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Challenge Problems1. A man has a daughter and a son. The son is three years older than the daughter. In one year, the man will be six times as old as the daughter is now. In ten years, the man would be fourteen years older than the combined ages of his children at that time. What is the present age of-the man.



Answer :

Step-by-step explanation:

Let the man's current age be given as M

Let his son's age be given as A

Let his daughter's age be given as D

From the question

S= 3+D.........(1)

In one year,

M+1 = 6D.....(2)

In sixteen years,

M+16 = (S+16) + (D+16).......(3)

From (1) to (3)

M+16 = (3+D+16) + (D+16)

M+16 = 2D + 35

M = 2D + 19........(4)

From (4) to (2)

(2D + 19) + 1 = 6D

2D + 20 = 6D

.....

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