Question

divide into 56 two parts such that three times the first part exceeds one third of the second by 48 the parts are​

Answer 1

Answer:

20 & 36

Step-by-step explanation:

Let the parts be a and b

Then,

a + b = 56 ......(1)

Also, according to question

3a = b/3 + 48

3a - b/3 = 48

(3a × 3/3) - b/3 = 48

9a/3 - b/3 = 48

(9a - b)/3 = 48

9a - b = 144 ......(2)

Add equations (1) and (2)

a + b + 9a - b = 56 + 144

10a = 200

a = 200/10

a = 20

If a = 20 then from equation (1)

a + b = 56

20 + b = 56

b = 56 - 20

b = 36