Three time the present age of a father is equal to eight times the present age of his son. Eight years hence the father will be twice as old as his son at that time. What are their present ages?
Let's use algebra to solve this problem. Let the present age of the father be "F" years and the present age of the son be "S" years.
We are given two pieces of information:
Three times the present age of the father is equal to eight times the present age of his son: 3F = 8S
Eight years hence (in the future), the father will be twice as old as his son at that time: (F + 8) = 2(S + 8)
Now, we have a system of two equations:
Equation 1: 3F = 8S Equation 2: F + 8 = 2(S + 8)
Let's solve this system of equations:
From Equation 1, we can express F in terms of S: 3F = 8S F = (8S)/3
Now, substitute this expression for F into Equation 2: (8S/3) + 8 = 2(S + 8)
Now, solve for S:
(8S/3) + 8 = 2S + 16
Multiply both sides by 3 to get rid of the fraction:
8S + 24 = 6S + 48
Subtract 6S from both sides:
2S + 24 = 48
Subtract 24 from both sides:
2S = 48 - 24 2S = 24
Divide by 2:
S = 24 / 2 S = 12
Now that we have found the age of the son (S), we can find the age of the father using Equation 1:
3F = 8S 3F = 8 * 12 3F = 96
Divide by 3:
F = 96 / 3 F = 32
So, the present age of the father (F) is 32 years, and the present age of the son (S) is 12 years.
Therefore, their present ages are 32 and 12 years, respectively.
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