A father is 3 times as old as his son. In 12 years he will be twice as old as his son. What is the son's present age?

12 years. With the father = 3s: 3s + 12 = 2(s + 12), so s = 12 years.

The present ages of A and B are in the ratio 4 : 5. Five years ago the ratio was 7 : 9. What is A's present age?

40 years. Let the ages be 4x and 5x. Then 4x - 55x - 5 = 79, so 36x - 45 = 35x - 35 and x = 10. A's age is 4 10 = 40 years.

The sum of the present ages of a mother and her daughter is 60 years. Six years ago the mother was five times as old as the daughter. What is the daughter's present age?

14 years. Let the daughter be d; the mother is 60 - d. Six years ago: (60 - d) - 6 = 5(d - 6), i.e. 54 - d = 5d - 30, so 6d = 84 and d = 14 years.

A is 2 years older than B, who is twice as old as C. If the total of their ages is 27 years, how old is B?

10 years. With A = B + 2 and C = B2: (B + 2) + B + B2 = 27, so 5B2 = 25 and B = 10 years.

Ten years ago the ages of a man and his son were in the ratio 3 : 1. Ten years from now the ratio will be 2 : 1. What are their present ages?

70 and 30 years. Ten years ago let the ages be 3x and x. Ten years from now they are 3x + 20 and x + 20, with 3x + 20x + 20 = 2. So x = 20, and the present ages are 3x + 10 = 70 and x + 10 = 30.

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Quantitative Aptitude · Averages, Ratios & Percentages

Problems on Ages — Questions & Answers

Age problems give relationships between people's ages at different times. Let the present age be a variable, write 'years ago' as a subtraction and 'years hence' as an addition, then solve.

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Key formulas included

Key formulas

If present age is $x$ , then $n$ years ago it was $x-n$ and $n$ years hence it will be $x+n$
The difference between two people's ages stays constant over time
A ratio of ages plus a time shift usually gives one linear equation in one unknown

5 solved Problems on Ages questions

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01
A father is 3 times as old as his son. In 12 years he will be twice as old as his son. What is the son's present age?
5 steps · reveal one at a time
1. Step 1: Set up variables
  Let $s$ represent the son's present age. The problem states the father is 3 times as old as his son, so the father's present age can be written as $3s$ .
2. Step 2: Ages in 12 years
  In 12 years, the son's age will be $s + 12$ . The father's age will be $3s + 12$ . We are told that in 12 years, the father will be twice as old as his son.
3. Step 3: Build the equation
  We can express the relationship in 12 years as an equation: the father's future age equals twice the son's future age. This gives us $3s + 12 = 2(s + 12)$ .
4. Step 4: Solve for son's age
  Now, let's solve the equation for $s$ : $3s + 12 = 2s + 24$ . Subtract $2s$ from both sides to get $s + 12 = 24$ . Then subtract $12$ from both sides, so $s = 12$ .
5. Step 5: The answer
  So the answer is (B) 12 years, because we found that the son's present age, $s$ , is 12 years.
02
The present ages of A and B are in the ratio 4 : 5. Five years ago the ratio was 7 : 9. What is A's present age?
5 steps · reveal one at a time
1. Step 1: Set up variables
  Let A's present age be $4x$ years and B's present age be $5x$ years. This directly reflects their current age ratio of $4:5$ .
2. Step 2: Ages five years ago
  Five years ago, A's age would have been $4x - 5$ and B's age would have been $5x - 5$ . We are told that their ratio at that time was $7:9$ .
3. Step 3: Formulate equation
  We can set up an equation using the ages from five years ago and their ratio: $\frac{4x - 5}{5x - 5} = \frac{7}{9}$ . Now, we need to solve for $x$ .
4. Step 4: Solve for x
  Cross-multiply to solve the equation: $9(4x - 5) = 7(5x - 5)$ . This simplifies to $36x - 45 = 35x - 35$ . Subtracting $35x$ from both sides gives $x - 45 = -35$ . Adding $45$ to both sides yields $x = 10$ .
5. Step 5: The answer
  So the answer is (B) 40 years. A's present age is $4x$ . Since we found $x = 10$ , A's age is $4 \times 10 = 40$ years.
03
The sum of the present ages of a mother and her daughter is 60 years. Six years ago the mother was five times as old as the daughter. What is the daughter's present age?
4 steps · reveal one at a time
1. Step 1: Define Present Ages
  Let the daughter's present age be $d$ years. Since the sum of their present ages is 60 years, the mother's present age can be expressed as $60 - d$ years. This sets up our variables.
2. Step 2: Ages Six Years Ago
  Next, let's determine their ages six years ago. The daughter's age then was $d - 6$ years. The mother's age was $(60 - d) - 6$ , which simplifies to $54 - d$ years. We'll use these past ages.
3. Step 3: Formulate and Solve
  The problem states that six years ago, the mother was five times as old as the daughter. So, we can write the equation: $54 - d = 5(d - 6)$ . Expanding this gives $54 - d = 5d - 30$ . Rearranging terms, we get $84 = 6d$ , which means $d = 14$ .
4. Step 4: The answer
  So the answer is (A) 14 years. The value we found for $d$ represents the daughter's present age, which is exactly what the question asked for.
04
A is 2 years older than B, who is twice as old as C. If the total of their ages is 27 years, how old is B?
5 steps · reveal one at a time
1. Step 1: Define ages
  Let B represent B's age. We know A is 2 years older than B, so $A = B + 2$ . We also know B is twice as old as C, which means C is half of B's age, so $C = B/2$ .
2. Step 2: Formulate equation
  The problem states that the total of their ages is 27 years. We can write this as an equation: $A + B + C = 27$ . Now, substitute the expressions for A and C in terms of B into this equation.
3. Step 3: Substitute and solve
  Substituting the expressions gives us $(B + 2) + B + (B/2) = 27$ . Combine the terms with B: $2B + B/2 + 2 = 27$ . This simplifies to $(5B/2) + 2 = 27$ .
4. Step 4: Calculate B's age
  Subtract 2 from both sides of the equation: $(5B/2) = 25$ . To find B, multiply both sides by $2/5$ : $B = 25 \times (2/5) = 50/5 = 10$ years.
5. Step 5: The answer
  So the answer is (A) 10 years, because we calculated B's age to be 10 years by setting up and solving the age equation.
05
Ten years ago the ages of a man and his son were in the ratio 3 : 1. Ten years from now the ratio will be 2 : 1. What are their present ages?
4 steps · reveal one at a time
1. Step 1: Set up variables
  Let the man's age ten years ago be $3x$ and his son's age be $x$ . This represents the given ratio of $3:1$ at that specific time.
2. Step 2: Calculate future ages
  To find their ages ten years from now, we add 20 years to their ages from ten years ago (10 years to reach the present, then another 10 years into the future). So, their ages will be $3x + 20$ and $x + 20$ .
3. Step 3: Form and solve equation
  We are told that ten years from now, the ratio of their ages will be $2:1$ . We can write this as $\dfrac{3x + 20}{x + 20} = 2$ . Solving for $x$ gives $3x + 20 = 2(x + 20)$ , which simplifies to $3x + 20 = 2x + 40$ , so $x = 20$ .
4. Step 4: The answer
  So the answer is (C) 70 and 30 years. Their present ages are found by adding 10 years to their ages from ten years ago: man's age is $3x + 10 = 3(20) + 10 = 70$ and son's age is $x + 10 = 20 + 10 = 30$ .

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