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Math Expert V
Joined: 02 Sep 2009
Posts: 53734

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Difficulty:   15% (low)

Question Stats: 73% (00:59) correct 27% (01:05) wrong based on 129 sessions

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If $$x=(\sqrt{5}-\sqrt{7})^2$$, then the best approximation of $$x$$ is:

A. $$0$$
B. $$1$$
C. $$2$$
D. $$3$$
E. $$4$$

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Math Expert V
Joined: 02 Sep 2009
Posts: 53734

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Official Solution:

If $$x=(\sqrt{5}-\sqrt{7})^2$$, then the best approximation of $$x$$ is:

A. $$0$$
B. $$1$$
C. $$2$$
D. $$3$$
E. $$4$$

$$x=(\sqrt{5}-\sqrt{7})^2=5-2\sqrt{35}+7=12-2\sqrt{35}$$.

Since $$\sqrt{35}\approx{6}$$, then $$12-2\sqrt{35}\approx{12-2*6}=0$$.

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Intern  Joined: 10 Sep 2014
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Schools: LBS '18

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Fractionalizing the result shows very clearly that the result is closer to 0 than to 1:
x=(√5−√7)^2 = [(5-7)/(√5+√7)]^2
=(-2)^2 / (√5+√7)]^2

= _4_
12+ √35

Now, √35 is close to 6, but undeniably between 5 and 6.
Even if you use both choices for the denominator's range:
12+2(5) = 22 --> x = 4/22 = 2/11
12+2(6) = 24 --> x = 4/24 = 1/6

1/6 < x < 2/11

With x(max) < 2/11, it is abundantly clear--with no room for error--that x is closer to 0.

Cheers!
Intern  B
Joined: 04 Sep 2018
Posts: 39
GPA: 3.33

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Hi, I didn't get this question at all or the solutions presented?

Can someone please explain this to me? Bought the gmat club tests, and I am working on building my concepts by solving the untimed quizzes.
Intern  B
Joined: 14 Jan 2018
Posts: 45
Location: India
Concentration: General Management, Entrepreneurship
GMAT 1: 660 Q50 V29 GPA: 3.8
WE: Engineering (Manufacturing)

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x=(5√−7√)2
So when we use (a-b)square formula which expands to a2 +b2 -2ab
we have 5+7-2*underrt(35)
We know under root (36) =6 so under rt (35) will be near to 6 or less than

So , X = 12-2*(Less than 6)
which gives us 0
Intern  B
Joined: 24 Dec 2013
Posts: 2
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Concentration: General Management, Marketing

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If you approximate Root 5 and Root 7 you will get numbers between 2 - 3 and if you subtract one from another then you will definitely get less than 1 and any fraction if you Square then it will tend towards Zero hence its Zero
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ganeshj07 wrote:
If you approximate Root 5 and Root 7 you will get numbers between 2 - 3 and if you subtract one from another then you will definitely get less than 1 and any fraction if you Square then it will tend towards Zero hence its Zero

This is how I solved it more or less. Just wanted to point out that "if you square any fraction it will tend towards zero hence it's zero" is a bit misleading IMHO. While it is true that x^2 is less than x for all x whose absolute value between 0 and 1, and in that sense "tends" towards 0, this question specifically asked whether x^2 is closer to 0 or 1. There are plenty of examples, x = 0.75, x = 0.9, etc. where x^2 would be closer to 1 than to 0.
Intern  B
Joined: 24 Dec 2013
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mattlang1 wrote:
ganeshj07 wrote:
If you approximate Root 5 and Root 7 you will get numbers between 2 - 3 and if you subtract one from another then you will definitely get less than 1 and any fraction if you Square then it will tend towards Zero hence its Zero

This is how I solved it more or less. Just wanted to point out that "if you square any fraction it will tend towards zero hence it's zero" is a bit misleading IMHO. While it is true that x^2 is less than x for all x whose absolute value between 0 and 1, and in that sense "tends" towards 0, this question specifically asked whether x^2 is closer to 0 or 1. There are plenty of examples, x = 0.75, x = 0.9, etc. where x^2 would be closer to 1 than to 0.

Yes, You are right.... Square of anything more than 0.75 tend towards 1 Re: D01-45   [#permalink] 16 Feb 2019, 06:48
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