Question

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

Answer 1

Let the 1st part be x.
Therefore, 2nd part=56-x
3 times first part=3*x=3x
1/3rd of 2nd part=(56-x)/3
According to the question,
3x-(56-x)/3=48
or, 9x-56+x=144
or,10x=144+56
or 10x=200 or, x=20
Therefore, 1st part=20 and
2nd part=(56-20)=36.
Ans:The parts are 20 and 36.

Answer 2

Let consider the first part =P

∴ 2nd part = 56 - P

3 times of first part = 3P

According to the question :-

3P - ( 56 - P) ×$\frac{1}{3}$ = 48

9P -56 + P = 48 ×3

10P -56= 144

10P = 144 + 56= 200

P = $\frac{200}{10}$ = $20$

∴First part = 20

2nd part = 56 - 20= 36

Ans :- The parts are 20 and 36.