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Age problems are easy algebra wearing a confusing costume. Nothing in them is harder than a two-step linear equation. The whole difficulty sits in four small words — 'ago', 'hence', 'was' and 'will be' — and in the habit of subtracting years from one person and forgetting to subtract them from the other.
That is the one rule the chapter really tests. Time moves for everybody at the same rate. If you go back 5 years, every person in the question goes back 5 years. If you go forward 12, everyone goes forward 12.
So the method never changes. Name the present age with one letter, build a small table of past, present and future, translate each sentence into one equation, and solve. One variable is almost always enough; reaching for two is what turns a 40-second question into a three-minute one.
SSC CGL and CHSL, RRB NTPC and Group D, and BPSC TRE-4 elementary mathematics all carry one or two of these in every paper.
1. A father is 3 times as old as his son. After 12 years, the father will be twice as old as the son. What is the son's present age?
2. The ratio of the present ages of A and B is 4 : 3. After 6 years the ratio becomes 6 : 5. What is A's present age?
3. The difference between the ages of two persons is 10 years. Fifteen years ago, the elder was twice as old as the younger. What is the elder person's present age?
4. The sum of the present ages of a mother and her daughter is 60 years and the difference between them is 26 years. What is the daughter's present age?
5. Ten years ago, P was half of Q's age. If the ratio of their present ages is 3 : 4, what is the sum of their present ages?
6. A is 5 years older than B. In 8 years, the sum of their ages will be 45 years. What is B's present age?
7. A is twice as old as B. Five years ago, A was three times as old as B. What is A's present age?
8. The ratio of the present ages of a man and his wife is 4 : 3. After 4 years the ratio will be 9 : 7. If at the time of their marriage the ratio was 5 : 3, how many years ago were they married?