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Question

How many pairs of positive integers m, n satisfy (1m)+(4n)=(112), where n is an odd integer less than 60? (CAT 2007)

A
4
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B
7
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C
5
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D
3
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E
6
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Solution

The correct option is D 3

Soln:d.
Given conditions :
(1m)+(4n)=(112)
M and n are positive integers
N is an odd integer less than 60.
On solving the equation, we get m=12n(n48);
Since m and n are positive integers, n>48, and from the question, 48<n<60.
Hence possible values on n=49,51,53,55,57,59.
For n=49,51 and 57 we get integral values of m. Hence the required no of integral pairs = 3


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