The sum of a father’s and son’s ages is (56) years. Six years ago, the father’s age was (3) times the son’s age. What is the father’s present age?
Answer and explanation
Correct answer: (45) years
Let the present ages of the father and son be \(x\) and \(y\) years, respectively. Then \(x+y=56\). Six years ago, \(x-6=3(y-6)\), which gives \(x=3y-12\). Substituting this in the first equation, \(3y-12+y=56\), so \(4y=68\), \(y=17\), and \(x=39\). Therefore, the father’s present age is (39) years.
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What is the correct answer to this question?
(45) years
Why is this the correct answer?
Let the present ages of the father and son be \(x\) and \(y\) years, respectively. Then \(x+y=56\). Six years ago, \(x-6=3(y-6)\), which gives \(x=3y-12\). Substituting this in the first equation, \(3y-12+y=56\), so \(4y=68\), \(y=17\), and \(x=39\). Therefore, the father’s present age is (39) years.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Pair of Linear Equations in Two Variables. Topic: Algebraic methods: Substitution method and Elimination method..
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