Question

If sum of two numbers is 1215 and their HCF is 81, then the possible number of pairs
of such numbers are
(a) 2
(b) 3
(c) 4
(d) 5

Answer 1

Answer:

It is given that the sum of two numbers is

1215 and their H.C.F is 81

Let the two numbers be x and y.

Now,

81x+81y=1215

81(x+y)=1215

x+y=15

So

For,x = 1, y = 14,the numbers are

1 x 81 +14 x 81 =81 + 1134 = 1215

For, x = 7, y = 8,the numbers are

7 x 81 +8 x 81 = 567 +648 = 1215

For, x= 2y=13,the numbers are

2 x 81 +13 x 81= 162 +1053=1215

For, x= 4y = 11,the numbers are

4 x 81 + 11 x 81= 324 + 891 = 1215

Therefore, the numbers of such pairs are

4

.

Answer 2

Given: Sum of the two numbers is 1215 and their HCF is 81.

To find: We have to find the possible number of pairs of such numbers.

Solution:

The highest common factor or HCF of the two numbers is 81.

So, 81 will be multiplied by the two numbers.

Let the numbers be 81x and 81y.

The sum of the two numbers is 1215.

So, we can write-

81(x+y)=1215

x+y=15

Now the two numbers can be (81×14,81×1)

(81×13,81×2),(81×7,81×8),(81×11,81×4).

So, possible pairs are 4.

The correct option is c.