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25 years ago a man was 87 times as old as his son. After 5 years the father will be 3 time as old as his son. Find their present age.

$25$ years ago a man was $87$ times as old as his son. After $5$ years the father will be $3$ time as old as his son. Find their present age.

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Exponential growth and decay: word problems

Full solution

Q.

25

years ago a man was

87

times as old as his son. After

5

years the father will be

3

time as old as his son. Find their present age.

Denote Ages: Let's denote the present age of the father as $F$ and the son as $S$ . $25$ years ago, the father's age was $F - 25$ and the son's age was $S - 25$ . The equation based on the condition $25$ years ago is: $F - 25 = 87(S - 25)$ .
Simplify Equation: Simplify the equation: $\newline$ $F - 25 = 87S - 2175$ , $\newline$ $F = 87S - 2175 + 25$ , $\newline$ $F = 87S - 2150$ .
Future Ages: Now, consider the condition that in $5$ years, the father will be $3$ times as old as his son: $\newline$ $F + 5 = 3(S + 5)$ , $\newline$ $F + 5 = 3S + 15$ , $\newline$ $F = 3S + 15 - 5$ , $\newline$ $F = 3S + 10$ .
Combine Equations: We now have two equations: $\newline$ $1$ . $F = 87S - 2150$ , $\newline$ $2$ . $F = 3S + 10$ . $\newline$ Set them equal to each other to find $S$ : $\newline$ $87S - 2150 = 3S + 10$ , $\newline$ $84S - 2150 = 10$ , $\newline$ $84S = 2160$ , $\newline$ $S = \frac{2160}{84}$ , $\newline$ $S = 25.71$ .

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