PROBLEMS ON AGE
Q1–10 of 30The present ages of three persons are in the proportion 4 : 7 : 9. Eight years ago, the sum of their ages was 56 years. The present age of the eldest person is
Present ages ratio = $4 : 7 : 9$, so let ages be $4x$, $7x$, $9x$.
$8$ years ago, sum of ages = $(4x-8) + (7x-8) + (9x-8) = 56 \Rightarrow 20x - 24 = 56 \Rightarrow 20x = 80 \Rightarrow x = 4$.
Present age of eldest = $9x = 9 \times 4 = 36$ years.
Answer: (b) 36 years
In 10 years, A will be twice as old as B was 10 years ago. If A is now 9 years older than B, the present age of B is
Let B's age be $x$, so A's age = $x + 9$.
In $10$ years: $(x + 9 + 10) = 2(x - 10) \Rightarrow x + 19 = 2x - 20 \Rightarrow x = 39$.
Answer: (c) 39 years
Reenu's father was 38 years of age when she was born while her mother was 36 years old when her brother 4 years younger to her was born. What is the difference between the ages of her parents?
Father's age when Reenu was born = $38$.
Brother was born $4$ years after Reenu, so father was $38 + 4 = 42$ years old then.
Mother's age when brother was born = $36$.
Difference = $42 - 36 = 6$ years.
Answer: (c) 6 years
The sum of the ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?
Let ages be $x, x+3, x+6, x+9, x+12$.
Sum = $5x + 30 = 50 \Rightarrow 5x = 20 \Rightarrow x = 4$.
Answer: (a) 4 years
A man was asked to state his age in years. His reply was, "Take my age 3 years hence, multiply it by 3 and then subtract 3 times my age 3 years ago and then you will know how old I am." What is the age of the man?
Let age be $x$.
Equation: $3(x + 3) - 3(x - 3) = x \Rightarrow 3x + 9 - 3x + 9 = x \Rightarrow x = 18$.
Answer: (a) 18 years
The sum of the ages of Jayant, Prem and Paras is 93 years. Ten years ago, the ratio of their ages was 2 : 3 : 4. What is the present age of Paras?
Sum of present ages = $93$.
$10$ years ago, sum of ages = $93 - 30 = 63$.
Ratio $10$ years ago = $2 : 3 : 4 \Rightarrow$ Paras's age then = $\frac{4}{9} \times 63 = 28$.
Paras's present age = $28 + 10 = 38$ years.
Answer: (e) 38 years
The sum of the ages of a man and his son is 45 years. Five years ago, the product of their ages was 34. The man's age is
Let present ages be M and S.
M+S=45
Five years ago:
(M−5)(S−5)=34
Also,
(M−5)+(S−5)=35
Two numbers with sum 35 and product 34 are:
1, 34
So five years ago, man's age =34
Present man's age: 39 years
✅ Answer: 39 years
The ratio of a man's age and his son's age is 7 : 3 and the product of their ages is 756. The ratio of their ages after 6 years will be
Let ages be $7x$ and $3x$. Product = $21x^2 = 756 \Rightarrow x^2 = 36 \Rightarrow x = 6$.
Present ages: $42$ and $18$.
Ratio after $6$ years: $(42+6) : (18+6) = 48 : 24 = 2 : 1$.
Answer: (b) 2 : 1
Sonal is 40 years old and Nitya is 60 years old. How many years ago was the ratio of their ages 3 : 5?
Let $x$ years ago ratio was $3 : 5$.
$\frac{40 - x}{60 - x} = \frac{3}{5} \Rightarrow 200 - 5x = 180 - 3x \Rightarrow 2x = 20 \Rightarrow x = 10$.
Answer: (b) 10 years
The ratio between the present ages of A and B is 5 : 3 respectively. The ratio between A's age 4 years ago and B's age 4 years hence is 1 : 1. What is the ratio between A's age 4 years hence and B's age 4 years ago?
Present ages: $A = 5x, B = 3x$.
$\frac{5x - 4}{3x + 4} = 1 \Rightarrow 5x - 4 = 3x + 4 \Rightarrow 2x = 8 \Rightarrow x = 4$.
Present ages: $A = 20, B = 12$.
Ratio of $(A+4)$ to $(B-4) = (20+4) : (12-4) = 24 : 8 = 3 : 1$.
Answer: (b) 3 : 1