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M02-13

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M02-13  [#permalink]

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New post 16 Sep 2014, 00:17
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A
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  55% (hard)

Question Stats:

69% (02:41) correct 31% (02:53) wrong based on 230 sessions

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Re M02-13  [#permalink]

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New post 16 Sep 2014, 00:17
Official Solution:

\(x\) and \(y\) are positive integers. If \(y = \sqrt{64}\) and \(x^2 - 10x = (-4y^3 + 64y)*\frac{1}{96}\), what is the minimum possible value of \(x\)?

A. 2
B. 4
C. 8
D. 12
E. 16


We know that \(y = 8\).

Plugging this into the second equation gives:
\(x^2-10x = \frac{-4*8^3 + 8^3}{8*4*3} = \frac{-3*8^3}{8*4*3} = -16\)
\(x^2-10x = -16\)
\(x^2-10x +16 = 0\)
\((x-8)(x-2)=0\)

The possible values of \(x\) are 8 and 2. Therefore the minimum value of \(x\) is 2.


Answer: A
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Re: M02-13  [#permalink]

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New post 31 Mar 2015, 08:08
Bunel,

I am a bit lost to the answer. what did u do with the 64Y? i see you have x-10x= -4*8^3 + 8^3.
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M02-13  [#permalink]

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New post 31 Mar 2015, 09:46
floody84 wrote:
Bunel,

I am a bit lost to the answer. what did u do with the 64Y? i see you have x-10x= -4*8^3 + 8^3.


Steps:
  • \sqrt{64} is 8. That is y=8
  • Insert 8 in quadratic formula and calculate the left hand side
    • Factor 96: 96 = 8*4*3 (denominator of the fraction)
    • Cancel out all possible factors to get -16 for the left hand side
  • Use -16 in quadratic expression to get the roots 8 and 2
  • Solution is 2 because the question asked for the min. value
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M02-13  [#permalink]

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New post Updated on: 22 Apr 2015, 20:38
Couldn't Y also equal -8? I know that when you plug in y=-8 you end up getting x^2 - 10x -16 = 0 which you can't factor out, but I didnt realize that until after I plugged in -8 (I plugged in -8 since we were trying to find the smallest value of X, I figured a negative Y would give us smaller values for X)

Originally posted by healthjunkie on 04 Apr 2015, 11:52.
Last edited by healthjunkie on 22 Apr 2015, 20:38, edited 1 time in total.
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Re: M02-13  [#permalink]

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New post 05 Apr 2015, 05:31
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healthjunkie wrote:
Couldn't Y also equal -8? I know that when you plug in y=-8 you end up getting x^2 - 10x -16 = 0 which you can't factor out, but I didnt realize that until after I plugged in -8 (I plugged in -8 since we were trying to find the smallest value of X, I figured a negative Y would give us smaller values for X)


No.

When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).

Theory on Number Properties: math-number-theory-88376.html
Tips on Numper Properties: number-properties-tips-and-hints-174996.html

All DS Number Properties Problems to practice: search.php?search_id=tag&tag_id=38
All PS Number Properties Problems to practice: search.php?search_id=tag&tag_id=59


Hope it helps.
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Re: M02-13  [#permalink]

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New post 09 Aug 2015, 07:08
Bunuel wrote:
healthjunkie wrote:
Couldn't Y also equal -8? I know that when you plug in y=-8 you end up getting x^2 - 10x -16 = 0 which you can't factor out, but I didnt realize that until after I plugged in -8 (I plugged in -8 since we were trying to find the smallest value of X, I figured a negative Y would give us smaller values for X)


No.

When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).

Theory on Number Properties: math-number-theory-88376.html
Tips on Numper Properties: number-properties-tips-and-hints-174996.html

All DS Number Properties Problems to practice: search.php?search_id=tag&tag_id=38
All PS Number Properties Problems to practice: search.php?search_id=tag&tag_id=59


Hope it helps.


Hi Bunuel ,

Please correct me if i amwrong

Does this mean when gmat gives sqare root sign itself in question stem , then it means we should consider only positive root
and on the other hand gmat gave me a quadratic eqaution in question stem and when i solve it i should consider both +ve and - ve root ?

right ?
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Re: M02-13  [#permalink]

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New post 17 Aug 2015, 03:27
adityadon wrote:
Bunuel wrote:
healthjunkie wrote:
Couldn't Y also equal -8? I know that when you plug in y=-8 you end up getting x^2 - 10x -16 = 0 which you can't factor out, but I didnt realize that until after I plugged in -8 (I plugged in -8 since we were trying to find the smallest value of X, I figured a negative Y would give us smaller values for X)


No.

When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).

Theory on Number Properties: math-number-theory-88376.html
Tips on Numper Properties: number-properties-tips-and-hints-174996.html

All DS Number Properties Problems to practice: search.php?search_id=tag&tag_id=38
All PS Number Properties Problems to practice: search.php?search_id=tag&tag_id=59


Hope it helps.


Hi Bunuel ,

Please correct me if i amwrong

Does this mean when gmat gives sqare root sign itself in question stem , then it means we should consider only positive root
and on the other hand gmat gave me a quadratic eqaution in question stem and when i solve it i should consider both +ve and - ve root ?

right ?


\(\sqrt{}\) sign always means non-negative root.

In contrast x^2 = 4 gives two solutions: 2 and -2.
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Re M02-13  [#permalink]

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New post 10 Sep 2016, 21:29
I think this is a high-quality question and I agree with explanation.
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Re: M02-13  [#permalink]

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New post 23 Mar 2017, 17:06
Bunuel wrote:
Official Solution:

\(x\) and \(y\) are positive integers. If \(y = \sqrt{64}\) and \(x^2 - 10x = (-4y^3 + 64y)*\frac{1}{96}\), what is the minimum possible value of \(x\)?

A. 2
B. 4
C. 8
D. 12
E. 16


We know that \(y = 8\).

Plugging this into the second equation gives:
\(x^2-10x = \frac{-4*8^3 + 8^3}{8*4*3} = \frac{-3*8^3}{8*4*3} = -16\)
\(x^2-10x = -16\)
\(x^2-10x +16 = 0\)
\((x-8)(x-2)=0\)

The possible values of \(x\) are 8 and 2. Therefore the minimum value of \(x\) is 2.


Answer: A


How did you get The numerator to -3*8^3? ->>> \(x^2-10x = \frac{-4*8^3 + 8^3}{8*4*3} = \frac{-3*8^3}{8*4*3} = -16\)
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Re: M02-13  [#permalink]

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New post 24 Mar 2017, 04:55
Prostar wrote:
Bunuel wrote:
Official Solution:

\(x\) and \(y\) are positive integers. If \(y = \sqrt{64}\) and \(x^2 - 10x = (-4y^3 + 64y)*\frac{1}{96}\), what is the minimum possible value of \(x\)?

A. 2
B. 4
C. 8
D. 12
E. 16


We know that \(y = 8\).

Plugging this into the second equation gives:
\(x^2-10x = \frac{-4*8^3 + 8^3}{8*4*3} = \frac{-3*8^3}{8*4*3} = -16\)
\(x^2-10x = -16\)
\(x^2-10x +16 = 0\)
\((x-8)(x-2)=0\)

The possible values of \(x\) are 8 and 2. Therefore the minimum value of \(x\) is 2.


Answer: A


How did you get The numerator to -3*8^3? ->>> \(x^2-10x = \frac{-4*8^3 + 8^3}{8*4*3} = \frac{-3*8^3}{8*4*3} = -16\)


It's basic manipulation: \(-4*8^3 + 8^3 = 8^3(-4+1) = 3*8^3\). The same way -4x + x = -3x.
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Re: M02-13  [#permalink]

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New post 15 Feb 2019, 14:22
Did anyone solve this faster than 4 mins or faster than 3 mins? If so, will be grateful for tips!
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Re: M02-13  [#permalink]

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New post 26 Apr 2019, 22:47
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medic19 wrote:
Did anyone solve this faster than 4 mins or faster than 3 mins? If so, will be grateful for tips!


This reply might be little late but here is the shortcut I found.

If you plug in the value of y=8 in the expression on the right handside, that expression will become an integer value. That integer value has to be a multiple of two numbers when added together gives you -10x. Looking at the answer choices, the smallest value is 2(Answer choice A). (-2) + (-8) = -10, and (-2) * (-8) = 16. So, the expression on the right-hand side is equal to 16.

Correct answer is (A).
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Re: M02-13  [#permalink]

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New post 27 Apr 2019, 23:40
by simple plugging in, we can solve it easily and faster..

as the ques stem said x,y are positive integers... which means y=8

substituting y in the RHS gives us the value to be -16

by plugging the choices(start vth the lowest as the question asked abt minimum possible value) in the LHS..

plugging 2 in LHS gives -16.

thus the answer is option A
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Re: M02-13   [#permalink] 27 Apr 2019, 23:40
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