Question 1188408: 5 years ago, Fred was one-third Joe’s age. In 10 years, he will be two-thirds of Joe’s age. What is Fred’s age now?
Found 2 solutions by MathLover1, ikleyn: Answer by MathLover1(20850) (Show Source): Answer by ikleyn(52800) (Show Source):
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5 years ago, Fred was one-third Joe’s age.
In 10 years, he will be two-thirds of Joe’s age.
What is Fred’s age now?
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I read, understand and interpret the problem differently from the other tutor,
so my solution and my answer are different.
Let Ferd's age be F and Jow's age be J.
First equation is
F - 5 =
I simplify it to
3F - 15 = J - 5
3F - J = 10. (1)
Second equation is
F + 10 = .
I simplify it to
3F + 30 = 2J + 20
3F - 2J = -10. (2)
So, I have now this system of two equations in two unknowns
3F - J = 10 (1)
3F - 2J = - 10 (2)
To solve it, apply the Eliminatiom method: from equation (1) subtract equation (2). You will get then
-J - (-2J) = 10 - (-10)
J = 20.
Then from equation (1), 3F - 20 = 10, 3F = 10 + 20 = 30, F = 30/3 = 10.
ANSWER. Fred's age is 10 years; Joe's age is 20 years.
CHECK. 5 years ago, they there 5 and 15 years old.
In 10 years, their ages will be 20 and 30 years,
so, the problem's conditions are satisfied.
Solved.
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