Question
How many pairs of positive integers m, n satisfy (1m)+(4n)=(112), where n is an odd integer less than 60? (CAT 2007)
Answer: Option D
:
D
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(n−48);
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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:
D
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(n−48);
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
Was this answer helpful ?
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