Problem: The sum of the age of 3 persons is 150 year. 10 years ago their ages were in the ratio 7:8:9. Find their present ages.

Solution:

Let the present ages of the three persons be a, b and c.

We know that the sum of their ages is 150, so we can write:

a + b + c = 150

10 years ago, their ages were (a-10), (b-10) and (c-10) respectively.

We are given that their ages 10 years ago were in the ratio 7:8:9, so we can write:

(a-10) : (b-10) : (c-10) = 7 : 8 : 9

Multiplying both sides by a common factor, we get:

7a - 70 : 8b - 80 : 9c - 90 = 7 : 8 : 9

Simplifying the above equation, we get:

7a - 8b + b - 70 - 80 = 0

8b - 9c + c - 80 - 90 = 0

Adding the above two equations, we get:

7a - c - 240 = 0

Substituting a + b + c = 150 in the above equation, we get:

7a - 150 - 240 = 0

7a = 390

a = 55

Substituting a = 55 in a + b + c = 150, we get:

b + c = 95

We know that (a-10) : (b-10) : (c-10) = 7 : 8 : 9

Substituting a = 55, we get:

45 : 46 : 47

Adding 10 to each of the above values, we get:

55 : 56 : 57

Therefore, the present ages of the three persons are 55, 56 and 57 years.

Answer: The present ages of the three persons are 55, 56 and 57 years.