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m and n are positive integers, each greater than 1. If 7(m 1) = 5(n
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27 Aug 2024, 02:46
27 Aug 2024, 02:46
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Question Stats:
80% (03:10) correct
20% (03:01) wrong based on 5 sessions
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m and n are positive integers, each greater than 1. If 7(m – 1) = 5(n – 1), which of the following options could be true?
A. m + n = 14
B. m – n = –2
C. n – m = –2
D. m + n = 12
A. m + n = 14
B. m – n = –2
C. n – m = –2
D. m + n = 12
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m and n are positive integers, each greater than 1. If 7(m 1) = 5(n
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31 Aug 2024, 23:57
31 Aug 2024, 23:57
\(7(m-1) = 5(n-1)\)
\(7m = 5n+2\)
Adding \(7n\) on both sides,
\(7(m+n) = 12n+2\)
\(m+n = \frac{12n+2}{7}\)
Now, check option A and D.
A: \(m+n = 14, \frac{12n+2}{7} = 14, n = 8\). This satisfies the condition that n is a positive integer greater than 1. So, A is correct.
D: \(m+n = 12, \frac{12n+2}{7} = 12, n = 6.83333\). So, n is not an integer, so D is wrong.
From \(7m = 5n+2\), if we subtract 7n on both sides, we get: \(m-n = \frac{-2n+2}{7}\)
B: \(\frac{-2n+2}{7}=-2, n = 8\). B is correct.
C: \(n-m = \frac{2n-2}{7} = -2, n=-6\). n is not positive, so C is wrong.
Hence, A and B are correct.
\(7m = 5n+2\)
Adding \(7n\) on both sides,
\(7(m+n) = 12n+2\)
\(m+n = \frac{12n+2}{7}\)
Now, check option A and D.
A: \(m+n = 14, \frac{12n+2}{7} = 14, n = 8\). This satisfies the condition that n is a positive integer greater than 1. So, A is correct.
D: \(m+n = 12, \frac{12n+2}{7} = 12, n = 6.83333\). So, n is not an integer, so D is wrong.
From \(7m = 5n+2\), if we subtract 7n on both sides, we get: \(m-n = \frac{-2n+2}{7}\)
B: \(\frac{-2n+2}{7}=-2, n = 8\). B is correct.
C: \(n-m = \frac{2n-2}{7} = -2, n=-6\). n is not positive, so C is wrong.
Hence, A and B are correct.
gmatclubot
m and n are positive integers, each greater than 1. If 7(m 1) = 5(n [#permalink]
31 Aug 2024, 23:57
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