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M02-30

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M02-30  [#permalink]

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New post 16 Sep 2014, 00:18
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A
B
C
D
E

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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Re M02-30  [#permalink]

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New post 16 Sep 2014, 00:18
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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).


Answer: A
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Re: M02-30  [#permalink]

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New post 23 Apr 2015, 05:32
it is A. Solved it with help the of back solving. This problem screams substitution.
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Re: M02-30  [#permalink]

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New post 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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Re: M02-30  [#permalink]

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New post 13 Sep 2015, 08:36
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Re: M02-30  [#permalink]

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