Question

Divide 3000 into two parts such that half of one part may be less than the other by 48.

Answer 1

Q) Divide 300 into two parts such that half of one part may be less than the other by 48. (Its 300 not 3000)


Solution: Let x + y = 300, where x and y are two (parts)numbers
Given, x - y/2 = 48
=> x = 48 + y/2
Hence, 48 + y/2 + y = 300
3y/2 = 252
y = 2 X 252 / 3 = 504 / 3 = 168
x = 300 - 168 = 132
Hence, the two parts (numbers)are 132 and 168.