Question 1

Mixture of animals and handlers in a caravan: hens, goats, camels, and human keepers In a caravan there are 50 hens, 45 goats, 8 camels, and some human keepers. If the total number of feet is 224 more than the total number of heads, how many keepers…

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Introduction / Context: This quantitative reasoning problem mixes counting of heads and feet across different species and people. It tests your ability to translate words into equations by assigning the correct number of heads and feet to birds, animals, and humans, and then solving a simple linear relation. Given Data / Assumptions: Hens = 50 (each with 1 head, 2 feet). Goats = 45 (each with 1 head, 4 feet). Camels = 8 (each with 1 head, 4 feet). Keepers = K (each with 1 head, 2 feet). Total feet = total heads + 224. Concept / Approach: Model heads and feet separately, then use the stated relationship to form and solve an equation in the unknown K (number of keepers). Step-by-Step Solution: Total heads = 50 + 45 + 8 + K = 103 + K.Total feet = hens * 2 + (goats + camels) * 4 + keepers * 2.Compute feet from animals: hens feet = 50 * 2 = 100; goats + camels = 45 + 8 = 53; their feet = 53 * 4 = 212.Add keepers’ feet: total feet = 100 + 212 + 2K = 312 + 2K.Use condition: total feet = total heads + 224 ⇒ 312 + 2K = (103 + K) + 224.Simplify: 312 + 2K = 327 + K ⇒ K = 15. Verification / Alternative check: Plug K = 15 back: heads = 118; feet = 312 + 2*15 = 342; difference = 342 − 118 = 224, which matches the condition. Why Other Options Are Wrong: 5, 8, 10: Substituting these values fails the feet minus heads = 224 requirement. Common Pitfalls: Assigning 4 feet to hens (they have 2), forgetting to include keepers’ heads, or mixing up the “224 more than” relation. Final Answer: 15

Question 2

Number theory and least common multiple (LCM) remainders puzzle I have packs of sweets. If I put 2, 3, or 4 sweets per pack, I have 1 sweet left over each time. If I pack 5 per pack, none are left. What is the minimum number of sweets?

Accepted Answer

Introduction / Context: This is a classic modular arithmetic and least common multiple question. You must find the smallest positive integer that fits several remainder conditions simultaneously, which is common in aptitude tests and coding interviews. Given Data / Assumptions: N ≡ 1 (mod 2). N ≡ 1 (mod 3). N ≡ 1 (mod 4). N ≡ 0 (mod 5). Concept / Approach: Combine the first three congruences using the least common multiple. If N leaves remainder 1 when divided by 2, 3, and 4, then N = 12k + 1 because lcm(2, 3, 4) = 12. Then enforce the multiple-of-5 condition to find the smallest k that satisfies all requirements. Step-by-Step Solution: Let N = 12k + 1.Impose N ≡ 0 (mod 5): 12k + 1 ≡ 0 (mod 5).Since 12 ≡ 2 (mod 5), we get 2k + 1 ≡ 0 ⇒ 2k ≡ 4 (mod 5).Therefore k ≡ 2 (mod 5). Take the smallest such k = 2.Compute N = 12 * 2 + 1 = 25. Verification / Alternative check: 25 divided by 2, 3, or 4 leaves remainder 1; 25 divided by 5 leaves remainder 0. All constraints check out. Why Other Options Are Wrong: 37, 54, 65: At least one of the given modular conditions fails for each of these numbers. Common Pitfalls: Forgetting to use lcm for the combined remainder-1 constraints or misapplying the modular equivalence when reducing 12 modulo 5. Final Answer: 25

Question 3

Exponential growth with folding: 1 mm thick paper doubles in thickness each fold If a sheet of paper is 1 mm thick and thickness doubles with every fold, what is the approximate thickness after 50 folds (expressed in kilometres)?

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Introduction / Context: This problem highlights exponential growth: each fold doubles thickness. After many folds, values explode in size, demonstrating how powers of 2 scale. The twist is expressing millimetres in kilometres. Given Data / Assumptions: Initial thickness t0 = 1 mm. Each fold doubles thickness: after n folds, thickness tn = t0 * 2^n. We need t50 in kilometres. Unit relation: 1 km = 10^6 mm. Concept / Approach: Compute 2^50 and convert millimetres to kilometres. Use approximate values for powers of two to match the nearest option. Step-by-Step Solution: Thickness after 50 folds: t50 = 1 mm * 2^50.Use known magnitudes: 2^10 ≈ 1.024 * 10^3, 2^50 = (2^10)^5 ≈ (1.024 * 10^3)^5 ≈ 1.126 * 10^15.Therefore t50 ≈ 1.126 * 10^15 mm.Convert to km: t50_km = (1.126 * 10^15) / 10^6 = 1.126 * 10^9 km.That is approximately 1.1 billion kilometres. Verification / Alternative check: Another route: 2^50 ≈ 1.1259 * 10^15; dividing by 10^6 gives ≈ 1.1259 * 10^9 km. This corroborates the “about 1 billion km” choice. Why Other Options Are Wrong: 100 km, 1000 km: Far too small; exponential doubling produces astronomically large values. 1 million km: Still three orders of magnitude too small compared with about 1.1 billion km. Common Pitfalls: Confusing “area halved” with thickness behavior—folding halves area but doubles thickness; also mixing up unit conversions between mm and km. Final Answer: 1 billion km

Question 4

Coin-mix linear equations: 324 coins (20p and 25p) total Rs. 71 – find count of 25p coins

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Introduction / Context: Mixed-coin problems are standard linear equation applications. Here, two denominations add to a known count and a known total value. The goal is to solve for the number of higher-value coins. Given Data / Assumptions: Total coins = 324. Denominations: 20 paise and 25 paise. Total value = Rs. 71 = 7100 paise. Concept / Approach: Let x be the count of 25p coins and y be the count of 20p coins. Translate the two constraints—total pieces and total value—into equations, then solve. Step-by-Step Solution: x + y = 324.25x + 20y = 7100.Substitute y = 324 − x into the value equation: 25x + 20(324 − x) = 7100.25x + 6480 − 20x = 7100 ⇒ 5x = 620 ⇒ x = 124.Therefore, the number of 25 paise coins is 124. Verification / Alternative check: Then y = 324 − 124 = 200 coins of 20p. Value check: 12425 + 20020 = 3100 + 4000 = 7100 paise (Rs. 71). Correct. Why Other Options Are Wrong: 120, 144, 200: None satisfy both the piece-count and value equations simultaneously. Common Pitfalls: Forgetting to convert rupees to paise, or mixing x and y when substituting back into equations. Final Answer: 124

Question 5

Comparative marksmanship with different firing rates: when B has missed 27 times, how many birds has A killed?

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Introduction / Context: This is a proportion and rate problem. Two shooters A and B fire in a fixed shot ratio and have different per-shot success rates. You must infer A’s kills when B’s number of misses is known. Given Data / Assumptions: Shot ratio: A:B = 5:3. A kills once in 3 shots ⇒ success rate per shot = 1/3. B kills once in 2 shots ⇒ success rate per shot = 1/2 (miss = 1/2). B has missed 27 times. Concept / Approach: Interpret the “kills once in n shots” as a steady rate. For B, misses:kills ≈ 1:1 because success = 1/2. Use B’s misses to deduce B’s shots, then apply the shot ratio to find A’s shots and thereby A’s kills. Step-by-Step Solution: B’s miss probability per shot = 1/2, so expected misses equal expected kills.Given misses = 27, expected kills ≈ 27. Therefore, B’s shots ≈ 54.Shot ratio A:B = 5:3, hence A’s shots = (5/3) * 54 = 90.A’s kills = A’s shots * (1/3) = 90 * (1/3) = 30. Verification / Alternative check: Think in pairs of B’s shots: every 2 shots, B averages 1 kill and 1 miss. With 27 misses, there are 27 such pairs → 54 shots, aligning with the calculation. Why Other Options Are Wrong: 60, 72, 90 birds: These assume incorrect conversions between misses, shots, and kills or ignore the 5:3 firing ratio. Common Pitfalls: Mixing up kills with misses for B, or applying the 5:3 ratio to kills instead of shots. Final Answer: 30 birds

Question 6

Age puzzle using present age difference: “I was your present age when you were born.” Father is 36 now—boy’s age five years ago?

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Introduction / Context: Classic age-word problem: a statement relates the father’s age at the son’s birth to the son’s present age. Translate into an equation using the constant age difference between parent and child. Given Data / Assumptions: Father’s current age = 36. Let the son’s current age = s. Father says: “I was of your present age when you were born.” Find the son’s age 5 years ago = s − 5. Concept / Approach: At the son’s birth, the father’s age equals the son’s current age s. The present age difference equals the father’s age at that time: father_now − son_now = s. Solve for s directly. Step-by-Step Solution: Set up equation: 36 − s = s.Solve: 36 = 2s ⇒ s = 18.Five years ago, the boy’s age = 18 − 5 = 13. Verification / Alternative check: If the son is 18 now, then 18 years ago (at son’s birth) the father was 18 (matching the son’s current age). This confirms the condition. Why Other Options Are Wrong: 15, 17, 20: These correspond to current ages 20, 22, 25 respectively, which contradict the father’s statement when checked. Common Pitfalls: Interpreting the quote as a sum rather than a difference, or forgetting that the parent-child age gap is constant. Final Answer: 13

Question 7

Quality control on garments: sleeve and body alterations across 60 shirts Every 2 out of 3 shirts need sleeve alterations and every 4 out of 5 need body alterations. For 60 shirts, how many total alterations are required (counting sleeve and body as…

Accepted Answer

Introduction / Context: Garment alteration planning often tracks different alteration types separately. Here, sleeve and body alterations have different rates. The question asks for the total number of alterations, not the number of shirts. Given Data / Assumptions: Total shirts = 60. Sleeve alterations needed in 2 out of 3 shirts. Body alterations needed in 4 out of 5 shirts. Each alteration type is counted once per shirt (if needed). Concept / Approach: Compute the count for each alteration type independently using the given fractions, then sum them. Shirts needing both contribute 2 alterations to the total. Step-by-Step Solution: Sleeves: (2/3) * 60 = 40 sleeve alterations.Body: (4/5) * 60 = 48 body alterations.Total alterations = 40 + 48 = 88. Verification / Alternative check: Whether or not some shirts overlap does not matter since the question counts alterations, not altered shirts. Overlap only affects how many shirts are touched, not the sum of alteration operations. Why Other Options Are Wrong: 123, 133, 143: These would require different rates or double-counting errors. Common Pitfalls: Interpreting the problem as asking for number of shirts altered at least once, or subtracting the overlap. The prompt clearly wants total operations. Final Answer: 88

Question 8

Carbon paper trick with folding: three sheets and two carbons to get more copies A uses three sheets with two carbon papers to make two copies. Then he folds the set so the upper half lies over the lower half and types again. How many carbon copies …

Accepted Answer

Introduction / Context: This puzzle relies on how carbon paper transfers impressions through layered sheets. Normally, with three sheets and two carbons you obtain two carbon copies beneath the top original. Folding the set cleverly allows the keystroke to produce duplicate transfers on the upper and lower halves, increasing the number of carbon copies. Given Data / Assumptions: Initial stack (top to bottom): sheet, carbon, sheet, carbon, sheet. Typically yields 2 carbon copies under the top original. Stack is folded so the upper half overlays the lower half before typing. Concept / Approach: Folding aligns layers so each keystroke passes through two thicknesses of the set across the fold, creating duplicate contact sequences of sheet–carbon–sheet in both folded halves. Each such sequence yields a transferred copy, effectively doubling the number of carbon copies. Step-by-Step Solution: Unfolded: 2 carbon copies.Folded arrangement duplicates the effective transfer path on both halves.Thus, copies ≈ 2 * 2 = 4 carbon copies. Verification / Alternative check: Practical office folklore puzzle: the fold creates two parallel stacks during impact, each reproducing the normal two-copy path, giving four carbon copies (original still on the top sheet). Why Other Options Are Wrong: 1, 2, 3: These ignore the doubling effect of the fold on the transfer paths. Common Pitfalls: Counting the original as a “copy” or assuming folding reduces transfer; in fact, it replicates it. Final Answer: 4

Question 9

Card suits and counts puzzle: 13 cards (7 black, 6 red) with suit ratios – find the number of clubs

Accepted Answer

Introduction / Context: This is a suit-distribution puzzle on a 13-card hand. You are given totals by color and proportional relations among suits. Translate the relationships into equations and solve for the unknown suit count (clubs). Given Data / Assumptions: Total cards = 13; blacks = 7; reds = 6. Let spades = s (black), clubs = c (black), diamonds = d (red), hearts = h (red). Twice as many diamonds as spades: d = 2s. Twice as many hearts as diamonds: h = 2d = 4s. Concept / Approach: Use color totals and the ratio constraints to determine s, then compute c from the total. Keep black vs. red counts consistent with the given 7 and 6. Step-by-Step Solution: Total = s + c + d + h = s + c + 2s + 4s = 7s + c = 13.Red total = d + h = 2s + 4s = 6s = 6 ⇒ s = 1.Then 7s + c = 13 ⇒ 7*1 + c = 13 ⇒ c = 6.Thus clubs = 6. Verification / Alternative check: Check color totals: black = s + c = 1 + 6 = 7; red = d + h = 2 + 4 = 6; total = 13. All conditions satisfied. Why Other Options Are Wrong: 4, 5, 7: These break either the color totals or the 2x and 4x suit-ratio constraints. Common Pitfalls: Forgetting reds sum to 6, or mixing up which suits are black (spades, clubs) and which are red (diamonds, hearts). Final Answer: 6

Question 10

Counting numbers not “connected with 6” from 1 to 60 (inclusive).  A number is defined as connected with 6 if it is divisible by 6, or if the sum of its digits equals 6, or if the digit 6 appears anywhere in the number. How many integers from 1 to 6…

Accepted Answer

Introduction / Context: This reasoning problem tests careful counting with inclusion–exclusion style thinking. A number is “connected with 6” if it meets any of three conditions: divisible by 6, has digit sum 6, or contains the digit 6. We must count how many numbers from 1 to 60 do not meet any of these conditions and then select the correct total. This is a common set-theory counting skill used in aptitude tests and quantitative exams. Given Data / Assumptions: Range of integers: 1 to 60 (inclusive). Connected with 6 if: (i) divisible by 6, or (ii) digit sum equals 6, or (iii) digit '6' occurs. We seek numbers that satisfy none of the above. Concept / Approach: We can mark numbers that are connected and then subtract from the total 60. Overlaps occur (for example, 24 is divisible by 6 and has digit sum 6 is false, but 36 is divisible by 6 and contains digit 6), so we must avoid double counting when making a running list. A safe approach is to enumerate systematically by property or by direct scan, ensuring each number is classified once. Step-by-Step Solution: Step 1: Count numbers divisible by 6 in 1..60. They are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60 → 10 numbers.Step 2: Count numbers with digit sum 6: 6, 15, 24, 33, 42, 51 → 6 numbers. Some overlap with Step 1 (24, 42).Step 3: Count numbers containing the digit 6: 6, 16, 26, 36, 46, 56, and 60 → 7 numbers. Some are already counted as divisible by 6 or digit sum 6 (e.g., 6, 36, 60).Step 4: To avoid inclusion–exclusion complexity errors, the robust method is to list all numbers 1..60 and mark those that satisfy any condition. After marking, the remaining unmarked count equals numbers not connected with 6.Carrying out that classification yields 43 numbers not connected with 6. Verification / Alternative check: Use a quick code-like or tabulation check: scan each decade (1–10, 11–20, …, 51–60), marking divisible-by-6 values, sums-to-6 values (6, 15, 24, 33, 42, 51), and any number containing '6'. Counting unmarked values across blocks confirms 43. Why Other Options Are Wrong: 18, 22, 40: These totals underestimate the many values unaffected by the three conditions. 42: Close but off by one; misses that among 1..60 exactly 17 are connected, leaving 43 not connected. Common Pitfalls: Double counting overlap when adding property counts directly. Forgetting that 60 contains digit 6 and is divisible by 6. Missing digit-sum-equals-6 cases such as 15 or 51. Final Answer: 43

Question 11

A book's pages are numbered starting from 1. If the printer used 3189 individual digits to print all page numbers, how many pages does the book contain?

Accepted Answer

Introduction / Context: This is a classic digit-counting problem on page numbers. We calculate how many digits are needed to print page numbers 1 to N and match the total 3189 to find N. Such questions strengthen place-value reasoning and arithmetic series logic. Given Data / Assumptions: Pages start at 1 and go consecutively up to N. Total digits used to print all page numbers = 3189. We count digits by ranges of lengths: 1-digit, 2-digit, 3-digit, etc. Concept / Approach: Break page numbers into intervals with constant digit lengths, then sum: 1–9 (9 pages, 1 digit each), 10–99 (90 pages, 2 digits each), 100–999 (900 pages, 3 digits each), and so on. Accumulate digits until we reach or exceed 3189, then solve for the remainder in the last interval. Step-by-Step Solution: Digits for 1–9: 9 * 1 = 9.Digits for 10–99: 90 * 2 = 180. Cumulative = 9 + 180 = 189.Remaining to reach 3189: 3189 - 189 = 3000 digits.Now use 3-digit pages from 100 upward: each page consumes 3 digits.Number of 3-digit pages required = 3000 / 3 = 1000 pages.Thus pages run from 100 to 1099 inclusive (1000 pages). Final page N = 1099.But verify cumulative: 1–99 used 189 digits; 100–1099 uses 1000 * 3 = 3000 digits; total = 3189 digits. Hence N = 1099. Wait—this contradicts options; reconcile carefully.Recheck: 100–999 is 900 pages * 3 = 2700 digits. Add to 189 → 2889. Remaining = 3189 - 2889 = 300 digits. These 300 digits belong to pages from 1000 onward (4-digit numbers, 4 digits each).Pages with 4 digits used = 300 / 4 = 75 pages.Therefore last page = 999 + 75 = 1074. Verification / Alternative check: Compute exact totals: 1–9 → 9; 10–99 → 180; 100–999 → 2700; subtotal = 2889; add 1000–1074 → 75 * 4 = 300; total = 3189. Matches perfectly. Why Other Options Are Wrong: 1000: Would require only 2889 + 1 * 4 = 2893 digits, far short of 3189. 1075: Would use 304 digits beyond 999, totaling 3193 digits, too many. 1080 and 1150: Exceed 3189 digits when computed precisely. Common Pitfalls: Jumping directly from 3-digit to a large 4-digit span without careful subtraction. Arithmetic slips in subtracting 2889 from 3189. Final Answer: 1074

Question 12

Two elevators move in opposite directions in the same building. David starts at the 11th floor going up at 57 floors per minute, and Albert starts at the 51st floor going down at 63 floors per minute. At which floor do they meet?

Accepted Answer

Introduction / Context: This is a relative speed problem on a linear scale (floors). With uniform speeds and opposite directions, the meeting point is found by equating positions as functions of time. Such problems are common in aptitude tests under the topic of time, speed, and distance. Given Data / Assumptions: David: starts at floor 11, moves upward at 57 floors/min. Albert: starts at floor 51, moves downward at 63 floors/min. They start simultaneously and move uniformly. Concept / Approach: Set positions versus time t (in minutes). Meeting occurs when David’s floor equals Albert’s floor. Because they move in opposite directions, their relative speed is the sum of the magnitudes. A simple linear equation suffices. Step-by-Step Solution: Let t be minutes after start.David’s floor: F_D = 11 + 57 * t.Albert’s floor: F_A = 51 - 63 * t.At meeting, 11 + 57 * t = 51 - 63 * t.Rearrange: 57 * t + 63 * t = 51 - 11 → 120 * t = 40.Solve: t = 40 / 120 = 1/3 minute.Meeting floor: F = 11 + 57 * (1/3) = 11 + 19 = 30. Verification / Alternative check: Relative speed method: distance between them initially = 51 − 11 = 40 floors; combined rate = 57 + 63 = 120 floors/min; time = 40 / 120 = 1/3 min; distance David ascends = 57 * (1/3) = 19 floors; 11 + 19 = 30th floor. Consistent. Why Other Options Are Wrong: 19, 28, 34, 37: Do not satisfy both motion equations at the same time t; plugging t back fails. Common Pitfalls: Using difference of speeds instead of sum for opposite directions. Arithmetic errors when solving for t. Final Answer: 30

Question 13

Find the number x such that, when x is added to itself 13 times (that is, 14x altogether), the result is 112. What is x?

Accepted Answer

Introduction / Context: This arithmetic problem examines precise interpretation of phrasing and linear equations. “Added to itself 13 times” means the original number plus 13 more copies of the same number, yielding 14 copies in total. We then solve for the unknown x. Given Data / Assumptions: The phrase implies total copies = 14. Total sum of these copies = 112. Concept / Approach: Translate the statement into a simple algebraic equation. If a number x is added to itself 13 times, there are 14x altogether. Set 14x = 112 and solve using basic division. Step-by-Step Solution: Let x be the number.“Added to itself 13 times” → total copies = 1 + 13 = 14, so sum = 14 * x.Given: 14 * x = 112.Divide both sides by 14: x = 112 / 14 = 8. Verification / Alternative check: Compute 14 * 8 = 112. The equality holds, confirming x = 8. Why Other Options Are Wrong: 7, 9, 11, 6: 14 times any of these does not produce 112 (98, 126, 154, 84 respectively). Common Pitfalls: Misreading “added to itself 13 times” as 13x instead of 14x, which would yield a non-integer not among options. Final Answer: 8

Question 14

A city taxi fare consists of a fixed charge plus a per-kilometre rate. If 16 km cost Rs 156 and 24 km cost Rs 204, what will be the fare for 30 km?

Accepted Answer

Introduction / Context: This linear model problem involves finding a fixed fee and a variable per-kilometre rate using two data points. Such fare questions model cost = fixed component + (rate * distance). Given Data / Assumptions: Total fare for 16 km = Rs 156. Total fare for 24 km = Rs 204. Fare = F + r * d, where F is fixed charge and r is Rs/km. Concept / Approach: Set up two linear equations in F and r using the two trips. Solve simultaneously, then compute the fare for 30 km by substitution. Step-by-Step Solution: Equation 1: F + r * 16 = 156.Equation 2: F + r * 24 = 204.Subtract Eq1 from Eq2: (F cancels) → r * (24 − 16) = 204 − 156 → 8r = 48 → r = 6 Rs/km.Back-substitute into Eq1: F + 6 * 16 = 156 → F + 96 = 156 → F = 60.Fare for 30 km: F + r * 30 = 60 + 6 * 30 = 60 + 180 = Rs 240. Verification / Alternative check: Check with 24 km: 60 + 6 * 24 = 60 + 144 = 204 (matches data). Model is consistent. Why Other Options Are Wrong: Rs. 236, Rs. 248, Rs. 252, Rs. 258: Do not satisfy the linear model with F = 60 and r = 6. Common Pitfalls: Mistaking the difference in costs per extra km (48 over 8 km) as the fixed part. Arithmetic slips in subtraction or multiplication. Final Answer: Rs. 240

Question 15

Nitin's age puzzle: last year his age was a perfect square, next year it will be a perfect cube. How many years from now until his age is again a perfect cube?

Accepted Answer

Introduction / Context: This number puzzle blends consecutive integers with perfect squares and perfect cubes. We are told Nitin’s current age sits exactly between a square (last year) and a cube (next year). We must deduce his present age and then the waiting time until his age becomes a cube again. Given Data / Assumptions: Last year: age = square of an integer. Next year: age = cube of an integer. Ages are positive integers. Concept / Approach: If current age is n, then n − 1 is a perfect square and n + 1 is a perfect cube. Search small squares and cubes that differ by 2. This is rare; we test candidates around typical small square–cube values. Step-by-Step Solution: Consider square 25 (5^2) and cube 27 (3^3): they are exactly 2 apart.Thus n − 1 = 25 and n + 1 = 27 → n = 26.Current age = 26. Last year 25 (a square), next year 27 (a cube) meets the conditions.Next time Nitin’s age is a cube: the next cube after 27 is 64 (4^3).Waiting time = 64 − 26 = 38 years. Verification / Alternative check: Check immediate neighbors of other small squares or cubes (e.g., 16 and 27; 36 and 27): only 25 and 27 straddle an integer. So 26 is unique in this range. Why Other Options Are Wrong: 10, 27, 39, 64 years: Do not correspond to the gap between 26 and the next cube 64. 64 years is itself a cube value, not a waiting time. Common Pitfalls: Assuming current age equals a square or a cube, instead of lying between them. Forgetting that “next year” must be the immediately next integer, hence 27 must be the cube. Final Answer: 38 years

Question 16

A student attempted 48 sums. He got twice as many sums wrong as he got right. How many sums did he solve correctly?

Accepted Answer

Introduction / Context: This algebra word problem checks ratio translation into equations. The statement “twice as many wrong as right” indicates a 2:1 ratio between wrong and right answers, with a fixed total of 48 attempts. Given Data / Assumptions: Total attempted = 48. Wrong = 2 * Right. Concept / Approach: Let R be the number correct (right). Then wrong = 2R. Since total = right + wrong, we get a single linear equation in R. Step-by-Step Solution: Let R = number right, W = number wrong.Given W = 2R and R + W = 48.Substitute: R + 2R = 48 → 3R = 48 → R = 16.Therefore, he solved 16 sums correctly. Verification / Alternative check: Compute wrong: W = 2 * 16 = 32; total = 16 + 32 = 48. Checks out. Why Other Options Are Wrong: 12, 18, 20, 24: Do not satisfy the ratio W = 2R with total 48 when tested. Common Pitfalls: Interpreting “twice as many wrong as right” backwards (setting R = 2W). Arithmetic mistakes when dividing by 3. Final Answer: 16

Question 17

Family ages puzzle: A father is three times his daughter's age. The mother is 9 years younger than the father. The son's age is half that of his mother, and the brother is 7 years older than his sister. What is the mother's age?

Accepted Answer

Introduction / Context: This is a multi-equation age problem. We translate relationships into algebra and solve simultaneously. Such questions teach consistent variable assignment and systematic elimination. Given Data / Assumptions: Daughter’s age = d. Father’s age = 3d. Mother’s age = Father − 9 = 3d − 9. Son’s age = Mother / 2 = (3d − 9) / 2. Brother is 7 years older than sister: Son = d + 7. Concept / Approach: Use the last relation to connect son and daughter, substitute the son’s expression from the mother’s relation, and solve for d. Then compute the mother’s age. Step-by-Step Solution: Equation from siblings: (3d − 9) / 2 = d + 7.Multiply by 2: 3d − 9 = 2d + 14.Solve: 3d − 2d = 14 + 9 → d = 23.Mother’s age = 3d − 9 = 69 − 9 = 60 years. Verification / Alternative check: Compute all ages: Daughter 23; Father 69; Mother 60; Son = Mother/2 = 30; Brother older than sister by 7 → 23 + 7 = 30; all relations satisfied. Why Other Options Are Wrong: 40, 45, 50, 54: None fit the system while keeping all relationships true. Common Pitfalls: Mistaking “mother is 9 years younger than father” as father = mother − 9. Forgetting to double both sides when clearing the fraction. Final Answer: 60 years

Question 18

Children's Day distribution: Sweets were to be equally shared among 175 children. On the day, 35 children were absent, and each present child received 4 extra sweets. How many sweets were there in total?

Accepted Answer

Introduction / Context: This is a proportion problem with equal distribution and attendance change. The key is relating the per-child share before and after some students are absent, then solving for the total number of sweets. Given Data / Assumptions: Planned recipients = 175 children. Actual recipients = 175 − 35 = 140 children. Each actual recipient gets 4 more sweets than originally planned. Total sweets = S (constant). Concept / Approach: If each child originally would have got S/175 sweets, but actually got S/140, then the difference per child equals 4. Set up a simple equation and solve for S. Step-by-Step Solution: Original per-child share = S / 175.Actual per-child share = S / 140.Difference = S/140 − S/175 = 4.Compute left side: S * (1/140 − 1/175) = S * ((175 − 140) / (140 * 175)) = S * (35 / 24500) = S / 700.So S / 700 = 4 → S = 4 * 700 = 2800. Verification / Alternative check: Original plan: 2800 / 175 = 16 sweets per child. Actual: 2800 / 140 = 20 sweets per child. Difference = 4, matches the condition. Why Other Options Are Wrong: 2400, 2480, 2520, 2680: Do not yield a per-child increase of exactly 4 when divided by 175 and 140. Common Pitfalls: Reversing the subtraction (using original minus actual). Arithmetic errors in fraction subtraction. Final Answer: 2800

Question 19

Nasir's birthday is on Thursday 18th May. On what day of the week will be Rehan's Birthday in the same year if Rehan was born on 19th August?

Accepted Answer

Saturday

Question 20

What are the Multiples of 18?

Accepted Answer

The Multiples of 18 are 18, 36, 54, 72, 90, 108, 126, .... Multiples of 18 are nothing but numbers which are divisible by 18 leaving remainder 0. An infinite number of multiples of 18 can be achieved by adding 18 to each subsequent multiple.

Arithmetic Reasoning Questions