Question 1

Linked ages with time shifts: Solve present ages from future and past relations.  A is three times as old as B. Four years ago, C was twice as old as A was then. In four years' time, A will be 31. Find the present ages of B and C. Choose the correct…

Accepted Answer

Compute A first A in 4 years = 31 ⇒ A now = 31 − 4 = 27. Find B A = 3B ⇒ 27 = 3B ⇒ B = 9. Find C Four years ago: A was 23 ⇒ C was twice that: 2 × 23 = 46. So C now = 46 + 4 = 50. Check Relations satisfied: A=27, B=9, C=50 ✔

Question 2

Family ages with mixed offsets: Determine parents' ages at a child's birth using given gaps.  Brother is 3 years older than Ravi. Father was 28 when sister was born. Mother was 26 when Ravi was born. Sister was 4 when brother was born. Find the fath…

Accepted Answer

Timeline setup Take the brother's birth as time T = 0. Father's age at brother's birth Sister was 4 at T = 0. Father was 28 when sister was born ⇒ 4 years later at T = 0, Father = 28 + 4 = 32. Mother's age at brother's birth Brother is 3 years older than Ravi ⇒ Ravi is born 3 years after T = 0. Mother was 26 when Ravi was born ⇒ at T = 0 she was 26 − 3 = 23.

Question 3

Shared-expense algebra: Determine the attending headcount from a per-person increase.  Total planned spend on eatables = ₹96. Four friends did not turn up. Remaining attendees each paid ₹4 extra than originally planned. Find the number of friends wh…

Accepted Answer

Let n = original number of friends, so original share = 96/n. After 4 drop New number = n − 4; new share = 96/(n − 4). Increase per person = 4 ⇒ 96/(n − 4) − 96/n = 4. Solve 96·(n − (n − 4)) / [n(n − 4)] = 4 ⇒ 96·4 / [n(n − 4)] = 4 n(n − 4) = 96 ⇒ n² − 4n − 96 = 0 ⇒ (n − 12)(n + 8) = 0 ⇒ n = 12. Attended = n − 4 = 8.

Question 4

Two-condition age puzzle (father and son): Solve simultaneous relations to find the son's age.  Now: Father's age = 3 × son's age. Five years ago: Father's age = 4 × son's age. Find the son's present age (in years). Choose the correct value.

Accepted Answer

Let Son = s, Father = 3s. Translate the past condition Five years ago: (3s − 5) = 4(s − 5). Solve 3s − 5 = 4s − 20 ⇒ −5 + 20 = 4s − 3s ⇒ s = 15. Check Now: Father = 45, Son = 15. Five years ago: 40 and 10 ⇒ 40 = 4 × 10 ✔

Question 5

Mixed-fruit purchase (linear Diophantine check): Find how many pineapples were bought.  Pineapple costs Rs. 7 each; watermelon costs Rs. 5 each; total spend = Rs. 38. Assume only whole fruits can be purchased; determine the pineapple count. Choose t…

Accepted Answer

Given data Pineapple price = Rs. 7 each. Watermelon price = Rs. 5 each. Total amount spent = Rs. 38. Concept/Approach Let p be pineapples and w be watermelons (integers ≥ 0). Solve the linear Diophantine equation: 7p + 5w = 38. Step-by-step calculation Reduce mod 5: 7p ≡ 38 (mod 5) ⇒ 2p ≡ 3 (mod 5) ⇒ p ≡ 4 (mod 5).Smallest feasible p = 4. Then cost from pineapples = 4 × 7 = 28.Remaining for watermelons = 38 − 28 = 10 ⇒ w = 10 / 5 = 2 (valid integer). Verification 7(4) + 5(2) = 28 + 10 = 38 (matches total). Final Answer 4

Question 6

Bus headcount equality after a stop: Find the initial total number of passengers.  Initially, women count = half the men count. At city Y, 10 men alight and 5 women board; then men and women become equal. Compute the number of passengers at the begi…

Accepted Answer

Variables Let initial men = m, women = w with w = m/2. After city Y: men = m − 10, women = w + 5 = m/2 + 5; these are equal. Equation m − 10 = m/2 + 5 ⇒ m/2 = 15 ⇒ m = 30Then w = m/2 = 15. Initial total m + w = 30 + 15 = 45. Final Answer 45

Question 7

Sports participation in a class: Use ratios and non-participant difference to find girls in class.  Total students = 60; participant ratio (boys : girls) = 3 : 2. Girls not participating are 5 more than boys not participating. If boys participating …

Accepted Answer

Translate ratio to counts Boys participating = 15 corresponds to the “3” part ⇒ 3k = 15 ⇒ k = 5. Girls participating = 2k = 10. Non-participants relationship Let total boys = B, girls = G (B + G = 60).Girls not participating = G − 10; boys not participating = B − 15.Given: G − 10 = (B − 15) + 5 ⇒ G = B. Compute totals If G = B and B + G = 60 ⇒ B = G = 30. Final Answer 30

Question 8

Cake sharing puzzle: Father takes 1/4 of cake and has triple each other member's share.  Father's portion = 1/4 of the whole cake. This equals 3 times the share of any other single member; all others share the remainder equally. Find the total numbe…

Accepted Answer

Let members = n Father takes 1/4. Leftover for (n − 1) others = 3/4 equally ⇒ each other gets (3/4)/(n − 1). Equation (father has triple any other) 1/4 = 3 × [(3/4)/(n − 1)] = (9/4)/(n − 1)Multiply both sides by 4: 1 = 9/(n − 1) ⇒ n − 1 = 9 ⇒ n = 10. Final Answer 10

Question 9

Word logic — “All but nine died”: Interpret the phrasing to find survivors.  A shepherd had 17 sheep; all but nine died. How many sheep remained alive? Choose the correct number.

Accepted Answer

Interpretation “All but nine died” means all except nine died. Computation Survivors = 9. Final Answer 9

Question 10

Ages — future condition: When will Manick be twice Rahul's age?  Currently, Manick is 12 years old and is three times Rahul's age. Find Manick's age when he becomes exactly twice Rahul's age (both ages advance together). Choose the correct age.

Accepted Answer

Now Manick M = 12. M = 3R ⇒ Rahul R = 12/3 = 4. After x years 12 + x = 2(4 + x) ⇒ 12 + x = 8 + 2x ⇒ x = 4.Manick's age then = 12 + 4 = 16 years. Final Answer 16 years

Question 11

Population literacy composition: Combine adult/child ratios and percentages to find overall literacy.  Adults : children = 2 : 3 in the city. Adults are 40% illiterate (so 60% literate); children are 85% literate. What percentage of the total popula…

Accepted Answer

Assume a convenient base Let adults = 2k and children = 3k ⇒ total = 5k. Literate counts Adults literate = 60% of 2k = 1.2k.Children literate = 85% of 3k = 2.55k.Total literate = 1.2k + 2.55k = 3.75k. Overall percentage (3.75k / 5k) × 100% = 75%. Final Answer 75%

Question 12

Income mix (salary + tips): What fraction of weekly income came from tips?  In one week, tips were 5/4 of the waiter's fixed salary for that week. Find the fraction of the total income represented by tips. Choose the correct fraction.

Accepted Answer

Let salary = S Tips = (5/4)S. Total income = S + (5/4)S = (9/4)S. Fraction from tips [(5/4)S] / [(9/4)S] = 5/9. Final Answer 5/9

Question 13

Two-destination ticket pricing: Solve simultaneous equations for fares to B and C.  2 tickets A→B + 3 tickets A→C cost Rs. 77. 3 tickets A→B + 2 tickets A→C cost Rs. 73. Find the individual fares from A to B and from A to C. Choose the correct order…

Accepted Answer

Let fares be x = fare A→B y = fare A→C Form equations 2x + 3y = 77 … (1)3x + 2y = 73 … (2) Solve (1)×3 ⇒ 6x + 9y = 231(2)×2 ⇒ 6x + 4y = 146Subtract ⇒ 5y = 85 ⇒ y = 17.From (2): 3x + 2(17) = 73 ⇒ 3x = 39 ⇒ x = 13. Final Answer Rs. 13, Rs. 17

Question 14

Work-rate proportionality (cats and mice): Find days for 4 cats to kill 4 mice.  Given: 100 cats kill 100 mice in 100 days (assume proportional, independent rates). Determine the number of days for 4 cats to kill 4 mice. Choose the correct duration.

Accepted Answer

Interpretation If 100 cats kill 100 mice in 100 days, then each cat kills 1 mouse in 100 days (rates scale linearly with both cats and mice). Apply proportionality 4 cats killing 4 mice ⇒ effectively 1 cat per mouse, each requiring 100 days. Final Answer 100 days

Question 15

Arithmetic word problem (currency denominations): Find the total number of notes when counts per denomination are equal.  A man has ₹480 in one-rupee, five-rupee, and ten-rupee notes. The number of notes of each denomination is the same.  Calculate …

Accepted Answer

Given data Total money = ₹480. Denominations: ₹1, ₹5, ₹10. Equal number of notes of each denomination ⇒ let that number be x. Concept/Approach Total value = x·1 + x·5 + x·10 = 16x ⇒ solve 16x = 480. Step-by-step calculation 16x = 480 ⇒ x = 480 ÷ 16 = 30. Total notes = 3x = 3 × 30 = 90. Common pitfalls Adding denominations but forgetting to multiply each by the same count x.

Question 16

Algebraic puzzle (money comparison and trebling): Determine the boys' original combined amount.  Mac initially has £3 more than Ken. Ken then trebles his money and ends up with £2 more than the original combined total. Find the amount the two boys h…

Accepted Answer

Let Ken = k, Mac = k + 3. Original total T = 2k + 3. Condition after winning Ken trebles his money ⇒ new Ken = 3k. Given: 3k = T + 2 = (2k + 3) + 2 = 2k + 5. Solve 3k = 2k + 5 ⇒ k = 5. Original total T = 2k + 3 = 10 + 3 = £ 13. Common pitfalls Adding the £2 to Ken's old amount instead of to the original combined total.

Question 17

Family ages (marriage before birth): Compute the father's age at marriage from the given age gaps.  Ayush was born two years after his father's marriage. Mother is five years younger than father and 20 years older than Ayush. Ayush is 10 years old. …

Accepted Answer

Given data Ayush = 10 years. Mother = Ayush + 20 = 30 years. Mother is 5 years younger than father ⇒ Father = 30 + 5 = 35 years. Ayush was born 2 years after father's marriage. Compute father's marriage age Time since marriage = Ayush's age + 2 = 10 + 2 = 12 years. Father's age at marriage = 35 − 12 = 23 years.

Question 18

Digit counting (1–100): Determine the frequency of a specific digit in a range.  Write all integers from 1 to 100 inclusive. How many times does the digit 3 appear across all these numbers? Choose the correct value.

Accepted Answer

Approach Count occurrences in the tens place and in the ones place separately. Step-by-step counting Tens place: 30–39 ⇒ 10 occurrences. Ones place: numbers ending in 3 ⇒ 3, 13, 23, 33, 43, 53, 63, 73, 83, 93 ⇒ 10 occurrences. Total occurrences = 10 + 10 = 20 (33 contributes twice, once for each place).

Question 19

Box distribution (system of equations): Find the number of balls in the green box.  Three boxes: Red (R), Green (G), Blue (B) with a total of 108 balls. G + R is twice B. B is twice R.  Compute the number of balls in the green box. Choose the correc…

Accepted Answer

Translate to equations G + R = 2B, and B = 2R. Solve Substitute B = 2R into G + R = 2B ⇒ G + R = 4R ⇒ G = 3R. Total: R + G + B = R + 3R + 2R = 6R = 108 ⇒ R = 18. Then G = 3R = 54. Note Since 54 is not among the numeric options, the correct choice by label would be "None of these", but the actual value is 54.

Question 20

Counting independent routes (fundamental principle of multiplication): Compute total ways across segments.  Routes known: Bristol → Birmingham: 4 routes Birmingham → Sheffield: 3 routes Sheffield → Carlisle: 2 routes  Assuming independence, find tot…

Accepted Answer

Concept When choices are independent across stages, multiply the counts. Calculation Total routes = 4 × 3 × 2 = 24.

Arithmetic Reasoning Questions