if sum of two numbers is 1215 and their HCF is 81, then the possible number of pairsof such numbers are 

 

It is given that the sum of two numbers is 1215 and their H.C.F is 81

Since the H.C.F of two numbers is 81, the two numbers are 81x and 81y.

Now, 81x + 81y = 1215 [Sum of two numbers is 1215]

or, 81 (x + y) = 1215

or, (x + y) = 15

 

Such pairs of numbers whose sum is 15 are (1, 14), (2, 13), (4, 11), (7, 8)

For x = 1, y = 14; in this case the numbers are 1 × 81 = 81 and 14 × 81 = 1134

For x = 2, y = 13; in this case the numbers are 2 × 81 = 162 and 13 × 81 = 1053

For x = 4, y = 11; in this case the numbers are 4 × 81 = 324 and 11 × 81 = 891

For x = 7, y = 8; in this case the numbers are 7 × 81 = 567 and 8 × 81 = 648

Hence four pairs are possible.