A father is 3 times as old as his son. In 10 years’ time, his age will be double his son’s age. Find their present ages.
Answers
Answer:
son is 10 years and father is 30 years old
Step-by-step explanation:
method 1 :
A father is three times old as his son. In 10 years time, he will be double his son's age. What are their present ages?
Not sure what the ratio method is, but if you set the father’s age as y and the son’s age as x then you can write the equations y= 3x (bc the father is currently three times as old as the son) and y+10=2(x+10) (bc in ten years the father will be twice as old as the son). If you swap the Y in the second equation with 3x (bc as we proved in the first equation 3x=y) and distribute the 2 to the x and the 10 you get 3x+10=2x+20. Now subtract 10 from both side and 2x from both sides and you get x=10. Plug 10 in for x in the first equation (y=3x) and you get y=30. So now you know the son is 10 and the father is 30.
method 2:
Let say:
F= Father’s age
S = Son's age
For (Eq1) Father is three times as old as his son:
F = 3S
For (Eq2) In 10 year time the Father age will be twice as old as his son:
F + 10 = 2(S + 10)
F=2S+20–10
F=2S+10 (Eq2)
Then, lets us substitute the (Eq1) into the (Eq2)
3S=2S+10
3S-2S=10
S=10
Then son is 10 years old.
and Father
F=3S so 3 x 10
Father age is 30 years old.
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Step-by-step explanation:
Let say:
F= Father’s age
S = Son's age
For (Eq1) Father is three times as old as his son:
F = 3S
For (Eq2) In 10 year time the Father age will be twice as old as his son:
F + 10 = 2(S + 10)
F=2S+20–10
F=2S+10 (Eq2)
Then, lets us substitute the (Eq1) into the (Eq2)
3S=2S+10
3S-2S=10
S=10
Then son is 10 years old.
and Father
F=3S so 3 x 10
Father age is 30 years old.
