Question
Two positive integers differ by 4 and sum of their reciprocals is 1021. Then one of the numbers is: (CAT 1995)
Answer: Option A
:
A
Ans: (a)
Let one number be x, then second number will be (x+4);
Thus :1x+(1(x+4))=1021;
(x+x+4)(x(x+4))=1021;
(2x+4)x(x+4)=1021;
x=3
Elimination strategy:
By looking at the answer option as the sum of reciprocal of the numbers will have to give 1021The number has to be a factor or multiple of 21. So directly option (c) can be eliminated.
Now the numbers that we will have to check will be 3 & 7 , 1 & 5, 17 & 21, 21 &25
The only possibility from here is 3 & 7. Hence option (a).
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:
A
Ans: (a)
Let one number be x, then second number will be (x+4);
Thus :1x+(1(x+4))=1021;
(x+x+4)(x(x+4))=1021;
(2x+4)x(x+4)=1021;
x=3
Elimination strategy:
By looking at the answer option as the sum of reciprocal of the numbers will have to give 1021The number has to be a factor or multiple of 21. So directly option (c) can be eliminated.
Now the numbers that we will have to check will be 3 & 7 , 1 & 5, 17 & 21, 21 &25
The only possibility from here is 3 & 7. Hence option (a).
Was this answer helpful ?
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