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Question Stats:
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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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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: \(x^210x = \frac{4*8^3 + 8^3}{8*4*3} = \frac{3*8^3}{8*4*3} = 16\) \(x^210x = 16\) \(x^210x +16 = 0\) \((x8)(x2)=0\) The possible values of \(x\) are 8 and 2. Therefore the minimum value of \(x\) is 2. Answer: A
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Re: M0213
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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 x10x= 4*8^3 + 8^3.



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floody84 wrote: Bunel,
I am a bit lost to the answer. what did u do with the 64Y? i see you have x10x= 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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M0213
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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: M0213
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05 Apr 2015, 05:31
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 \(x=\sqrt{9}=3\). Hope it helps.
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Re: M0213
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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 \(x=\sqrt{9}=3\). 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: M0213
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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 \(x=\sqrt{9}=3\). 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 nonnegative root. In contrast x^2 = 4 gives two solutions: 2 and 2.
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Re M0213
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10 Sep 2016, 21:29
I think this is a highquality question and I agree with explanation.



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Re: M0213
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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: \(x^210x = \frac{4*8^3 + 8^3}{8*4*3} = \frac{3*8^3}{8*4*3} = 16\) \(x^210x = 16\) \(x^210x +16 = 0\) \((x8)(x2)=0\) 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? >>> \(x^210x = \frac{4*8^3 + 8^3}{8*4*3} = \frac{3*8^3}{8*4*3} = 16\)



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Re: M0213
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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: \(x^210x = \frac{4*8^3 + 8^3}{8*4*3} = \frac{3*8^3}{8*4*3} = 16\) \(x^210x = 16\) \(x^210x +16 = 0\) \((x8)(x2)=0\) 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? >>> \(x^210x = \frac{4*8^3 + 8^3}{8*4*3} = \frac{3*8^3}{8*4*3} = 16\) 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: M0213
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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: M0213
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26 Apr 2019, 22:47
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 righthand side is equal to 16. Correct answer is (A).



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Re: M0213
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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










