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27 May 2023, 07:37
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
46% (02:33) correct
54%
(02:07)
wrong
based on 13
sessions
History
Date
Time
Result
Not Attempted Yet
If x and y are positive integers satisfying the equation x^2+y^2=85, what could be the value of x^2-y^2?
(a) 53
(b) 65
(c) -58
(d) -92
(e) 98
(a) 53
(b) 65
(c) -58
(d) -92
(e) 98
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27 May 2023, 11:00
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Jarvis07
Given
- \(x^2 + y^2 = 85\) --------(1)
- x and y are positive integers
Question
\(x^2 - y^2 =\) ?
Let's assume
\(x^2 - y^2 = z\) --------(2)
Adding (1) and (2)
\(2x^2 = 85+z\)
\(x^2 = \frac{85+z}{2}\)
The first thing to notice is that 85+z must be an even integer, hence z must be odd. So we can rule out C, D, and E.
We can substitute the values from Option A and Option B to check which value results in an integer value of x.
A. 53
\(x^2 = \frac{85 + 53 }{ 2}\)
\(x^2 = \frac{138 }{ 2} = 69\)
\(x = \sqrt{69}\) ⇒x is not an integer. Eliminate A
B. 65
\(x^2 = \frac{85 + 65 }{ 2}\)
\(x^2 = \frac{150 }{ 2} = 75\)
\(x = \sqrt{75}\) ⇒x is not an integer
IMO the question is flawed
28 May 2023, 13:50
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The only way of getting 85 by adding two squares of integers is to have x=9 and y=2 (or x=2 and y=9). So x^2-y^2 = 81-4=77 or x^2-y^2=4-81 = -77. There is a mistake in the question.
