Problem on ages is one of the easiest topics of the Quantitative Aptitude section. Today, we will discuss the basic approach and some previous year questions of Age-based problems.
In this topic, you need to know the basics of linear equations and ratios. You may across many formulas in this topic, but if you understand the basics well, you won’t need them. So, make sure that you clear your basics of the above two topics and practice well.
The main difficulty in this chapter is that students are not able to express given data in a proper manner, so solving these problems becomes a bit difficult.
Note - Try to make only one base variable and express other variables in terms of it.
Read questions carefully and convert the given information into a linear equation.
We will discuss various questions asked in previous year exams.
Example 1. The ratio of A’s age 3 years ago and B’s age 5 years ago is 4:5. If A is 4 years younger than B then what is the present age of B?
APPROACH 1: Let assume at present A’s age is X and B’s age is Y.
Given, (X-3):(Y-5) = 4:5 and X = Y-4
Solve both equations and then get X and Y.
Now, we have two equations and two variables. Generally, this type of approach consumes more time and is not preferable.
APPROACH 2: In this type of question, assume only one variable i.e. called a base variable. Let A’s present age is X, then B’s present age is X+4. Students often commit silly mistakes, they tick the option which has X value but the present age of B is asked in the question.
So, (A’s 3yrs ago):(B’s 5yrs ago) = 4:5
(X-3): (X+4-5) = 4:5
(X-3): (X-1) = 4:5
5X-15 = 4X-4
X= 11 i.e. A’s Present age
and B’s present age is X+4 = 11+4 = 15yrs.
2. A is 3 years older than B while B is 2 years older than C. The ratio of age of A 4 years hence and B 3years ago is 5:3. What was the age of C 6 years ago?
Solution: There are three variables in this question and students are often confused about which variables should be assumed as a base variables. You can see in question relation of B with both A and C is mentioned. So, you can assume B as the base variable. If you assume another one as a base variable, there is a chance that calculation may become a little harder.
Let present age of B’s is X, then A’s age = X+3 and C’s age = X-2
(Age of A 4yrs hence): (Age of B 4yrs ago) = 5:3
(X+3+4): (X-3) = 5:3
(X+7): (X-3) = 5:3
3X+21 = 5X-15
2X = 36
X = 18
present age of C is X-2 = 18-2 = 16yrs
age of C 6yrs ago = 16-6 = 10yrs
Shortcut: (X+7): (X-3) = 5:3
difference of 5 and 3 is directly proportional to difference of (X+7)and(X-3)
So,(5-3) ∝ (X+7)-(X-3)
2 ∝ 10
1 ∝ 5
We can write X+7 = 5×5 , X = 18
Hence, age of C 6yrs ago = 10yrs
3. The average age of Atul, Jatin and Sonu is 24years. 2 years ago, the average age of Atul and Sonu was 23yrs. 2yrs hence average age of Jatin and Sonu is 26 years. Find the present age of Sonu?
Solution: Given, the average age of Atul, Jatin and Sonu = 24 yrs
∴ total age of Atul, Jatin and Sonu = 24×3 = 72 yrs ….(1)
2yrs ago, the average age of Atul and Sonu = 23
2yrs ago, total age of atul and sonu = 46
∴ present total age of atul and sonu = 50 ….(2)
2yrs hence, average age of Jatin and Sonu = 26
2yrs hence, total age of Jatin and Sonu = 52
∴ present total age of Jatin and Sonu = 48 …….(3)
From equation 1, 2 and 3
Present age of Sonu is 26 years.
4. The average age of A and B is 25 years. If C were to replace A, the average would be 24 and if C were to replace B, the average would be 26. What are the ages of A, B and C respectively?
Solution: Given, avg. of (A+B) = 25
∴ (A+B) = 50
Avg. of (C+B) = 24
∴ (C+B) = 48
Avg. of (A+C) = 26
∴ (A+C) = 52
Now , A+B+C = (50+48+52)/2
A+B+C = 75
C = 25, A = 27, B = 23
5. The ages of Sona and Mona are in the ratio of 15:17 respectively. After 6 years, the ratio of their ages will be 9:10. What will be the age of Mona after 6 years?
Solution: Let present age of Sona is 15X, then Mona = 17X
Given, (15X+6) : (17X+6) = 9 : 10
150X+60 = 153X+54
3X = 6
X = 2 yrs
Age of Mona after 6yrs is (17×2+6) = 40yrs.
Now, we will discuss questions that are asked by some of our students.
1. The present ages of A,B and C in proportion 4:7:9, eight years ago, the sum of their ages was 56. find their respective ages (in years)
Solution: Let ratio of present ages of A,B and C is 4x:7x:9x
sum of their ages eight years ago = 56
(4x-8)+(7x-8)+(9x-8) = 56
20x = 80
x = 4
A = 4x = 16 years
B = 7x = 28 years
C = 9x = 36 years
2. The sum of the ages of Rinku and Gopal is 40 years. 5 years hence ratio of their ages will be 3:7. Find the age of Rinku?
Solution: Given 5yrs hence ratio of ages of Rinku and Gopal is 3:7
let 5yrs hence, age of Rinku = 3X and Gopal = 7X
So, present age of Rinku = 3X-5 and Gopal = 7x-5
Sum of present ages of Rinku and Gopal = 40
(3X-5)+(7X-5) = 40
10X = 50
X = 5
present age of Rinku = 3X-5 = 10yrs
3. The present age of Romila is one-fourth that of her father. After 6 years the father’s age will be twice the age of Kapil. If Kapil celebrated fifth birthday 8 years ago. What is Romila’s present age?
Solution: Let present age of Romila is X, then Father’s age = 4X
6 years hence,
father’s age = 4X+6
2 (Age of Kapil) = 4X+6
Age of Kapil = 2X+3
Present age of Kapil = 2X+3-6 = 2X-3
Kapil celebrated his 5th bairthday 8 years ago
So, Present age of Kapil is 5+8 = 13 years
2X-3 = 13
2X = 16
X = 8years.
Shortcut approach: Kapil celebrated his 5th birthday 8 years ago.
Present age of Kapil = 13
After 6years, father’s age will be twice of the Kapil.
2x(13+6) = 4X+6
X= 8 years
4. A man’s age is 133(1/3)% of what it was 8 years ago, but 80% of what it will be after 8 years. What is his present age?
Solution: Let the present age be X years.
Then 133(1/3)% of (X-8) = X and 80%(X+8) = X
So, 133(1/3)% of (X-8) = 80%(X+8)
4(X-8)/3 = 4(X+8)/5
5(X-8) = 3(X+8)
2X = 64
X = 32
Shortcut: You don’t need to solve both equations. Solve any equation you will get the answer.
133(1/3)% of (X-8) = X
4(X-8)/3 = X
4X-32 = 3X
X = 32 years.
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