Question

divide 56 into 2parts such that 3 times the first part exceeds one third of the second by 48 .find the parts

Answer 1

first number is 20 and second number is 36

Answer 2

Let the first part be x.
then, second part = (56 - x)

According to question, first part exceeds second part. It means that the difference between the first part and second part is 48. So,

3x + $\frac{1}{3}$ × (56 - x) = 48

Take L.C.M

$\frac{9x - (56 - x)}{3}$ = 48
9x - 56 + x = 48 × 3
10x = 144 - 56
x = $\frac{200}{10}$ = 20

∴ First part = x = 20
   Second part = 56 - x = 36