Problems on Ages
- 1. 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 :
Options- A. 19
- B. 29
- C. 39
- D. 49 Discuss
Correct Answer: 39
Explanation:
Let B's present age = x years. Then, A's present age = (x + 9) years.
(x + 9) + 10 = 2(x - 10)
=> x + 19 = 2x - 20
=> x =39.
- 2. The total age of A and B is 12 years more than the total age of B and C. C is how many years younger than A ?
Options- A. 12
- B. 13
- C. 14
- D. 15 Discuss
Correct Answer: 12
- 3. The age of a man is 4 times his son. Five years ago, the man was nine times old as his son was at that time. The present age of man is?
Options- A. 24 years
- B. 32 years
- C. 40 years
- D. 44 years Discuss
Correct Answer: 32 years
Explanation:
Let the son's be x years, then father's age = 4x years ∴ (4x - 5) = 9(x - 5) ⇒ 5x = 40 ⇒ x = 8 ∴ Present age of the man = 32 years
- 4. The ratio of the present ages of P and Q is 3 : 4. Five years ago, the ratio of their ages was 5 : 7. Find their present ages.
Options- A. 40, 30
- B. 30, 50
- C. 30, 40
- D. 50, 30 Discuss
Correct Answer: 30, 40
Explanation:
As the ratio of their present ages is 3: 4, let their present ages be 3x and 4x. So, 5 years ago, as the ratio of their ages was 5 : 7, =>(3x-5) : (4x-5) = 5 : 7. Solving, we get X = 10. Hence, their present ages are 3x = 30 and 4x = 40
- 5. The sum of the present ages of a father and his son is 60 years. five years ago, father's age was four times the age of the son. so now the son's age will be:
Options- A. 5
- B. 10
- C. 15
- D. 20 Discuss
Correct Answer: 15
Explanation:
Let the present ages of son and father be x and (60 -x) years respectively.
Then, (60 - x) - 5= 4(x - 5)
55 - x = 4x - 20
5x = 75 => x = 15
- 6. The average age of a group of 10 students is 15 years. When 5 more students join the group, the average age increase by 1 year. The average age of the new students is?
Options- A. 18 years
- B. 21 years
- C. 19 years
- D. 20 years Discuss
Correct Answer: 18 years
Explanation:
Total age of 10 students = 150 years Total age of 15 students = 240 years Total age of 5 new students = 240 - 150 = 90 years Therefore, average age of 5 new students = 90/5 = 18 years
- 7. "I am five times as old as you were, when I was as old as you are", said a man to his son. Find out their present ages, if the sum of their ages is 64 years ?
Options- A. Father = 50; Son =14
- B. Father = 40; Son =24
- C. Father = 60; Son =4
- D. Father = 48; Son =16 Also asked in: AIEEE, Bank Exams, CAT, GATE, GRE, Analyst, Bank Clerk, Bank PO
Correct Answer: Father = 40; Son =24
Explanation:
Let the present age of the man be 'P' and son be 'Q',
Given, P + Q = 64 or Q = (64 - P)
Now the man says "I am five times as old as you were, when I was as old as you are",
So, P = 5[B - (P - Q)]
We get 6P = 10Q,
Substitute value for Q,
6P = 10(64 - P),
Therefore P = 40, Q = 24.
- 8. Six years ago Anita was P times as old as Ben was. If Anita is now 17 years old, how old is Ben now in terms of P ?
Options- A. 11/P + 6
- B. P/11 +6
- C. 17 - P/6
- D. 17/P Also asked in: GRE
Correct Answer: 11/P + 6
Explanation:
Let Ben?s age now be B
Anita?s age now is A.
(A - 6) = P(B - 6)
But A is 17 and therefore 11 = P(B - 6)
11/P = B-6
(11/P) + 6 = B
- 9. The ratio of the father's age to the son's age is 3: 1. The product of their ages is 147. The ratio of their ages after 5 years will be?
Options- A. 6:13
- B. 15:6
- C. 6:15
- D. 13:6 Discuss
Correct Answer: 13:6
Explanation:
Let the son's age = x. therefore, the father's age be 3x, Given the product of their ages = 147. => x * 3x = 147 => 3x² = 147. => x² = (147/3) => x² = 49 => x = 7. Therefore, present age of son = x = 7 years. Present age of father = 3x = 21. Age of father after 5 years = 21+5 = 26 years. Age of son after 5 years = 7+5 = 12 years. Ratio=13 : 6
- 10. Sachin is younger than Rahul by 7 years. If the ratio of their ages is 7:9, find the age of Sachin
Options- A. 24.5
- B. 25.5
- C. 26.5
- D. 27.5 Discuss
Correct Answer: 24.5
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