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# M01-19

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Math Expert
Joined: 02 Sep 2009
Posts: 51185
Re: M01-19  [#permalink]

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25 Jun 2018, 01:51
sanjay1810 wrote:
Hi Bunuel,

Can you please elaborate on how st 2 is not sufficient ?

x - y = 9
( \sqrt{x} + \sqrt{y} ) ( \sqrt{x} - \sqrt{y} ) = 9 = 9 . 1 OR 3. 3
Since it's given the x> y >0, sum of two numbers cannot be equal to difference of the same two numbers. So 3,3 is out. So 9, 1 is still left.
Both numbers positive, one greater than the other,
so
\sqrt{x} + \sqrt{y} = 9
\sqrt{x} - \sqrt{y} = 1
this can be solved and we get the ans which obviously i wrong as per OA.

What's wrong here and why is 2 not sufficient?

Thanks.

You are assuming that $$\sqrt{x} + \sqrt{y}$$ and $$\sqrt{x} - \sqrt{y}$$ are integers, which is not given. Why cannot $$\sqrt{x} + \sqrt{y}$$ be 81 and $$\sqrt{x} - \sqrt{y}$$ be 1/9? x - y = 9 has infinitely many solutions, so you cannot get the single numerical value of $$\sqrt{x} - \sqrt{y}$$
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Re: M01-19  [#permalink]

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25 Jun 2018, 03:01
Bunuel wrote:

You are assuming that $$\sqrt{x} + \sqrt{y}$$ and $$\sqrt{x} - \sqrt{y}$$ are integers, which is not given. Why cannot $$\sqrt{x} + \sqrt{y}$$ be 81 and $$\sqrt{x} - \sqrt{y}$$ be 1/9? x - y = 9 has infinitely many solutions, so you cannot get the single numerical value of $$\sqrt{x} - \sqrt{y}$$

Fell for the trap. Thanks Bunuel as always for the quick response - eye opener!
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M01-19  [#permalink]

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Updated on: 01 Aug 2018, 04:18
How do we solve the equation from √x-√y=2

Originally posted by gsingh0711 on 31 Jul 2018, 15:20.
Last edited by gsingh0711 on 01 Aug 2018, 04:18, edited 1 time in total.
Math Expert
Joined: 02 Sep 2009
Posts: 51185
Re: M01-19  [#permalink]

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31 Jul 2018, 19:22
gsingh0711 wrote:
How do we solve the quation from √−y√=2

Please learn how to format math formulae: https://gmatclub.com/forum/rules-for-po ... l#p1096628
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Re: M01-19  [#permalink]

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01 Aug 2018, 04:23
Bunuel wrote:
gsingh0711 wrote:
How do we solve the quation from √−y√=2

Please learn how to format math formulae: https://gmatclub.com/forum/rules-for-po ... l#p1096628

Thanks for sharing the link. Hope the expression is decipherable now. please clarify my doubt pertaining to the question.
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Re: M01-19  [#permalink]

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02 Sep 2018, 10:49
Hi Bunuel,
for the second statement, could you please let me know how the below approach is incorrect
Statement 2: x -y =9
This can be expanded as
$$(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})$$ = 9
The Right Hand Side can be either -3 * -3, -1 * -9, 1*9, 3*3
Since it is already given that both x and y are positive, we are left with only 3*3 and 1*9. Also, since the sum and difference of two numbers cannot be the same unless one or both are 0, 3*3 is also eliminated. Hence, we are left with 9*1 thereby, giving us a solution
$$(\sqrt{x} + \sqrt{y})$$ = 9 and
$$(\sqrt{x} - \sqrt{y})$$ = 1
Math Expert
Joined: 02 Sep 2009
Posts: 51185
Re: M01-19  [#permalink]

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02 Sep 2018, 20:28
aroraishita02 wrote:
Hi Bunuel,
for the second statement, could you please let me know how the below approach is incorrect
Statement 2: x -y =9
This can be expanded as
$$(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})$$ = 9
The Right Hand Side can be either -3 * -3, -1 * -9, 1*9, 3*3
Since it is already given that both x and y are positive, we are left with only 3*3 and 1*9. Also, since the sum and difference of two numbers cannot be the same unless one or both are 0, 3*3 is also eliminated. Hence, we are left with 9*1 thereby, giving us a solution
$$(\sqrt{x} + \sqrt{y})$$ = 9 and
$$(\sqrt{x} - \sqrt{y})$$ = 1

Check here: https://gmatclub.com/forum/m01-183530-20.html#p2083627
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Re: M01-19 &nbs [#permalink] 02 Sep 2018, 20:28

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# M01-19

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