Question

Divide 56 Into two parts such that three times the first part exceeds one third of the second by 48. The parts are

Answer 1

Answer:

20 and 36

Step-by-step explanation:

Let the two parts be x and y

Therefore,

$x + y = 56$

From the given condition,

$3x = (y \div 3) + 48$

Therefore,

$3x - (y \div 3) = 48 \\ 9x - y = 144$

Adding x+y=56 and 9x-y=144

$10x = 200 \\ x = 20$

Substituting x=20 in x+y=56

$x + y = 56 \\ 20 + y = 56 \\ y = 56 - 20 \\ y = 36$

Hope this helps!!