does \(y=0\) or/and \(x^2+100x-10,000=0\)?

(2) \(y(y+1)=1\). Here it's clear that \(y\neq{0}\), (substitute \(y=0\) in this equation: \(0(0+1)=0\neq{1}\)). So we know that \(y\neq{0}\), but don't know whether \(x^2+100x-10,000=0\)? Not sufficient.

To elaborate more: the problem with your solution is that you solved incorrectly \(y(y+1)=1\). \(y(y+1)=1\) --> \(y^2+y-1=0\) --> solving for \(y\): \(y=\frac{-1-\sqrt{5}}{2}\) or \(y=\frac{-1+\sqrt{5}}{2}\), so \(y\neq{0}\).

Answer: A.

Hope it's clear.

Can you explain this part \(\frac{x}{100}*\frac{x}{100}*y=y(1-\frac{x}{100})\)? --> my R.H.S of equation is coming as y-x/100 kindly correct me if i am wrong _________________

_______________________________________________________________________________________________________________________________ If you like my solution kindly reward me with Kudos.

Re: MGMAT test 4: X Percent of X Percent [#permalink]

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07 Jul 2012, 04:27

1

This post received KUDOS

Expert's post

riteshgupta wrote:

Bunuel wrote:

zisis wrote:

Is x% of x% of y equal to x% less than y ?

(1) x(x + 100) = 10,000 (2) y(y + 1) = 1

OA: A

x2y = 10,000y – 100xy [multiplying by the common denominator 10,000] x2y + 100xy – 10,000y = 0 [everything to one side, because it’s quadratic] y(x2 + 100x – 10,000) = 0 [factoring] Therefore, the answer to the prompt question is affirmative if either x2 + 100x – 10,000 = 0 or y = 0.

(1) SUFFICIENT: This statement rearranges to give = 0.

(2) INSUFFICIENT: y cannot be 0, but no information is provided about x, making it impossible to determine whether x2 + 100x – 10,000 = 0.

But, I believe D: A sufficient B: y=1 or y+1=1 therefore y=0 therefore B sufficient...

what am i missing? please help

Question: is \(\frac{x}{100}*\frac{x}{100}*y=y(1-\frac{x}{100})\)? --> is \(x^2y=100y(100-x)\)? --> is \(x^2y=y(10,000-100x)\) --> is \(y(x^2+100x-10,000)=0\)?

Basically question is does \(y=0\) or/and \(x^2+100x-10,000=0\)?

(2) \(y(y+1)=1\). Here it's clear that \(y\neq{0}\), (substitute \(y=0\) in this equation: \(0(0+1)=0\neq{1}\)). So we know that \(y\neq{0}\), but don't know whether \(x^2+100x-10,000=0\)? Not sufficient.

To elaborate more: the problem with your solution is that you solved incorrectly \(y(y+1)=1\). \(y(y+1)=1\) --> \(y^2+y-1=0\) --> solving for \(y\): \(y=\frac{-1-\sqrt{5}}{2}\) or \(y=\frac{-1+\sqrt{5}}{2}\), so \(y\neq{0}\).

Answer: A.

Hope it's clear.

Can you explain this part \(\frac{x}{100}*\frac{x}{100}*y=y(1-\frac{x}{100})\)? --> my R.H.S of equation is coming as y-x/100 kindly correct me if i am wrong

Consider this: 10% less than \(y\) is \(y*(1-\frac{10}{100})=y*0.9\), the same way "\(x%\) less than \(y\)": is \(y(1-\frac{x}{100})\).

Is x% of x% of y equal to x% less than y ? [#permalink]

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25 Aug 2012, 23:26

Hello Guys,

I am new to GMAT preparation so need some help from you veterans .I cannot understand this question.As per my understanding x% of x% of any number is always less than x% of that number.For example, 10% of 10% of 100 : Since there is no bracket ,I can start from the left hand side .10% of 10% is 1% . 1% of 100 is 1. Now to the second part ...x% less than y,which means 10% less than y,which is nothing but 90.So 1 is always less than 90 . Obvously I have got it all wrong ,but I cannot interpret it in any other manner .Please help .

Regards, arijitb1980 _________________

arijitb1980 Its nice to day dream sometimes,if you chase it the rest of the times.

Re: Is x% of x% of y equal to x% less than y ? (1) x(x + 100) = [#permalink]

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25 Aug 2012, 23:56

arijitb1980 wrote:

Hello Guys,

I am new to GMAT preparation so need some help from you veterans .I cannot understand this question.As per my understanding x% of x% of any number is always less than x% of that number.For example, 10% of 10% of 100 : Since there is no bracket ,I can start from the left hand side .10% of 10% is 1% . 1% of 100 is 1. Now to the second part ...x% less than y,which means 10% less than y,which is nothing but 90.So 1 is always less than 90 . Obvously I have got it all wrong ,but I cannot interpret it in any other manner .Please help .

Regards, arijitb1980

From your computations it is clear that \(x\) cannot be 10. But it doesn't mean that there is no value of \(x\) for which the equality holds.

Translating the question into an equation: is \(\frac{x}{100}*\frac{x}{100}y=(1-\frac{x}{100})y\)?

(1) If \(y=0,\) then the equality certainly holds. Dividing through by \(y,\) the equation becomes \(x^2+x-10,000=0.\) You can solve this quadratic and the positive root is approximately \(61.8.\) So, here is the answer: for \(x=61.8\), \(x%\) of \(x%\) of \(y\) is \(x%\) less than \(y.\) Sufficient.

(2) We can deduce that \(y\neq0\) but we don't know anything about \(x.\) Not sufficient.

Answer A _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: Is x% of x% of y equal to x% less than y ? (1) x(x + 100) = [#permalink]

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09 Dec 2012, 07:09

Expert's post

morfin wrote:

I do not understand the question. Why did you rephrase the question FROM "Is x% of x% of y equal to x% less than y ?" TO Is y = 0 ?

When I read the question, I wrote down (x%((x%)y)) = x%<y

Please help. thanks in advance.

"Is x% of x% of y equal to x% less than y?" means is \(\frac{x}{100}*\frac{x}{100}*y=y(1-\frac{x}{100})\)?

If you manipulate with this expression as shown above the questions becomes: is \(y(x^2+100x-10,000)=0\)? So, the question basically asks whether \(y=0\) or/and \(x^2+100x-10,000=0\)?

Re: Is x% of x% of y equal to x% less than y ? [#permalink]

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17 Oct 2013, 06:31

Expert's post

adg142000 wrote:

my approach for the above was i reduced the word statement to equation with statement 1 as :

\(xy/(x+100)=y(1-x/100)\)? ,,and took the values x=10 and y =10 for which 100/110< 9

but for x=10 and y=-10

-100/110 is not less than -9.

I got this problem wrong for this reason. Do for percentage problems -ve sign has to be ignored???

Plugging-in is not the correct approach for this question, as because the values of x and y are fixed by the fact statements.

We have to prove whether \(\frac{x}{100}*\frac{x}{100}*y = y*(1-\frac{x}{100})\) or not --> \(\frac{x}{100}*y[\frac{x}{100}*+1] = y\) \(\to\)After re-arranging we get

\(x*(x+100)*y = 10000y \to Is y*[x*(x+100)-10000]=0?\)

From F.S 1, we know that x*(x+100) = 10000, thus Sufficient.

From F.S 2, we know only the value of y, and nothing about x. Insufficient

A.

I believe negative percentage makes sense when there is a decrease . So it will not be wrong to say for example that the decrease in the value was 5% or the percentage change was -5%. _________________

Classroom Centre Address: GMATinsight 107, 1st Floor, Krishna Mall, Sector-12 (Main market), Dwarka, New Delhi-110075 ______________________________________________________ Please press the if you appreciate this post !!

Classroom Centre Address: GMATinsight 107, 1st Floor, Krishna Mall, Sector-12 (Main market), Dwarka, New Delhi-110075 ______________________________________________________ Please press the if you appreciate this post !!

Re: Is x% of x% of y equal to x% less than y ? (1) x(x + 100) = [#permalink]

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26 Jul 2015, 10:57

Bunuel wrote:

prashantbacchewar wrote:

Is x% of x% of y equal to x% less than y ?

(1) x(x + 100) = 10,000 (2) y(y + 1) = 1

Question: is \(\frac{x}{100}*\frac{x}{100}*y=y(1-\frac{x}{100})\)? --> is \(x^2y=100y(100-x)\)? --> is \(x^2y=y(10,000-100x)\) --> is \(y(x^2+100x-10,000)=0\)?

Basically question is \(y=0\) or/and \(x^2+100x-10,000=0\)?

(2) \(y(y+1)=1\). Here it's clear that \(y\neq{0}\), (substitute \(y=0\) in this equation: \(0(0+1)=0\neq{1}\)). So we know that \(y\neq{0}\), but don't know whether \(x^2+100x-10,000=0\)? Not sufficient.

Answer: A.

Hope it's clear.

Hello Bunuel, I could not understand one thing that why did you not eliminate y from \(x^2y=100y(100-x)\) and kept till the end ?. As far as I know percentages dont have sign.

Re: Is x% of x% of y equal to x% less than y ? (1) x(x + 100) = [#permalink]

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26 Jul 2015, 11:17

1

This post received KUDOS

Expert's post

anurag356 wrote:

Bunuel wrote:

prashantbacchewar wrote:

Is x% of x% of y equal to x% less than y ?

(1) x(x + 100) = 10,000 (2) y(y + 1) = 1

Question: is \(\frac{x}{100}*\frac{x}{100}*y=y(1-\frac{x}{100})\)? --> is \(x^2y=100y(100-x)\)? --> is \(x^2y=y(10,000-100x)\) --> is \(y(x^2+100x-10,000)=0\)?

Basically question is \(y=0\) or/and \(x^2+100x-10,000=0\)?

(2) \(y(y+1)=1\). Here it's clear that \(y\neq{0}\), (substitute \(y=0\) in this equation: \(0(0+1)=0\neq{1}\)). So we know that \(y\neq{0}\), but don't know whether \(x^2+100x-10,000=0\)? Not sufficient.

Answer: A.

Hope it's clear.

Hello Bunuel, I could not understand one thing that why did you not eliminate y from \(x^2y=100y(100-x)\) and kept till the end ?. As far as I know percentages dont have sign.

1. We are concerned about the sign when we are dealing with inequalities, not equations. 2. Have you read the highlighted part??? y = 0 also satisfies the equation hence we cannot reduce by it because division by 0 is not allowed. _________________

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