Question 1204000: Ten years ago, Pepito's age was three times the age of his son, Chito. After ten years, Pepito’s age will be twice that of his son. Find the ratio of their present ages
Found 3 solutions by josgarithmetic, ikleyn, greenestamps:
Answer by josgarithmetic(39616)   (Show Source):
Answer by ikleyn(52776)  (Show Source):
You can put this solution on YOUR website! .
Ten years ago, Pepito's age was three times the age of his son, Chito.
After ten years, Pepito’s age will be twice that of his son. Find the ratio of their present ages
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Let C be the Chito's age; P be the Pepito's age.
Then for their ages 10 years ago we have
P-10 = 3(C-10)
or P = 3(C-10) + 10 = 3C -20. (1)
For their ages in ten years, we have this equation
P + 10 = 2(C+10),
or P = 2C + 10. (2)
From (1) and (2), we have
3C - 20 = 2C + 10,
3C - 2C = 20 + 10,
C = 30.
So, the son is 30 years old; the father is 3C - 20 = 3*30 - 20 = 70.
The ratio of their ages is P/C =  =  . ANSWER
Solved.
Answer by greenestamps(13198)  (Show Source):
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