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# M02-30

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Math Expert
Joined: 02 Sep 2009
Posts: 54401

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16 Sep 2014, 00:18
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Difficulty:

5% (low)

Question Stats:

93% (01:03) correct 7% (01:22) wrong based on 107 sessions

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If there is a right triangle with legs of integer lengths $$x$$ and $$y$$, satisfying $$x^2 + (y^2-4) = 21$$, which of the following can be true?

A. $$x = 3$$ and $$y = 4$$
B. $$x = 5$$ and $$y = 1$$
C. $$x = 4$$ and $$y = 1$$
D. $$x = 2$$ and $$y = 5$$
E. $$x = 1$$ and $$y = 3$$

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Math Expert
Joined: 02 Sep 2009
Posts: 54401

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16 Sep 2014, 00:18
1
Official Solution:

If there is a right triangle with legs of integer lengths $$x$$ and $$y$$, satisfying $$x^2 + (y^2-4) = 21$$, which of the following can be true?

A. $$x = 3$$ and $$y = 4$$
B. $$x = 5$$ and $$y = 1$$
C. $$x = 4$$ and $$y = 1$$
D. $$x = 2$$ and $$y = 5$$
E. $$x = 1$$ and $$y = 3$$

Solution (1): Backsolve. Plug answer choices into the original equation and see if it equals 21. (A) works.

Solution (2):
$$x^2 + y^2 -4 = 21$$
$$x^2 + y^2 = 25$$
$$x^2 + y^2 = 5^2$$

If the largest side in a right triangle equals 5, then the other sides equal 3 and 4. Therefore (A).

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Joined: 22 Apr 2015
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23 Apr 2015, 05:32
it is A. Solved it with help the of back solving. This problem screams substitution.
MBA Blogger
Joined: 19 Apr 2014
Posts: 82
Location: India
Concentration: Strategy, Technology
Schools: NTU '19
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13 Sep 2015, 08:31
57 seconds to solve this question. I used the second method. Good timing or bad timing?
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Warm Regards.
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Math Expert
Joined: 02 Sep 2009
Posts: 54401

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13 Sep 2015, 08:36
scofield1521 wrote:
57 seconds to solve this question. I used the second method. Good timing or bad timing?

____________
Very good timing.
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Joined: 21 Oct 2015
Posts: 8

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09 Apr 2019, 07:09
This question is easy if you know your pythagorean triples, unfortunately l didn't immediately recognize it!
Re: M02-30   [#permalink] 09 Apr 2019, 07:09
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# M02-30

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