Latest MBA Aptitude Question SOLUTION: The sum of two numbers is 462 and their HCF is 22. What is the maximum possible number of pairs that satisfy these conditions?

1) 1
2) 3
3) 2
4) 6
5)

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The sum of two numbers is 462 and their HCF is 22. What is the maximum possible number of pairs that satisfy these conditions?

1) 1
2) 3
3) 2
4) 6
5) 8

Asked In MBA (12 years ago)

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sruthi kaipa  (12 years ago)

22(x+y) = 462
x+y will be 21
the pairs which will satisfy 22 as hcf are (1,20) (2,19) (4,17) (5,16) (8,13) (10,11)

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vivek kumar  (11 years ago)

if HCF is 22 numbers will be 22x and 22y
22x+22y=462
x+y=21
therefore the set of co-primes will be (1,20)(2,19)(4,17)(5,16)(8,13)(10,11)
therefore numbers will be(22,440)(44,418)(88,374)(110,352)(176,286)(220,242)
hence answer is 6.

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Piyush  (8 years ago)

First, let the numbers be 22a and 22b.

Therefore, 22(a+b) = 462.

or, a+b = 21. Therefore, the possible pairs are: 1+21, 4+17, 5+16, 8+13, 10+11, i.e. 5.

Remember, a and b are prime to each other, i.e. they do not have any common factor between them.
You cannot choose 3+18 as one of the pairs as 3 and 18 have 3 as a common factor. Similarly you cannot choose 6+15 as a possible pair. So only 5 pairs are possible.

If u multiply each of those 5 pairs by 22, their HCF will come as 22 and their sum will be 462.

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