Q1 — C · The series uses two alternating rules: add 1, then add 3. So 4 + 1 = 5, 5 + 3 = 8, 8 + 1 = 9, 9 + 3 = 12, 12 + 1 = 13. The next step is add 3, giving 13 + 3 = 16. Choosing 14 means applying add 1 again by mistake.
Q2 — D · 63, 72 and 81 are all multiples of 9 (9 x 7, 9 x 8, 9 x 9), but 92 is not, because 9 x 10 = 90 and 92 - 90 = 2. The trap is to see 63, 72, 81 climbing by 9 and accept 92 without checking: 81 + 9 = 90, not 92.
Q3 — C · The jumps alternate between 2 letters and 3 letters: A (+2) C (+3) F (+2) H (+3) K (+2) M. The next jump is +3, so M, skip N and O, gives P. Choosing O means applying +2 again instead of switching rules.
Q4 — B · Compare PEN with QGQ: P moves 1 step to Q, E moves 2 steps to G, N moves 3 steps to Q. So the rule shifts each letter by 1, 2, 3 in order. Applying it to BAG: B + 1 = C, A + 2 = C, G + 3 = J, giving CCJ. CBH is the trap of moving every letter 1 step; EDJ is the trap of moving every letter 3 steps.
Q5 — B · March has 31 days, so from 25 March to 31 March is 6 days, and 8 more days reach 8 April: 6 + 8 = 14 days in total. 14 is exactly 2 weeks, so 8 April falls on the same day, Friday. Choosing Thursday means treating March as a 30-day month (13 days instead of 14).
Q6 — D · Two rules alternate: multiply by 2, then subtract 1. So 3 x 2 = 6, 6 - 1 = 5, 5 x 2 = 10, 10 - 1 = 9, 9 x 2 = 18, 18 - 1 = 17. The next step is multiply by 2, giving 17 x 2 = 34. Choosing 16 means subtracting 1 again; 33 is the off-by-one slip 34 - 1.
Q7 — C · In DF, IK and SU the second letter comes 2 steps AFTER the first (D to F, I to K, S to U). In PN the second letter comes 2 steps BEFORE the first (N is 2 steps before P). The gap size is the same in all four pairs — the direction is what changes — so PN is the odd one out.
Q8 — C · Track each direction separately. North-south: 40 m north then 40 m south cancel out, so she is level with her start. East-west: 30 m east + 10 m east = 40 m east. So she finishes 40 m due east of her starting point. 120 m is the total distance walked, not the distance from the start; 30 m east forgets the final leg.
Q9 — D · First find Sana's position from the front: she is on the front side of Ravi with exactly one child between them, so she is 8 - 2 = 6th from the front. Then convert to a position from the back: in a queue of 20, the 6th from the front is 20 - 6 + 1 = 15th from the back. 14th is the trap of computing 20 - 6 and forgetting to add 1; 6th answers the wrong end of the queue.
Q10 — A · Chain the clues: Omar before Maya, Maya before Noor, Noor before Leila gives Omar, Maya, Noor, Leila. Kito is before Omar, so Kito must be first. The full order is Kito, Omar, Maya, Noor, Leila — so Maya finishes third. Omar is the trap for anyone who forgets to place Kito in front of him.
Q11 — C · One 6 is in the thousands place, so its value is 6,000. The other 6 is in the tens place, so its value is 60. The difference is 6,000 - 60 = 5,940. Choosing 0 treats the two digits as equal; 6,060 is their sum, not their difference.
Q12 — B · To be even the number must end in 0, 2 or 4. Ending in 0 lets the biggest digits fill the front: 9, 7, 4, 2 then 0 gives 97,420 — larger than 97,402 (ending in 2) or 97,204 (ending in 4). In 97,420 the tens digit is 2. Choosing 0 confuses the tens place with the ones place.
Q13 — D · Work backwards: the number equals divisor x quotient + remainder = 7 x 208 + 5. First 7 x 208 = 1,456, then 1,456 + 5 = 1,461. Check: 1,461 divided by 7 gives 208 remainder 5. Choosing 1,456 forgets to add the remainder; 1,451 subtracts it instead.
Q14 — C · First simplify the right side: 828 + 72 = 900. Then find the missing factor: 900 divided by 36 = 25, since 36 x 25 = 900. Choosing 23 comes from dividing 828 alone by 36 and ignoring the + 72.
Q15 — B · A number that is a multiple of both 4 and 6 must be a multiple of 12 (the smallest number in both lists: 12, 24, 36, 48, ...). Between 1 and 50 these are 12, 24, 36 and 48 — four numbers. Choosing 2 comes from wrongly using 4 x 6 = 24 as the common multiple (giving only 24 and 48); 8 counts the multiples of 6 alone.
Q16 — C · If 3/7 of the number is 27, then 1/7 of it is 27 divided by 3 = 9. So 5/7 of the number is 9 x 5 = 45. Check: the whole number is 9 x 7 = 63, and 5/7 of 63 is 45. Choosing 63 answers the whole number instead of 5/7 of it.
Q17 — B · The rectangle's perimeter is 2 x (14 + 6) = 2 x 20 = 40 cm. The square has the same perimeter, and a square's perimeter is 4 times its side, so the side is 40 divided by 4 = 10 cm. Choosing 20 stops at the half-perimeter (14 + 6); choosing 40 stops at the perimeter itself.
Q18 — C · Write them all with two decimal places: 6.20, 6.02, 6.22, 6.12. In order from largest: 6.22, 6.20, 6.12, 6.02. So the second largest is 6.2. The trap is thinking 6.2 is smaller than 6.12 because 2 is smaller than 12 — but 6.2 means 6.20, which is more than 6.12.
Q19 — D · Such a number is 3 more than a multiple of both 5 and 6, so 3 more than a multiple of 30. The candidates are 33, 63, 93, ..., and the smallest 2-digit one is 33. Check: 33 = 5 x 6 + 3 and 33 = 6 x 5 + 3. The traps each satisfy only one condition: 13 and 28 leave remainder 3 when divided by 5 but not by 6, and 18 leaves remainder 3 when divided by 5 but remainder 0 when divided by 6.
Q20 — A · List the 2-digit multiples of 7: 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98. Keep the odd ones: 21, 35, 49, 63, 77, 91. Now check digit differences: 21 gives 1, 35 gives 2, 49 gives 5, 63 gives 3, 77 gives 0, 91 gives 8. Only 35 satisfies all three conditions. 57 is the trap: it is odd with digit difference 2, but 57 = 3 x 19 is not a multiple of 7.
Q21 — B · The mangoes cost 35 x 3 = 105 shillings. Her change is 200 - 105 = 95 shillings. Choosing 105 stops at the cost of the mangoes; 165 is the mistake of subtracting the price of only one mango (200 - 35).
Q22 — C · Add 1 hour to 7:45 a.m. to get 8:45 a.m., then add 25 minutes: 8:45 + 15 minutes reaches 9:00, and the remaining 10 minutes give 9:10 a.m. Choosing 9:20 mishandles the carry over the hour.
Q23 — B · Work backwards from the change: the tarts together cost €50 - €18 = €32, so one tart costs €32 divided by 4 = €8. Choosing €12.50 divides the €50 note by 4 instead of the amount actually spent; €4.50 divides the change by 4.
Q24 — B · First find the total delivered: 4 kg 250 g = 4,250 g, and 4,250 x 6 = 25,500 g = 25 kg 500 g. Then subtract the amount sold: 25,500 - 15,500 = 10,000 g = 10 kg. Choosing 25 kg 500 g stops after the first step; 9 kg 750 g and 10 kg 750 g come from borrowing errors in the subtraction.
Q25 — C · The bottles take 750 x 12 = 9,000 mL = 9 L. The cooler held 20 L, so 20 - 9 = 11 L is left. Choosing 9 L stops at the amount poured out instead of the amount remaining.
Q26 — B · Convert to centimetres: 6 m 40 cm = 640 cm. Divide by 45: 45 x 14 = 630, and 45 x 15 = 675 is too much, so 14 full pieces are cut. The leftover is 640 - 630 = 10 cm. Choosing 14 confuses the number of pieces with the leftover; 40 cm ignores the 6 m part of the conversion.
Q27 — C · After spending 2/5, she has 3/5 of her money left, and that equals ¥720. So 1/5 of her money is 720 divided by 3 = ¥240, and the whole amount is 240 x 5 = ¥1,200. Check: 2/5 of 1,200 is 480, and 1,200 - 480 = 720. Choosing ¥1,800 treats ¥720 as 2/5 instead of 3/5; ¥288 computes 2/5 of 720.
Q28 — B · First train: 10:35 a.m. to 12:00 noon is 1 hour 25 minutes, and 12:00 to 2:10 p.m. is 2 hours 10 minutes, so the journey takes 3 hours 35 minutes. The second train takes 3 hours 20 minutes, so the first is 35 - 20 = 15 minutes slower. Choosing 35 stops at the minutes digit of the first journey; 25 comes from miscounting the time to noon.
Q29 — B · She saves 96 divided by 8 = 12 francs a week. She still needs 150 - 96 = 54 francs. Now 54 divided by 12 is 4 remainder 6, so after 4 more weeks she has only 96 + 48 = 144 francs — not enough. She needs a 5th week to pass 150 (96 + 60 = 156). Choosing 4 ignores the remainder; 12 is the weekly saving, not a number of weeks.
Q30 — D · Work backwards. Before buying the snack he had 35 + 25 = 60 pesos. That 60 pesos was the half left after the ride tickets, so at first he had 60 x 2 = 120 pesos. Check: half of 120 is 60, minus 25 leaves 35. Choosing 60 stops one step early; 70 doubles the 35 without first adding back the snack.
Q31 — C · The perimeter is 2 x (35 + 20) = 2 x 55 = 110 m. The gates remove 5 m, so 110 - 5 = 105 m needs fencing. The cost is 105 x 45 = 4,725 rand (100 x 45 = 4,500 and 5 x 45 = 225; 4,500 + 225 = 4,725). Choosing 4,950 fences the whole perimeter and forgets the gates; 4,500 uses 100 m instead of 105 m.
Q32 — B · The fractions spent are 1/4 + 1/3 = 3/12 + 4/12 = 7/12, so the money left is 5/12 of the total. Since 5/12 equals £25, 1/12 equals £5 and the total is £60. The paints cost 1/3 of £60 = £20. Check: book £15 + paints £20 + £25 left = £60. Choosing £60 stops at the total; £15 answers the book instead of the paints.
Q33 — A · Undo each step in reverse order. Before the division: 9 x 6 = 54. Before the addition: 54 - 14 = 40. Before the doubling: 40 divided by 2 = 20. Check forwards: 20 x 2 = 40, 40 + 14 = 54, 54 divided by 6 = 9. Choosing 40 stops one step early; 27 comes from halving 54 before removing the 14.
Q34 — B · If all 15 animals were hens there would be 15 x 2 = 30 legs. There are 46 - 30 = 16 extra legs, and each goat adds 2 extra legs compared with a hen, so there are 16 divided by 2 = 8 goats. Check: 8 goats have 32 legs and 7 hens have 14 legs; 32 + 14 = 46. Choosing 7 gives the number of hens; 16 stops at the extra legs.
Q35 — B · Divide 250 by 12: 12 x 20 = 240, so there are 20 full boxes and 250 - 240 = 10 loose sandwiches. The boxes bring in 20 x $30 = $600 and the loose sandwiches 10 x $2 = $20, so the total is $600 + $20 = $620. Choosing $600 forgets the leftovers; $630 wrongly rounds up to 21 boxes.