In order to crack the Quantitative Aptitude (Math) questions asked in Banking, SSC and other Competitive Examinations, conceptual clarity is a must. It is often seen that students are able to solve the calculation based questions comfortably. However, they get stuck while attempting word problems asked in these exams.
To address this concern, Vidya Guru, which is regarded as the Centre for Best SSC CGL Coaching in Delhi region, has started a series on word problems taken from different topics of Quantitative Aptitude. In this part, we discuss some quality age-based problems. These problems will not only improve your understanding of the topic, but also provide you with some practice.
Question-1. Three years ago, a mother age was 4 times as old as her daughter. Their combined ages three years from now will be 67. What is the Mother’s present age?
Solution: Let the daughter’s age three years ago be x. So, the mother’s age three years ago will be 4x.
Daughter’s present age = x+3 & Mother’s present age = 4x+3
Three years from now, Daughter’s age will be = x+6 & Mother’s age will be = 4x+6.
(x+6) + (4x+6) = 67
5x + 12 = 67
x = 11. So, the daughter’s age three years ago was 11 and mother’s age was 4×11 = 44.
So, mother’s present age = 44+3 = 47 years.
Question-2. A grandfather is 3 times as old as his grandson. In 20 years, the grandfather will be twice as old as his grandson. How old are they now?
Solution: Let the grandson’s age be x. Then the grandfather’s age is 3x.
In 20 years, the grandson’s age will be: x + 20, and the grandfather’s age will be: 3x + 20.
According to the problem, the grandfather will be twice as old as the grandson in 20 years, i.e.,
3x + 20 = 2(x + 20)
Simplifying, we get:
x = 20
So the grandson’s age is 20, and the grandfather’s age is 3x = 60.
Question-3. The ratio of the ages of a man and his wife is 4:3. After 5 years, the ratio will be 5:4. What are their current ages?
Solution: Let the current ages of the man and his wife be 4x and 3x, respectively.
After 5 years, their ages will be 4x + 5 and 3x + 5, respectively.
According to the problem, the ratio of their ages after 5 years will be 5:4, i.e.,
(4x + 5)/(3x + 5) = 5/4
Simplifying, we get:
x = 5
So the man’s current age is 4x = 20, and his wife’s current age is 3x = 15.
Question-4. Ratio of ages of two friends Ravi & Amit is 5:4. After 4 yrs, the ratio will be 11:9. The ratio was 7:5 when they first met. How many years ago did they first met?
Solution: Let Ravi’s age be 5x and Amit’s age be 4x.
After 4 years, their ages will be 5x+4 and 4x+4 respectively.
5x+4/4x+4 = 11/9.
Solving for x, we get x = 8.
Thus, Ravi’s age = 5x = 40 years and Amit’s age = 4x = 32 years.
Let the first time they met be y years go.
So, 40-y/32-y = 7/5.
Solving for y, we get y = 12.
So, they first met 12 years ago.
Question-5. A father is 42 years old and his son is 10 years old. In how many years will the father be twice as old as his son?
Solution: Let the number of years be x.
After x years, the father’s age will be 42 + x, and the son’s age will be 10 + x.
According to the problem, the father will be twice as old as his son after x years, i.e., 42 + x = 2(10 + x)
Simplifying, we get: x = 22
So the father will be twice as old as his son in 22 years.
Question-6. Two years ago, Ruby was 6 times as old as her son. Eleven years hence, Ruby’s age will exceed her son’s age by twenty five years. What is the Ratio of present ages of Ruby and her son?
Solution: Two years ago, let the son’s age be x and Ruby’s age be 6x. Eleven years from now, the gap between their ages will be 25 years. This gap is constant. It was the same even 2 years ago.
6x – x = 25 (As per the question statement)
Solving for x, we get x = 5.
Two years ago, son’s age = x = 5 years and Ruby’s age = 6x = 30 years.
Thus, present ages of Ruby and her son are 32 (30+2) and 7 (5+2) respectively.
The final ratio is 32:7.
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