Question

there are 3 numbers in ascending order such that the difference of first two is 10 and that of last two is 12. if the sum of three numbers is 47, find the least number.​

Answer 1

5

Explanation:

Answer 2

17/3

Explanation:

x + y + z = 47

y - x = 10 x = y - 10

z - y = 10 z = 10 + y

y-10 + y + 10 + y = 47

3 y = 47

y = 47/3 x = 17/3 z = 77/3