Age Comparison and Multiple-Condition Problems: 12 Solved MCQs

Age comparison questions rarely need difficult arithmetic. The challenge is translating several time and comparison statements without changing a fixed age difference or shifting only one person. This set covers 12 solved MCQs on older-younger chains, data sufficiency, present-past-future equations, birth years, ratios, and family constraints, so attempt each question before reading its explanation. KnowledgeGate has about 40 published questions mapped to this exact subtopic; use the wider Aptitude route for related topics and Aptitude for Placement when you want to build the concepts in sequence.

Age comparison questions: four rules to write before solving

For calibration, let B =x and A =3x. If A is twice B after 6 years, 3x+6=2(x+6), so x=6. Their ages are 18,6 today, 24,12 after 6 years, and 30,18 after another 6. The difference stays 12; the ratio changes from 3 to 2 to 30/18=5/3.

On your scratch pad, name present-age variables, translate each statement at its stated time, then check every condition. The Aptitude for Placements: Section Topics and Weekly Plan gives a broader framework.

Older-younger chains and data-sufficiency MCQs

Comparison words create directed relations. Data sufficiency asks only whether the requested result is unique, even when some ages remain unknown.

Question 1

Who is the youngest in the group?

In a group of four people:

I. Amit is older than Mahesh and younger than Tanuja.

II. Rahim is younger than Mahesh.

• (a) Both the Statements, 1 and 2 are necessary.

• (b) Either statement I alone or statement II alone is sufficient.

• (c) Statement I alone is sufficient.

• (d) Statement II alone is sufficient.

Answer: (a). Statement I gives Tanuja > Amit > Mahesh, leaving Rahim unplaced. Statement II gives only Mahesh > Rahim. Together, Tanuja > Amit > Mahesh > Rahim, so Rahim alone is youngest. Both are necessary.

Question 2

Hari (H), Gita (G), Irfan (I) and Saira (S) are siblings (i.e. brothers and sisters). All were born on 1st january. The age difference between any two successive siblings (that is born one after another) is less than 3 years. Given the following facts:

i. Hari’s age + Gita’s age > Irfan’s age + Saira’s age

ii. The age difference between Gita and Saira is 1 year. However Gita is not the oldest and Saira is not the youngest.

iii. There are no twins.

In what order were they born (oldest first)?

• (a) HSIG

• (b) SGHI

• (c) IGSH

• (d) IHSG

Answer: (b) SGHI. Each gap is 1 or 2 years, and Gita and Saira are adjacent. Test S=10, G=9, H=8, I=6: the gaps are 1,1,2; neither boundary rule is broken; and H+G=17 > I+S=16. In (a), S and G are not adjacent. Positive 1-or-2-year gaps make H+G<I+S in (c) and (d).

Question 3

Direction: Each of the following problems consist of a question followed by two statements, I and II. You must determine whether the information given by the statements is sufficient to answer the question asked. In addition to the information provided in the statements, you should rely on your knowledge of mathematics and ordinary facts (such as the number of seconds in a minute).

Is Raja older than Ramu?

I. Sita is 4 years younger than Raja and 2 years younger than Ramu.

II. The average age Raja and Ramu is 21 years.

• (a) if the question can be answered by using only statement I

• (b) if the question can be answered by using only statement II.

• (c) if the question can be answered by using both statements together, but cannot be answered using either statement alone.

• (d) if the question cannot be answered even by using both statements together.

Answer: (a). For Sita’s age S, statement I gives Raja=S+4 and Ramu=S+2, so Raja is 2 years older. Statement II gives only Raja+Ramu=42; (22,20) and (19,23) reverse the comparison. Continue relation-chain practice with Logical Reasoning for Placement Tests: 3 Solving Methods.

Present and future age-multiplier MCQs

Add the same number of years to everyone before applying the new ratio. If a present age is 5x, after 5 years it is 5x+5, not 5(x+5).

Question 4

A man’s age is 5 times his son’s age. After 5 years, the man’s age will be three times his son’s age. Find their present ages.

• (a) 25, 5

• (b) 35, 7

• (c) 40, 8

• (d) 45, 9

Answer: (a) 25, 5. Son =x and man =5x. Then 5x+5=3(x+5), so 2x=10 and x=5; the man is 25. Check: 30=3×10 after five years.

Question 5

Sankar is four times the age of Parvati. If after 5 years, he would be threee times of Parvati’s age, then further after 5 years, how many times he would be of Parvati’s age?

• (a) 2 times

• (b) 2.5 times

• (c) 3 times

• (d) 3.5 times

Answer: (b) 2.5 times. Parvati =x, Sankar =4x. From 4x+5=3(x+5), x=10, giving present ages 10,40. “Further after 5 years” means 10 years from today, so they become 20,50 and 50/20=2.5.

Question 6

A father is twice as old as his son. If 30 years ago the age of the father was 5 times the age of the son, what is the present age of the father?

• (a) 80 years

• (b) 75 years

• (c) 60 years

• (d) 85 years

Answer: (a) 80 years. Son =s, father =2s. Then 2s-30=5(s-30), giving 120=3s, so s=40 and father =80. Check: 30 years ago, 50=5×10.

Family-age conditions: translate one relationship at a time

Write a small variable map before solving, for example son=x, father=4x, mother=4x-6, daughter=x+3. Do not combine every sentence mentally.

Question 7

A father tells his son,"I was of your present age when you were born." If the father is now 36 years of age, how old was the boy 5 years ago?

• (a) 13

• (b) 15

• (c) 17

• (d) None of these

Answer: (a) 13. Son =x. The father was x when the son was born x years ago, so 2x=36. The son is 18; five years ago he was 18-5=13.

Question 8

The sum of the present ages of Akshara and Deepa is 67 years. Roopa is 9 years younger than Deepa. If Akshara is 20 years old, then how old is Roopa ?

• (a) 35 years

• (b) 36 years

• (c) 37 years

• (d) 38 years

Answer: (d) 38 years. Deepa =67-20=47; Roopa =47-9=38. Check: 20+47=67, and 47-38=9.

Question 9

A couple has a son and a daughter. The age of the father is four times that of the son and the age of the daughter is one-third of that of her mother. The wife is 6 years younger to her husband and the sister is 3 years older than her brother. The mother’s age is –

• (a) 42 years

• (b) 48 years

• (c) 54 years

• (d) 63 years

Answer: (c) 54 years. Son =x, father =4x, mother =4x-6, daughter =x+3. Thus x+3=(4x-6)/3, so 3x+9=4x-6 and x=15. Mother =4(15)-6=54; daughter 18 is one-third of 54.

Birth years, changing ratios, and repeated time shifts

Under a question’s convention, age at the end of year Y is Y-birth year. Write each ratio at its stated time before moving to the present.

Question 10

At the end of 1986 , R was half as old as his grandmother . The sum of the years in which they were born is 3846 . How old R was at the end of 1998?

• (a) 54

• (b) 45

• (c) 100

• (d) 95

Answer: (a) 54. In 1986, R =k and grandmother =2k. Their birth years give (1986-k)+(1986-2k)=3846, or 3972-3k=3846, so k=42. By 1998, R is 42+12=54.

Question 11

One year ago the ratio between Samir and Ashok's age was 4 : 3. One year hence the ratio of their age will be 5 : 4. What is the sum of their present ages in years?

• (a) 12 years

• (b) 15 years

• (c) 16 years

• (d) Cannot be determined

• (e) Question not attempted

Answer: (c) 16 years. One year ago, ages were 4x,3x; one year hence, 4x+2,3x+2. Then (4x+2)/(3x+2)=5/4, so 16x+8=15x+10 and x=2. Present ages are 9,7, totalling 16.

Question 12

After 8 years, a man will be 3 times as much old as he is now. After how much time he will be 5 times as much old as now?

• (a) 16 years

• (b) 10 years

• (c) 12 years

• (d) 8 years

Answer: (a) 16 years. Present age =x. From x+8=3x, x=4. For time n, x+n=5x, so 4+n=20 and n=16. Check: 12=3×4 and 20=5×4.

Common traps these 12 age questions expose

• Multiplying the shift, as in 5(x+5), instead of writing 5x+5.

• Shifting only one person’s age when time moves for everyone.

• Reading “further after 5 years” as five years from today.

• Assuming a sum or average fixes which person is older. In Question 3, (22,20) and (19,23) both total 42 but give opposite answers.

• Treating a constant difference as a constant ratio. In Question 6, 80,40 today becomes 50,10 thirty years ago, so the difference stays 40 while the ratio changes.

Before selecting an option, take 20 seconds to check that ages are non-negative, everyone advances equally, and the option satisfies every condition.

Answer-key recap and the next practice set

1-a, 2-b, 3-a, 4-a, 5-b, 6-a, 7-a, 8-d, 9-c, 10-a, 11-c, 12-a. Redo any missed question by drawing a timeline and writing variables before reopening its explanation.

Use the Logical Reasoning guide above for relation chains. For adjacent work, try GATE Verbal Aptitude and Reasoning: Solved Examples. Continue with Aptitude for GATE Exam for competitive-exam coverage, or Aptitude for Placement for placement-focused coverage. Then return to the Age Problems slice for another timed set.