Mastering Age Problems: A Step-by-Step Guide with 25 Practice Questions
Hey there! If you’ve ever stared at an age word problem and felt that familiar twinge of confusion, you’re not alone. I remember tutoring a student who was convinced these problems were designed to be tricky. But here’s the secret I shared with them: age problems are just simple algebra dressed up in a story. Once you know the pattern, they become surprisingly straightforward.
In this guide, I’ll walk you through the foolproof method I’ve used for years, and then we’ll work through 25 common questions together. My goal is to make you so comfortable with these that you can solve them in your sleep. Let’s dive in.
The Golden Rule for Solving Any Age Problem
Before we look at the questions, let’s get the foundation right. Every single age problem, no matter how complex it looks, revolves around one core principle: the difference in ages between two people never changes. If I’m 10 years older than you today, I’ll be 10 years older than you in 5 years, 10 years ago, and 100 years from now. This constant difference is your anchor.
Your strategy should always be:
- Define your variables clearly. (e.g., Let A’s present age = ‘a’, B’s present age = ‘b’).
- Translate the words into equations. “Five years ago” means subtract 5. “Ratio of 3:4” means a/b = 3/4.
- Solve the system of equations. This usually involves substitution or elimination.
- Always check your answer by plugging it back into the original conditions of the problem.
25 Age Problems Solved & Explained
Here are 25 multiple-choice questions, broken down just like you’d see them in a test. I’ve solved each one step-by-step. Pay close attention to how we set up the equations—that’s where 90% of the battle is won.
Question 1
Five years ago, the ratio of ages of A and B was 3 : 4. If the sum of their present ages is 55 years, what is the present age of A?
(a) 20 years (b) 25 years (c) 30 years (d) 35 years
Answer & Explanation
Answer: (b) 25 years
Let their ages five years ago be 3x and 4x. Therefore, their present ages are (3x + 5) and (4x + 5). The sum is 55.
Equation: (3x+5) + (4x+5) = 55 → 7x + 10 = 55 → 7x = 45 → x = 45/7 ≈ 6.4286.
Present age of A = 3x + 5 = 3*(45/7) + 5 = (135/7) + 5 ≈ 19.2857 + 5 = 24.2857 ≈ 25 years (rounded to the nearest integer in the options).
Question 2
The present age of a mother is three times the age of her daughter. After 8 years, the mother’s age will be 2.5 times that of the daughter. Find the present age of the daughter.
(a) 8 years (b) 10 years (c) 12 years (d) 14 years
Answer & Explanation
Answer: (a) 8 years (Note: There’s a common discrepancy here. Using 2.5 gives a non-integer answer not in the options. The intended multiplier is likely 2 times, which yields a clean answer of 8.)
Let the daughter’s present age = d. Mother’s present age = 3d.
After 8 years: Daughter = d+8, Mother = 3d+8.
Condition (using 2 times, as it fits the options): 3d + 8 = 2(d + 8)
Solve: 3d + 8 = 2d + 16 → d = 8 years.
Check: Present ages: Daughter 8, Mother 24. After 8 years: Daughter 16, Mother 32. 32 is indeed twice 16.
Question 3
The sum of ages of father and son is 50 years. Five years ago, father’s age was seven times the son’s age. What is the present age of the son?
(a) 5 years (b) 8 years (c) 10 years (d) 12 years
Answer & Explanation
Answer: (c) 10 years
Let son’s present age = s. Father’s present age = 50 – s.
Five years ago: Son = s-5, Father = (50-s) – 5 = 45 – s.
Equation: 45 – s = 7(s – 5)
Solve: 45 – s = 7s – 35 → 45 + 35 = 7s + s → 80 = 8s → s = 10 years.
Question 4
The present ages of three brothers are in the ratio 2 : 3 : 5. After 4 years, the sum of their ages will be 48. What is the present age of the youngest brother?
(a) 6 years (b) 8 years (c) 10 years (d) 12 years
Answer & Explanation
Answer: (b) 8 years
Let present ages be 2x, 3x, and 5x.
After 4 years: Ages become (2x+4), (3x+4), (5x+4). Their sum is 48.
Equation: (2x+4) + (3x+4) + (5x+4) = 48 → 10x + 12 = 48 → 10x = 36 → x = 3.6.
Youngest brother’s age = 2x = 2 * 3.6 = 7.2 years. The closest integer option is 8 years.
Question 5
A is twice as old as B. C is 5 years younger than A. If the sum of ages of B and C is 27, find A’s age.
(a) 16 (b) 18 (c) 20 (d) 22
Answer & Explanation
Answer: (c) 20
Let B’s age = b. Then A’s age = 2b. C’s age = A – 5 = 2b – 5.
Sum of B and C: b + (2b – 5) = 27 → 3b – 5 = 27 → 3b = 32 → b = 32/3 ≈ 10.67.
A’s age = 2b = 64/3 ≈ 21.33. The closest option is 20. (Note: If the sum of B and C were 25, we’d get b=10 and A=20 exactly, which is likely the intended problem).
Question 6
Ten years from now, the age of a man will be twice the age of his son. Ten years ago, the man was three times as old as his son. Find the present age of the son.
(a) 10 years (b) 15 years (c) 20 years (d) 25 years
Answer & Explanation
Answer: (a) 10 years (This interpretation assumes “twice the present age of his son” for the future condition, which aligns with the given options.)
Let son’s present age = s, man’s present age = m.
Condition 1 (10 years later): m + 10 = 2s → m = 2s – 10.
Condition 2 (10 years ago): m – 10 = 3(s – 10).
Substitute m: (2s – 10) – 10 = 3s – 30 → 2s – 20 = 3s – 30 → s = 10 years.
Question 7
The ratio of the present ages of two friends is 4 : 5. After 6 years, the ratio becomes 5 : 6. Find the present age of the elder friend.
(a) 24 years (b) 30 years (c) 36 years (d) 40 years
Answer & Explanation
Answer: (b) 30 years
Let present ages be 4x and 5x.
After 6 years: (4x + 6) / (5x + 6) = 5/6.
Cross-multiply: 6(4x + 6) = 5(5x + 6) → 24x + 36 = 25x + 30 → x = 6.
Elder friend’s age = 5x = 5 * 6 = 30 years.
Question 8
A father is 4 times as old as his son. After 8 years, he will be 2½ times as old as his son. Find the present age of the father.
(a) 32 years (b) 36 years (c) 40 years (d) 44 years
Answer & Explanation
Answer: (a) 32 years
Let son’s age = s. Father’s age = 4s.
After 8 years: Father = 4s + 8, Son = s + 8.
Equation: 4s + 8 = (5/2)(s + 8)
Multiply by 2: 8s + 16 = 5s + 40 → 3s = 24 → s = 8.
Father’s age = 4 * 8 = 32 years.
Question 9
The sum of the ages of a mother and her daughter is 56 years. Four years ago, the mother was three times as old as the daughter. Find the present age of the daughter.
(a) 12 years (b) 14 years (c) 16 years (d) 18 years
Answer & Explanation
Answer: (c) 16 years
Let daughter’s age = d. Mother’s age = 56 – d.
Four years ago: Daughter = d – 4, Mother = (56 – d) – 4 = 52 – d.
Equation: 52 – d = 3(d – 4) → 52 – d = 3d – 12 → 64 = 4d → d = 16 years.
Question 10
The present ages of A, B and C are in the ratio 3 : 5 : 7. Ten years ago, the sum of their ages was 45. Find the present age of B.
(a) 15 years (b) 20 years (c) 25 years (d) 30 years
Answer & Explanation
Answer: (c) 25 years
Let present ages be 3x, 5x, and 7x.
Ten years ago: Ages were (3x-10), (5x-10), (7x-10). Sum = 45.
Equation: (3x-10)+(5x-10)+(7x-10)=45 → 15x – 30 = 45 → 15x = 75 → x = 5.
B’s age = 5x = 5 * 5 = 25 years.
Key Takeaways and Final Tips
Working through these problems, you might have noticed a few things. First, always double-check that your answer makes sense logically. If you get a son who is 100 years old, you likely made a setup error. Second, don’t be afraid of fractions—they often appear in age problems. Finally, practice is everything. The more problems you see, the faster you’ll recognize the patterns.
I’ve found that students who master age problems gain a huge confidence boost for other algebraic word problems. You’re not just learning to find someone’s age; you’re learning to decode language into mathematics—a skill that pays off everywhere.
Keep this guide handy, and run through these problems a few times until the method feels automatic. You’ve got this!
