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# M15-05

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Math Expert
Joined: 02 Sep 2009
Posts: 47977

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16 Sep 2014, 00:55
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Difficulty:

55% (hard)

Question Stats:

61% (01:11) correct 39% (01:18) wrong based on 98 sessions

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If for any pair of two positive integers $$M$$ and $$N$$, their arithmetic mean $$A(M, N)$$ is defined as $$\frac{M + N}{2}$$ while their geometric mean $$G(M, N)$$ is defined as $$\sqrt{MN}$$, is $$M$$ larger than $$N$$?

(1) $$A(G(M, N), M) = M$$

(2) $$A(M, N) - G(M, N) = 0$$

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Joined: 02 Sep 2009
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16 Sep 2014, 00:55
1
Official Solution:

Statement (1) by itself is sufficient. $$A(G(M, N), M) = \frac{G(M, N) + M}{2} = \frac{\sqrt{MN} + M}{2}$$. From S1 it follows that $$\frac{\sqrt{MN} + M}{2} = M$$ or $$\frac{\sqrt{MN}}{2} = \frac{M}{2}$$ or $$\sqrt{M} \sqrt{N} = M$$ or $$\sqrt{M} = \sqrt{N}$$. Thus, $$M = N$$ and the answer to the question is "no". $$A(M, N) - G(M, N) = \frac{M + N}{2} - \sqrt{MN}$$.

Statement (2) by itself is sufficient. From S2 it follows that $$M + N - 2\sqrt{MN} = 0$$ or $$(\sqrt{M} - \sqrt{N})^2 = 0$$. Thus, $$M = N$$ and the answer to the question is "no".

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17 Oct 2015, 00:36
Hi,
Could you tell me if the method I used for calculating statement 2's sufficiency is valid, hmm foolproof ?
A(M,N)−G(M,N)=0,
M+N/2=SqRt MN, square both sides
(M+N/2)(M+N/2)= MN
2(M+N)= 2MN
M+N=MN
Since M and N are positive integers only two situations satisfy this equation- either both M and N are 1 or they are 2. In both these cases M=N, hence sufficient.
Is this correct method?
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GMAT 1: 660 Q49 V30
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22 Aug 2016, 11:36
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. m+n-2(sqrt of mn)=0

Can you please explain how you got this from statement 2?
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Joined: 14 May 2016
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23 Aug 2016, 10:52
tae808 wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. m+n-2(sqrt of mn)=0

Can you please explain how you got this from statement 2?

m + n - 2sqrt(mn) = 0

LHS follows the formula (a-b)^2 = a^2 + b^2 -2ab
So we compress that to get:
[sqrt(m) - sqrt(n)]^2 = 0

This translates to: sqrt(m) - sqrt(n) = 0
=> sqrt(m) = sqrt(n)
=> m = n

also, recall that both are positive integers...
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Joined: 02 Sep 2009
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24 Aug 2016, 05:04
tae808 wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. m+n-2(sqrt of mn)=0

Can you please explain how you got this from statement 2?

$$A(M, N) = \frac{M + N}{2}$$

$$G(M, N)=\sqrt{MN}$$

$$A(M, N) - G(M, N)=0$$

$$\frac{M + N}{2}-\sqrt{MN}=0$$

Multiply by 2: $$M + N - 2\sqrt{MN} = 0$$
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23 Mar 2017, 17:54
Bunuel wrote:
Official Solution:

Statement (1) by itself is sufficient. $$A(G(M, N), M) = \frac{G(M, N) + M}{2} = \frac{\sqrt{MN} + M}{2}$$. From S1 it follows that $$\frac{\sqrt{MN} + M}{2} = M$$ or $$\frac{\sqrt{MN}}{2} = \frac{M}{2}$$ or $$\sqrt{M} \sqrt{N} = M$$ or $$\sqrt{M} = \sqrt{N}$$. Thus, $$M = N$$ and the answer to the question is "no". $$A(M, N) - G(M, N) = \frac{M + N}{2} - \sqrt{MN}$$.

Statement (2) by itself is sufficient. From S2 it follows that $$M + N - 2\sqrt{MN} = 0$$ or $$(\sqrt{M} - \sqrt{N})^2 = 0$$. Thus, $$M = N$$ and the answer to the question is "no".

How do we reduce step by step from statement 1? I am particularly confused with - > $$\sqrt{M} \sqrt{N} = M$$ or $$\sqrt{M} = \sqrt{N}$$. Thus, $$M = N$$
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24 Mar 2017, 04:57
Prostar wrote:
Bunuel wrote:
Official Solution:

Statement (1) by itself is sufficient. $$A(G(M, N), M) = \frac{G(M, N) + M}{2} = \frac{\sqrt{MN} + M}{2}$$. From S1 it follows that $$\frac{\sqrt{MN} + M}{2} = M$$ or $$\frac{\sqrt{MN}}{2} = \frac{M}{2}$$ or $$\sqrt{M} \sqrt{N} = M$$ or $$\sqrt{M} = \sqrt{N}$$. Thus, $$M = N$$ and the answer to the question is "no". $$A(M, N) - G(M, N) = \frac{M + N}{2} - \sqrt{MN}$$.

Statement (2) by itself is sufficient. From S2 it follows that $$M + N - 2\sqrt{MN} = 0$$ or $$(\sqrt{M} - \sqrt{N})^2 = 0$$. Thus, $$M = N$$ and the answer to the question is "no".

How do we reduce step by step from statement 1? I am particularly confused with - > $$\sqrt{M} \sqrt{N} = M$$ or $$\sqrt{M} = \sqrt{N}$$. Thus, $$M = N$$

$$\sqrt{M} \sqrt{N} = M$$

$$\sqrt{M} \sqrt{N} = \sqrt{M}*\sqrt{M}$$

$$\sqrt{N} = \sqrt{M}$$
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Joined: 09 May 2016
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24 Mar 2017, 09:54
Bunuel wrote:
Official Solution:

Statement (1) by itself is sufficient. $$A(G(M, N), M) = \frac{G(M, N) + M}{2} = \frac{\sqrt{MN} + M}{2}$$. From S1 it follows that $$\frac{\sqrt{MN} + M}{2} = M$$ or $$\frac{\sqrt{MN}}{2} = \frac{M}{2}$$ or $$\sqrt{M} \sqrt{N} = M$$ or $$\sqrt{M} = \sqrt{N}$$. Thus, $$M = N$$ and the answer to the question is "no". $$A(M, N) - G(M, N) = \frac{M + N}{2} - \sqrt{MN}$$.

Statement (2) by itself is sufficient. From S2 it follows that $$M + N - 2\sqrt{MN} = 0$$ or $$(\sqrt{M} - \sqrt{N})^2 = 0$$. Thus, $$M = N$$ and the answer to the question is "no".

Bunuel For Statement 1 , I did this m=(MN)^1/2;Thus M-(MN)^1/2=0 and therefore m^1/2 = 0 or/and m=n.Please help me in understanding why this is wrong
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24 Mar 2017, 10:51
KARISHMA315 wrote:
Bunuel wrote:
Official Solution:

Statement (1) by itself is sufficient. $$A(G(M, N), M) = \frac{G(M, N) + M}{2} = \frac{\sqrt{MN} + M}{2}$$. From S1 it follows that $$\frac{\sqrt{MN} + M}{2} = M$$ or $$\frac{\sqrt{MN}}{2} = \frac{M}{2}$$ or $$\sqrt{M} \sqrt{N} = M$$ or $$\sqrt{M} = \sqrt{N}$$. Thus, $$M = N$$ and the answer to the question is "no". $$A(M, N) - G(M, N) = \frac{M + N}{2} - \sqrt{MN}$$.

Statement (2) by itself is sufficient. From S2 it follows that $$M + N - 2\sqrt{MN} = 0$$ or $$(\sqrt{M} - \sqrt{N})^2 = 0$$. Thus, $$M = N$$ and the answer to the question is "no".

Bunuel For Statement 1 , I did this m=(MN)^1/2;Thus M-(MN)^1/2=0 and thereforem^1/2 = 0 or/and m=n.Please help me in understanding why this is wrong

If for any pair of two positive integers M and N...
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24 Nov 2017, 22:44
This is a high quality question. Thanks Bunuel for the amazing solution, it took me quite sometime to see how statement 2 is sufficient.
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If you must err, err on the side of hope.

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06 Aug 2018, 09:54
Bunuel wrote:
KARISHMA315 wrote:
Bunuel wrote:
Official Solution:

Statement (1) by itself is sufficient. $$A(G(M, N), M) = \frac{G(M, N) + M}{2} = \frac{\sqrt{MN} + M}{2}$$. From S1 it follows that $$\frac{\sqrt{MN} + M}{2} = M$$ or $$\frac{\sqrt{MN}}{2} = \frac{M}{2}$$ or $$\sqrt{M} \sqrt{N} = M$$ or $$\sqrt{M} = \sqrt{N}$$. Thus, $$M = N$$ and the answer to the question is "no". $$A(M, N) - G(M, N) = \frac{M + N}{2} - \sqrt{MN}$$.

Statement (2) by itself is sufficient. From S2 it follows that $$M + N - 2\sqrt{MN} = 0$$ or $$(\sqrt{M} - \sqrt{N})^2 = 0$$. Thus, $$M = N$$ and the answer to the question is "no".

Bunuel For Statement 1 , I did this m=(MN)^1/2;Thus M-(MN)^1/2=0 and thereforem^1/2 = 0 or/and m=n.Please help me in understanding why this is wrong

If for any pair of two positive integers M and N...

Bunuel..
For 1st statement.. either m= 0 or m=n,
For m=0, we cannot be sure about relationship of m & n.. so not sufficient..
Is this not correct..??

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06 Aug 2018, 09:55
I think this the explanation isn't clear enough, please elaborate.
For 1st statement.. either m= 0 or m=n,
For m=0, we cannot be sure about relationship of m & n.. so not sufficient..
Is this not correct..??

Posted from my mobile device
Math Expert
Joined: 02 Sep 2009
Posts: 47977

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07 Aug 2018, 03:38
GAURAV1113 wrote:
I think this the explanation isn't clear enough, please elaborate.
For 1st statement.. either m= 0 or m=n,
For m=0, we cannot be sure about relationship of m & n.. so not sufficient..
Is this not correct..??

Posted from my mobile device

If for any pair of two positive integers M and N...
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Re: M15-05 &nbs [#permalink] 07 Aug 2018, 03:38
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# M15-05

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