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Pair of Linear Equations in Two Variables
CH03 Question Bank
CBSE_CH03_QUESTION_BANK
Grade 10
Question:
Ten years ago, a father was $12$ times as old as his son. Ten years from now, the father will be twice as old as the son will be then. Find their present ages.
Step-by-Step Solution
Key Concept: Let the present ages be $F$ and $S$; translate both time-shifted conditions into linear equations and solve by substitution.
Let the father's present age be $F$ years and the son's present age be $S$ years. [0.5 Mark]
\"Ten years ago, father was 12 times as old as son\": $F-10=12(S-10)\Rightarrow F=12S-110$ — (1). [1.5 Marks]
\"Ten years hence, father will be twice as old as son\": $F+10=2(S+10)\Rightarrow F=2S+10$ — (2). [1.5 Marks]
Equating (1) and (2): $12S-110=2S+10\Rightarrow 10S=120\Rightarrow S=12$. [1.0 Mark]
$F=2(12)+10=34$. So the father is $34$ years old and the son is $12$ years old. [0.5 Mark]
Correct Answer:
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