Aptitude : Problems on Ages

(16) Ten years ago, the age of a mother was six times the age of her daughter. If the mother is currently 40 years old, how old is the daughter now?

• (a) 4 years
• (b) 5 years
• (c) 6 years
• (d) 7 years

Answer

Correct Answer: (d) 7 years
Explanation:
Let the age of the daughter be \( D \) years. Ten years ago, the mother’s age was \( 40 – 10 = 30 \) years, and the daughter’s age was \( D – 10 \) years.
Given that the mother is currently 40 years old, we can write the equation \( 30 = 6(D – 10) \).
Solving for \( D \), we get \( D = 7 \) years.
30 = 6(D - 10)
D - 10 = \frac{30}{6}
D - 10 = 5
D = 5 + 10 = 7

(17) A family consists of a mother, father, and their two children. The sum of their ages is 100 years. If the mother is 5 years older than the father, how old is the father?

• (a) 35 years
• (b) 40 years
• (c) 45 years
• (d) 50 years

Answer

Correct Answer: (c) 45 years
Explanation:
Let the father’s age be \( F \) years. Therefore, the mother’s age is \( F + 5 \) years.
The sum of their ages is given as \( F + (F + 5) + 2C = 100 \), where \( C \) is the combined age of the two children.
Since there are two children, \( 2C \) represents their combined age.
Solving for \( F \), we get \( 2F + 5 + 2C = 100 \).
2F + 5 + 2C = 100
2F + 2C = 100 - 5
2F + 2C = 95
2(F + C) = 95
F + C = \frac{95}{2}
Since \( F \) and \( C \) are both integers, \( \frac{95}{2} \) must be an integer, so \( F + C \) must be an even number.
The only even number less than 100 that can be expressed as the sum of an odd number and an even number is 50.
Therefore, \( F + C = 50 \), and since \( C \) represents the combined age of the two children, we can assume \( C = 0 \).
Thus, \( F = 50 – C = 50 \) years.

(18) The sum of the present ages of Emily and her daughter is 45 years. If Emily is twice as old as her daughter, how old is Emily?

• (a) 25 years
• (b) 30 years
• (c) 35 years
• (d) 40 years

Answer

Correct Answer: (b) 30 years
Explanation:
Let Emily’s current age be \( E \) years and her daughter’s current age be \( D \) years.
Given that the sum of their ages is 45 years, we can write the equation \( E + D = 45 \).
Also, Emily is twice as old as her daughter, so \( E = 2D \).
We can now solve these equations simultaneously. Substituting \( E = 2D \) into \( E + D = 45 \), we get \( 2D + D = 45 \), which simplifies to \( 3D = 45 \).
Solving for \( D \), we find \( D = 15 \) years. Therefore, Emily’s age is \( E = 2D = 2 \cdot 15 = 30 \) years.

(19) The ratio of Sarah’s age to Michael’s age is 4:7. If Michael is currently 28 years old, how old is Sarah?

• (a) 14 years
• (b) 16 years
• (c) 18 years
• (d) 20 years

Answer

Correct Answer: (b) 16 years
Explanation:
Let Sarah’s current age be \( S \) years and Michael’s current age be \( M \) years.
Given that the ratio of their ages is 4:7, we can write the equation \(\frac{S}{M} = \frac{4}{7}\).
Also, we are given that Michael is currently 28 years old, so \( M = 28 \).
We can solve for \( S \) by substituting \( M = 28 \) into the ratio equation: \(\frac{S}{28} = \frac{4}{7}\).
Solving for \( S \), we find \( S = 16 \) years.

(20) Ten years ago, the age of a father was six times the age of his son. If the son’s current age is 12 years, how old is the father now?

• (a) 52 years
• (b) 62 years
• (c) 72 years
• (d) 82 years

Answer

Correct Answer: (b) 62 years
Explanation:
Let the son’s current age be \( S \) years and the father’s current age be \( F \) years.
Given that the son’s age is 12 years, we can write the equation \( S = 12 \).
We are also given that ten years ago, the father’s age was six times the son’s age, so \( F – 10 = 6(S – 10) \).
Substituting \( S = 12 \) into the equation, we get \( F – 10 = 6(12 – 10) \), which simplifies to \( F – 10 = 12 \).
Solving for \( F \), we find \( F = 62 \) years.