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# M02-13

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Math Expert
Joined: 02 Sep 2009
Posts: 50627

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15 Sep 2014, 23:17
00:00

Difficulty:

45% (medium)

Question Stats:

70% (02:08) correct 30% (02:31) wrong based on 165 sessions

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$$x$$ 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

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Math Expert
Joined: 02 Sep 2009
Posts: 50627

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15 Sep 2014, 23:17
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.

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Intern
Joined: 04 Sep 2014
Posts: 5

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31 Mar 2015, 07:08
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.
Intern
Joined: 29 Jan 2015
Posts: 5
Concentration: Technology, Finance

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31 Mar 2015, 08:46
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
Current Student
Joined: 14 Oct 2013
Posts: 45

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Updated on: 22 Apr 2015, 19:38
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, 10:52.
Last edited by healthjunkie on 22 Apr 2015, 19:38, edited 1 time in total.
Math Expert
Joined: 02 Sep 2009
Posts: 50627

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05 Apr 2015, 04:31
2
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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Joined: 17 Mar 2014
Posts: 228
Location: India
Concentration: Operations, Strategy
GMAT 1: 670 Q48 V35
GPA: 3.19
WE: Information Technology (Computer Software)

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09 Aug 2015, 06:08
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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Math Expert
Joined: 02 Sep 2009
Posts: 50627

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17 Aug 2015, 02:27
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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Intern
Joined: 01 Sep 2016
Posts: 12

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10 Sep 2016, 20:29
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 25 Jan 2016
Posts: 9
Location: United States (NJ)
GPA: 2.34
WE: Analyst (Commercial Banking)

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23 Mar 2017, 16:06
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.

How did you get The numerator to -3*8^3? ->>> $$x^2-10x = \frac{-4*8^3 + 8^3}{8*4*3} = \frac{-3*8^3}{8*4*3} = -16$$
Math Expert
Joined: 02 Sep 2009
Posts: 50627

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24 Mar 2017, 03:55
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.

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 &nbs [#permalink] 24 Mar 2017, 03:55
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# M02-13

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