Question

3. The sum of two numbers is 192 & their HCF is 24. The number of pairs of numbers

satisfying the above condition is-

a) 6 b) 5 c) 4 d) 2​

Answer 1

Answer:

6

Step-by-step explanation:

Let number 1 be x and number 2 be y.

x + y = 192

x = 16a

y = 16b

16a + 16b = 192

16(a + b) = 192

a + b = 12

possibilities for a & b not x & y:

0,12 (rejected)

1,11

2,10

3,9

4,8

5,7

6,6

Therefore, 6 pairs can be formed.

Answer 2

Answer:

6

Step-by-step explanation:

Let, the numbers are 24x and 24y

according to the question,

24x+24y = 192

x+y = 8

so, the possible pairs are

(1,7)

(2,6)

(3,5)

(4,4) rejected , because their HCF is not 24

(5,3)

(6,2)

(7,1)

so, their are 6 possibility