Question 1

Relative ages and ordering – Ram's age is double Shyam's age and half of Sohan's age. Shyam is older than Mohan. Based on these relations, who is the oldest among the four?

Accepted Answer

Introduction / Context:This is a classic “problems on ages” ordering question. We are not asked for numerical ages; instead, we must use the given multiplicative relationships to rank the people from oldest to youngest and identify the oldest individual. Given Data / Assumptions: Ram's age is double Shyam's age. Ram's age is half of Sohan's age. Shyam is older than Mohan. All ages are present ages and positive integers are not required; only ordering matters. Concept / Approach:Translate words to proportional relations. If Shyam = s, then Ram = 2s. If Ram is half of Sohan, then Sohan = 2 * Ram = 4s. The statement “Shyam is older than Mohan” only says Mohan is below Shyam; it does not change who is overall oldest because the Sohan–Ram–Shyam chain already fixes the top end of the order. Step-by-Step Solution:Let Shyam = s.Then Ram = 2s (given).Since Ram is half of Sohan, Sohan = 2 * (2s) = 4s.Therefore, in descending order: Sohan (4s) > Ram (2s) > Shyam (s) > Mohan (below Shyam).Hence, Sohan is the oldest. Verification / Alternative check:Pick a simple number: let Shyam = 10. Then Ram = 20 and Sohan = 40. Any Mohan that is younger than Shyam (say, 9) maintains the order. The oldest remains Sohan (40). Why Other Options Are Wrong: Ram and Shyam are strictly younger than Sohan by the relations. Mohan is explicitly younger than Shyam, so he cannot be the oldest. “None of these” is incorrect because Sohan is determinable. Common Pitfalls:Confusing “double” and “half,” or assuming “older than Mohan” implies Mohan could be the oldest; that statement only places Mohan below Shyam. Final Answer:Sohan

Question 2

Two years before the reference stage – A was 4 years older than B at a certain stage. Sixteen years after that stage, A will be thrice his present age and B will be five times his present age. How old were A and B two years before that initial stage?

Accepted Answer

Introduction / Context:This problem involves a “reference stage,” future projections from that stage, and then a request to report ages two years before that stage. Careful variable setup keeps the timeline consistent and prevents sign mistakes. Given Data / Assumptions: At the reference stage, A was 4 years older than B. Sixteen years after that stage, A will be thrice his age at the stage, and B will be five times his age at the stage. We must give A's and B's ages two years before the stage. Concept / Approach:Let A and B at the stage be a and b. From “older by 4,” a = b + 4. From “after 16 years,” a + 16 = 3a and b + 16 = 5b. Solve these to find a and b, then subtract 2 for the requested earlier time. Step-by-Step Solution:From a + 16 = 3a ⇒ 16 = 2a ⇒ a = 8.From b + 16 = 5b ⇒ 16 = 4b ⇒ b = 4.Check the 4-year difference: a − b = 8 − 4 = 4 (consistent).Two years before the stage: A = 8 − 2 = 6; B = 4 − 2 = 2. Verification / Alternative check:Move forward 16 years from the stage: A = 24 which is 3 × 8; B = 20 which is 5 × 4 — both conditions satisfied. Why Other Options Are Wrong: 10 & 6, 12 & 8, 8 & 4 conflict with the “16 years later” multipliers when traced back to the stage. “None of these” is unnecessary since 6 & 2 fits all constraints. Common Pitfalls:Applying the multipliers to future ages instead of the stage ages, or subtracting 2 at the wrong point in time. Final Answer:6 and 2 years

Question 3

Linear equations in ages – A is 16 years older than B, and half of B’s age equals one-third of A’s age. Find their present ages.

Accepted Answer

Introduction / Context:Here we mix an additive relation (“older by 16”) with a fractional comparison (“half equals one-third”). Such systems are best handled with two equations in two variables, then solved by substitution or elimination. Given Data / Assumptions: A = B + 16. (1/2) * B = (1/3) * A. All ages are present ages. Concept / Approach:Convert the fractional statement to an equation without fractions. From (1/2)B = (1/3)A, multiply both sides by 6 to get 3B = 2A. Substitute A = B + 16 into 3B = 2A and solve for B, then back-solve for A. Step-by-Step Solution:3B = 2A.With A = B + 16 ⇒ 3B = 2(B + 16) = 2B + 32.Thus B = 32, and A = B + 16 = 48. Verification / Alternative check:(1/2)B = 16 and (1/3)A = 16, so the fractional condition holds. The 16-year difference also holds: 48 − 32 = 16. Why Other Options Are Wrong: 40/32 does not satisfy 3B = 2A. 46/30 and 32/16 violate one or both constraints. “None of these” is not needed because 48/32 is consistent. Common Pitfalls:Misreading “half B equals one-third A,” or subtracting 16 in the wrong direction. Keep the algebraic mapping faithful to the words. Final Answer:A = 48, B = 32

Question 4

Birthday algebra – Today is Ketan’s birthday. One year from today he will be twice as old as he was 10 years ago. How old is Ketan today?

Accepted Answer

Introduction / Context:This single-variable age equation tests translating “one year from today” and “10 years ago” into algebra. Correctly placing the time shifts on the same unknown is the key step. Given Data / Assumptions: Let Ketan’s present age be x years. One year from now: x + 1. Ten years ago: x − 10. Condition: x + 1 is twice (x − 10). Concept / Approach:Write the equation (x + 1) = 2(x − 10) and solve for x. Ensure both references (future and past) are based on the same present variable. Step-by-Step Solution:x + 1 = 2x − 20.Bring terms together: 1 + 20 = 2x − x ⇒ 21 = x.Therefore, Ketan is 21 years old today. Verification / Alternative check:If x = 21, then one year later = 22. Ten years ago = 11. Twice 11 = 22, which matches. Why Other Options Are Wrong:Other numbers do not satisfy x + 1 = 2(x − 10) when substituted. Common Pitfalls:Applying the “twice” to the future age instead of the past age, or mixing up the 10-year offset signs. Final Answer:21 years

Question 5

Sum and difference of two ages – The combined age of a mother and her daughter is 60 years, and their age difference is 30 years. What is the mother’s age?

Accepted Answer

Introduction / Context:When a sum and a difference of two numbers are known, each number can be found using the standard half-sum formulas. This technique appears frequently in age problems involving a parent and child. Given Data / Assumptions: Total age (mother + daughter) = 60. Age difference = 30 (mother is older). Concept / Approach:Let M = mother, D = daughter. Then M + D = 60 and M − D = 30. Solve by adding the equations or using half-sum: M = ( (M + D) + (M − D) ) / 2. Step-by-Step Solution:M = (60 + 30) / 2 = 90 / 2 = 45.D = (60 − 30) / 2 = 30 / 2 = 15.Therefore, mother = 45 years. Verification / Alternative check:M + D = 45 + 15 = 60 and M − D = 45 − 15 = 30: both constraints satisfied. Why Other Options Are Wrong:Any value other than 45 fails one of the two simultaneous conditions. Common Pitfalls:Using 30 directly as the mother’s age or halving the wrong quantity. Always apply the half-sum formulas carefully. Final Answer:45 years

Question 6

Present ages from future sum – “X” and her grandfather differ by 50 years. Six years from now, their ages will sum to 152. What are their present ages?

Accepted Answer

Introduction / Context:Two linear relations are provided: a present-time difference and a future-time sum. Solving them together cleanly yields the present ages without trial-and-error. Given Data / Assumptions: Let x = X’s present age, g = grandfather’s present age. g − x = 50. (x + 6) + (g + 6) = 152 ⇒ x + g = 140. Concept / Approach:Solve the simple system: x + g = 140 and g − x = 50. Adding gives 2g = 190 ⇒ g = 95, then x = 45. Step-by-Step Solution:x + g = 140.g − x = 50.Add: 2g = 190 ⇒ g = 95.Then x = 140 − 95 = 45. Verification / Alternative check:Six years later: 51 + 101 = 152, as required; difference now: 95 − 45 = 50. Why Other Options Are Wrong:The other pairs do not simultaneously satisfy both equations. Common Pitfalls:Subtracting the equations in the wrong order or misreading “after 6 years” as “in 6 years total.” Final Answer:45, 95

Question 7

Mother–father–child relations – Your mother is 4 years younger than your father, and your father is 6 times as old as you. If you are 6 years old, what is your mother’s age?

Accepted Answer

Introduction / Context:This item mixes a multiplicative relation (father relative to child) with an offset relation (mother relative to father). It is a straightforward two-step computation once the child’s current age is known. Given Data / Assumptions: Child's present age = 6 years. Father = 6 × child = 36 years. Mother = father − 4 years. Concept / Approach:Compute the father’s age from the multiplier, then subtract 4 to get the mother’s age. Step-by-Step Solution:Father = 6 × 6 = 36 years.Mother = 36 − 4 = 32 years. Verification / Alternative check:Mother + 4 = 36 confirms she is 4 years younger than father. Why Other Options Are Wrong:They do not satisfy both constraints simultaneously. Common Pitfalls:Reading “six times older” incorrectly; in aptitude context it is taken as “six times as old.” Final Answer:32 years

Question 8

Father and two daughters – Father is three times the elder daughter now; five years ago he was eight times the younger daughter. If the daughters differ by 5 years, what is the father’s present age?

Accepted Answer

Introduction / Context:We are given two conditions involving the father's age relative to each of two daughters at different times, plus a fixed age gap between daughters. This leads to a compact system of equations. Given Data / Assumptions: Let elder daughter = E, younger daughter = Y (present ages). Father = F (present age). F = 3E (now). Five years ago: F − 5 = 8(Y − 5). E − Y = 5 (elder is older by 5). Concept / Approach:Express E as Y + 5, then F as 3(Y + 5). Substitute into the “five years ago” relation to solve for Y. Back-substitute to get E and then F. Step-by-Step Solution:E = Y + 5; F = 3E = 3(Y + 5) = 3Y + 15.Five years ago: F − 5 = 8(Y − 5) ⇒ (3Y + 15) − 5 = 8Y − 40.3Y + 10 = 8Y − 40 ⇒ 50 = 5Y ⇒ Y = 10.E = Y + 5 = 15; F = 3E = 45. Verification / Alternative check:Five years ago: F − 5 = 40 and Y − 5 = 5, and 40 = 8 × 5, confirmed. Why Other Options Are Wrong:Other numbers fail either the “now 3× elder” or the “5 years ago 8× younger” condition. Common Pitfalls:Mixing which girl is referenced in each condition, or using the present ages inside the “five years ago” condition without subtracting 5. Final Answer:45

Question 9

Backtracking from a fraction statement – Twenty years ago, Antony’s age was half of what his present age is. What is his current age?

Accepted Answer

Introduction / Context:This problem is a direct one-variable equation framed around a past fraction of the present age. Such prompts are staples of the “ages” topic and emphasize precise temporal translation. Given Data / Assumptions: Let Antony’s present age be A. Twenty years ago his age was A − 20, and it equaled (1/2)A. Concept / Approach:Set A − 20 = (1/2)A and solve for A. Do not confuse “half of present” with “half of past.” Step-by-Step Solution:A − 20 = (1/2)A.A − (1/2)A = 20 ⇒ (1/2)A = 20 ⇒ A = 40. Verification / Alternative check:Twenty years ago: 40 − 20 = 20, which indeed is half of 40. Why Other Options Are Wrong:Substituting 20, 35, 30, or 25 fails the defining equation. Common Pitfalls:Writing 20 years ago as “A/2 − 20” (incorrect) or misplacing the half on the wrong side of the timeline. Final Answer:40 years

Question 10

Father–son ratio over time – A father is five times as old as his 6-year-old son. After how many years will the father be four times as old as his son?

Accepted Answer

Introduction / Context:Here we are given a current multiplier and asked to find when a different multiplier will hold in the future. Setting up the linear-in-time equation for both ages and solving for the time variable gives the answer swiftly. Given Data / Assumptions: Son's present age S = 6. Father's present age F = 5 × 6 = 30. Find t such that F + t = 4(S + t). Concept / Approach:Use the standard “after t years” linear relation. Both people age at the same rate (+t), so only the intercepts differ. Step-by-Step Solution:30 + t = 4(6 + t) = 24 + 4t.30 − 24 = 4t − t ⇒ 6 = 3t ⇒ t = 2.Therefore, in 2 years, father will be four times the son’s age. Verification / Alternative check:After 2 years: father 32, son 8, and 32 = 4 × 8. Why Other Options Are Wrong:Other values of t do not satisfy the future 4× relation. Common Pitfalls:Forgetting that both ages increase by the same t or incorrectly scaling the future ages. Final Answer:2 years

Question 11

Family chain differences – Mahesh is 60. Ram is 5 years younger than Mahesh and 4 years older than Raju. Babu is 6 years younger than Raju. What is the age difference between Mahesh and Babu?

Accepted Answer

Introduction / Context:This is a chain of differences problem: each relation links to the next person. Compute each person step by step and then take the required difference at the end. Given Data / Assumptions: Mahesh = 60. Ram = Mahesh − 5 = 55. Ram is 4 years older than Raju ⇒ Raju = 55 − 4 = 51. Babu is 6 years younger than Raju ⇒ Babu = 51 − 6 = 45. Concept / Approach:Apply the offsets sequentially and preserve signs. Then compare the endpoint with the start point to get the requested difference. Step-by-Step Solution:Mahesh: 60.Ram: 55.Raju: 51.Babu: 45.Difference Mahesh − Babu = 60 − 45 = 15. Verification / Alternative check:Trace the offsets in a single expression: 60 − 5 − 4 − 6 = 45 for Babu; difference 15. Why Other Options Are Wrong:Other numbers misapply one of the offsets or reverse a sign. Common Pitfalls:Reading “older than” as “younger than” by mistake or skipping one link in the chain. Final Answer:15 years

Question 12

Mother–son ages from past ratio and present sum – Eight years ago, Mrs. Lata was three times as old as her son. Their present ages total 64. How old is Mrs. Lata now?

Accepted Answer

Introduction / Context:We combine a past-time multiplier relation with a present-time sum. Writing both equations in terms of present ages prevents confusion and makes elimination straightforward. Given Data / Assumptions: Let mother = M, son = S (present ages). Eight years ago: M − 8 = 3(S − 8). M + S = 64. Concept / Approach:From M − 8 = 3S − 24 ⇒ M = 3S − 16. Substitute into M + S = 64 and solve for S; then compute M. Step-by-Step Solution:M = 3S − 16.(3S − 16) + S = 64 ⇒ 4S = 80 ⇒ S = 20.M = 3 × 20 − 16 = 60 − 16 = 44. Verification / Alternative check:Eight years ago: M − 8 = 36 and S − 8 = 12; 36 = 3 × 12, satisfied. Present sum: 44 + 20 = 64, satisfied. Why Other Options Are Wrong:They fail either the 3× past relation or the present sum. Common Pitfalls:Applying the factor 3 to present ages or forgetting to subtract 8 on both sides of the relation. Final Answer:44

Question 13

Linking past and future ages – Ten years ago Hema was twice Geeta. Ten years from now Hema will be 40. How old is Geeta today?

Accepted Answer

Introduction / Context:We are given Hema’s future age and a past ratio between Hema and Geeta. Determine Hema’s present age first, then use the 10-years-ago relation to get Geeta’s present age. Given Data / Assumptions: Hema in 10 years = 40 ⇒ Hema now = 30. Ten years ago: Hema = 20; she was twice Geeta then. Concept / Approach:If Hema was 20 ten years ago and was twice Geeta’s age then, Geeta was 10 at that time. Therefore, Geeta is 20 now. Step-by-Step Solution:Hema now = 40 − 10 = 30.Ten years ago: Hema = 20, Geeta = 10 (since 20 = 2 × 10).Therefore, Geeta now = 10 + 10 = 20. Verification / Alternative check:Check the past relation: 30 − 10 = 20 and 20 − 10 = 10; 20 = 2 × 10 true. Why Other Options Are Wrong:They do not satisfy the past 2× relation when combined with Hema’s known future age. Common Pitfalls:Forgetting that both ages move together by the same 10-year offset when converting between times. Final Answer:20 years

Question 14

Sum after 5 years and fixed difference – Jais and his father differ by 35 years now. After 5 years, their ages will sum to 125. What will their ages be 12 years from now?

Accepted Answer

Introduction / Context:This problem provides a present-time difference and a future-time sum and asks for another future-time pair. Solve present ages first, then advance to the requested time. Given Data / Assumptions: Let J = Jais now, F = father now. F − J = 35. (J + 5) + (F + 5) = 125 ⇒ J + F = 115. Concept / Approach:From J + F = 115 and F − J = 35, add to get 2F = 150 ⇒ F = 75 and J = 40. Then add 12 years to both for the asked time. Step-by-Step Solution:F = 75, J = 40.After 12 years: J = 52, F = 87. Verification / Alternative check:After 5 years: J = 45, F = 80 → sum 125, as given. Difference now 75 − 40 = 35, as given. Why Other Options Are Wrong:They do not arise from the unique present pair (40, 75) advanced by 12 years. Common Pitfalls:Using the 35-year difference at the future time incorrectly (differences stay constant across equal increments, but sums do not unless specified). Final Answer:52 and 87

Question 15

Average and difference – The average of Amit’s and Anil’s present ages is 32 years. If Amit is 6 years older than Anil, what is Anil’s present age?

Accepted Answer

Introduction / Context:Average of two numbers and their difference uniquely determine each number. Convert the average to a sum and split using the half-difference idea. Given Data / Assumptions: (Amit + Anil) / 2 = 32 ⇒ Amit + Anil = 64. Amit = Anil + 6. Concept / Approach:Substitute into the sum to solve for Anil first, then compute Amit if needed. We only need Anil. Step-by-Step Solution:(Anil + 6) + Anil = 64.2·Anil + 6 = 64 ⇒ 2·Anil = 58 ⇒ Anil = 29. Verification / Alternative check:Amit = 35; average = (35 + 29) / 2 = 32, as stated. Why Other Options Are Wrong:They do not satisfy both the sum and the 6-year difference simultaneously. Common Pitfalls:Using 32 directly for one of the ages or halving the difference incorrectly. Final Answer:29 years

Question 16

Inferring parental age gap – A man was 32 when his first son was born. When that son was 7, his wife was 35. What is the age difference between father and mother?

Accepted Answer

Introduction / Context:Age differences between adults remain constant over time. The problem gives two snapshots (at son’s birth and when son reaches 7) that let us compute the constant father–mother gap quickly. Given Data / Assumptions: Father's age at son's birth = 32. When son is 7, mother is 35; thus 7 years have passed since the birth. Concept / Approach:At the time the son was 7, father's age = 32 + 7 = 39. Compare with mother's 35 to get the difference. That gap equals the father–mother age difference always. Step-by-Step Solution:Father at son age 7: 32 + 7 = 39.Mother at that time: 35.Difference = 39 − 35 = 4 years. Verification / Alternative check:Any later or earlier time will retain the 4-year gap; adding/subtracting the same number to both keeps their difference constant. Why Other Options Are Wrong:They contradict the snapshot calculations and the invariance of the age gap. Common Pitfalls:Trying to average or otherwise combine the numbers instead of using the constant-difference property. Final Answer:4 years

Question 17

Mother–daughter equation system – A mother is five times her daughter’s present age. After 5 years she will be three times her daughter’s age. Find the mother’s current age.

Accepted Answer

Introduction / Context:This is a standard two-equation ages setup with current and future ratios. Solving carefully yields exact present ages without guessing. Given Data / Assumptions: Let daughter = D, mother = M. M = 5D (now). M + 5 = 3(D + 5) (after 5 years). Concept / Approach:Substitute M from the first equation into the second and solve for D; then compute M. Step-by-Step Solution:5D + 5 = 3D + 15.2D = 10 ⇒ D = 5.M = 5D = 25. Verification / Alternative check:After 5 years: mother 30, daughter 10; 30 = 3 × 10 holds. Why Other Options Are Wrong:They do not meet both the present 5× and future 3× constraints simultaneously. Common Pitfalls:Adding 5 to only one side or mixing present and future ages in a single step. Final Answer:25 years

Question 18

Four children at 4-year intervals – The sum of their present ages is 60. What is the age of the youngest child?

Accepted Answer

Introduction / Context:When children are born at equal intervals, their ages form an arithmetic progression. Summation formulas for APs make such problems quick to solve. Given Data / Assumptions: Let youngest = x. Then ages are x, x + 4, x + 8, x + 12. Sum = 60. Concept / Approach:Add the four-term AP and solve for x. Equally spaced terms keep the algebra simple. Step-by-Step Solution:x + (x + 4) + (x + 8) + (x + 12) = 60.4x + 24 = 60 ⇒ 4x = 36 ⇒ x = 9. Verification / Alternative check:Actual ages: 9, 13, 17, 21; sum = 60. Why Other Options Are Wrong:They do not satisfy the sum with 4-year spacing. Common Pitfalls:Using 3 terms or spacing by 3 years accidentally. Always match the stated interval. Final Answer:9

Question 19

Siblings with fixed gap – Nikhil is 8 years younger than Rohan. How old will Rohan be when he is exactly twice as old as Nikhil?

Accepted Answer

Introduction / Context:We are asked for the age Rohan will be at the moment he is twice Nikhil, given a constant present gap. Modeling “t years from now” for both brothers and equating produces a single linear equation in t that directly yields Rohan’s target age. Given Data / Assumptions: Nikhil = Rohan − 8 (present ages). We want t such that Rohan + t = 2(Nikhil + t). Concept / Approach:Substitute Nikhil in terms of Rohan, solve for the calendar time t, then compute Rohan’s age at that time. Step-by-Step Solution:R + t = 2((R − 8) + t) = 2R − 16 + 2t.Bring terms together: R + t − 2R + 16 − 2t = 0 ⇒ −R − t + 16 = 0 ⇒ R + t = 16.Therefore, when the condition holds, Rohan will be 16 years old. Verification / Alternative check:If Rohan is 16, Nikhil is 8 (because the gap is 8). 16 = 2 × 8, so the condition is satisfied. Why Other Options Are Wrong:They do not meet the 2× condition alongside the fixed 8-year gap. Common Pitfalls:Solving for t and reporting t instead of Rohan’s age at that time. The question explicitly asks for Rohan’s age. Final Answer:16 years

Question 20

Sum now and ratio in the past – The sum of the present ages of a father and son is 60. Six years ago the father was five times the son. After 6 years, what will be the son’s age?

Accepted Answer

Introduction / Context:We have a present-time sum and a past-time ratio. Converting both into equations in present variables and solving yields the son’s current age, from which we can project six years into the future. Given Data / Assumptions: Let F = father, S = son (present ages). F + S = 60. Six years ago: F − 6 = 5(S − 6). Asked: S + 6. Concept / Approach:From the past ratio: F = 5S − 24. Substitute into the sum to solve for S, then add 6. Step-by-Step Solution:F = 5S − 24.(5S − 24) + S = 60 ⇒ 6S = 84 ⇒ S = 14.After 6 years: S + 6 = 20. Verification / Alternative check:Check the past relation: F = 46; six years ago 40 and 8 → 40 = 5 × 8, true. Present sum: 46 + 14 = 60, true. Why Other Options Are Wrong:They do not follow from the unique solution S = 14. Common Pitfalls:Dropping the −6 term on only one side of the past equation or forgetting to compute “after 6 years.” Final Answer:20

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