CAT
Exam
The product of two numbers is '2028' and their H.C.F.is '13'. The number of such pairs is:
A.1
B.2
C.3
D.4
Read Solution (Total 4)
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- ans=2
let the numbers be 13x and 13y
13x*13y=2028
xy=12
so the co-primes with their product 12 are(1,12)and(3,4)
so the required numbers are(13*1,13*12)and(13*3,13*4)
so, there are 2 such pairs. - 11 years agoHelpfull: Yes(2) No(0)
- ans: C
78*26
39*52
13*156 - 11 years agoHelpfull: Yes(0) No(0)
- l.c.m*h.c.f=product
l.c.m=2028/13=156
2028=156*13&52*39,so only two pairs are possible.ans=2(b) - 11 years agoHelpfull: Yes(0) No(0)
- 13x*13y=2028=>xy=12
3 possibilities:
1. 1,12
2. 3,4
3. 2,6 - 10 years agoHelpfull: Yes(0) No(0)
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