Section A
Sec A MCQs — 15 Questions
(+4)
1
arithmetic-series
medium
Find the sum: 2 + 8 + 12 + 18 + 22 + ⋯ + 92 + 98.
A
2000
B
900
C
800
D
1000
E
1100
Show answer
Answer: D
Pairing the first and last terms, second and second-last, and so on: (2+98) + (8+92) + (12+88) + ⋯ + (48+52) = 100 × 10 = 1000 (there are 10 such pairs). Independently verified by listing every term of the sequence (the gaps alternate +6, +4, +6, +4, ...): 2, 8, 12, 18, 22, 28, 32, 38, 42, 48, 52, 58, 62, 68, 72, 78, 82, 88, 92, 98 — 20 terms whose average is (2+98)/2 = 50, giving sum = 20 × 50 = 1000.
2
algebra-word-problem
easy
A two-digit number is doubled and added to 25. The new number is three times the original number. What is the original number?
A
20
B
60
C
50
D
75
E
25
Show answer
Answer: E
Let the original number be x. Doubling and adding 25 gives 2x + 25, and this equals three times the original: 2x + 25 = 3x, so x = 25. Independently verified: 25 is a two-digit number, 2(25)+25 = 75, and 3(25) = 75 — both sides match.
3
calendar-days-of-week
medium
Suppose today is 15th August, Friday. How many Sundays will there be as of 30th October?
A
10
B
11
C
8
D
12
E
9
Show answer
Answer: B
From 15th August to 30th October is 16 (remaining days of August) + 30 (September) + 30 (days 1-30 of October) = 76 days later. 76 ÷ 7 = 10 remainder 6, i.e. 10 full weeks plus 6 extra days, giving 10 + 1 = 11 Sundays. Independently verified via day-of-week arithmetic: if 15 Aug (day 0) is a Friday, then day d is a Sunday exactly when d ≡ 2 (mod 7), i.e. d = 2, 9, 16, ..., up to the largest such d ≤ 76, which is d = 72 (since 2+7×10=72). The count of Sundays is (72−2)/7 + 1 = 11, confirming the source's answer.
4
factors
medium
Find the difference between two factors of 72 such that their sum is 40.
A
60
B
32
C
36
D
9
E
24
Show answer
Answer: B
The factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72. Checking pairs of these factors for a sum of 40, only 4 and 36 work (4+36=40). Their difference is 36−4 = 32. Independently verified by checking every other candidate pair (e.g. 8+32, 12+28, 16+24, 18+22 — none of the second numbers are factors of 72), confirming (4,36) is the unique such pair.
5
distributive-property
easy
76 × 8 + 12 × 8 + 2 × 8
A
170
B
800
C
900
D
720
E
810
Show answer
Answer: D
Factoring out the common 8: 76×8 + 12×8 + 2×8 = (76+12+2)×8 = 90×8 = 720. Independently verified by direct computation: 76×8=608, 12×8=96, 2×8=16, and 608+96+16 = 720.
6
estimation-multiplication
easy
The product of 58 × 24 is
A
More than 1500
B
Between 1000 and 1500
C
Between 500 and 1000
D
Between 100 and 500
E
Less than 100
Show answer
Answer: B
Using an approximation/bounding strategy: 60×25=1500 and 50×20=1000, so 58×24 must lie between 1000 and 1500. Independently verified by exact computation: 58×24 = 1392, which indeed lies between 1000 and 1500.
7
rate-word-problem
medium
Louis was riding his bike around the park. He rode 5 and a half rounds before he stopped to rest. Assuming he rode his bike at a constant speed, if he completed 2200 metres, what is the distance covered by him after one round, in metres?
A
450
B
250
C
400
D
500
E
350
Show answer
Answer: C
5.5 rounds correspond to 2200 metres. Doubling both quantities, 11 rounds correspond to 4400 metres, so one round is 4400 ÷ 11 = 400 metres. Independently verified: 400 × 5.5 = 2200, matching the given distance.
8
digit-puzzle
hard
If 5A8 + B2 adds up to 600, what is the largest possible difference between digit A and digit B?
Source PDF defect: the Questions page prints only 4 answer choices for this question (7, 6, 3, 1) instead of the usual 5. Independent re-derivation shows the full set of achievable |A−B| values is {7, 5, 3, 1} (from valid non-zero digit pairs (1,8), (2,7), (3,6), (4,5) satisfying A+B=9); the printed option 'B: 6' does not correspond to any valid digit pair, and the genuinely missing value, 5, was reconstructed here as option E. The correct answer (7) was already present as option A, so this defect does not affect the credited answer.
A
7
B
6
C
3
D
1
E
5
Show answer
Answer: A
Here 5A8 is read as 500 + (a two-digit number 'A8' = 10A+8), so the equation is 500 + (10A+8) + (10B+2) = 600, giving 10A+10B = 90, i.e. A+B = 9. For 'A8' and 'B2' to be genuine two-digit numbers, A and B must be non-zero, so the valid digit pairs (A,B) are (1,8), (2,7), (3,6), (4,5) and their reverses. The largest possible difference is |1−8| = 7. Independently verified: 518 + 82 = 600 ✓, and no other valid pair gives a larger difference than 7 (the next largest is |2−7|=5).
9
divisibility-lcm
medium
When we subtract 10 from a number, the result is divisible by 2, 5 and 7. Find the smallest such number.
A
35
B
60
C
70
D
45
E
80
Show answer
Answer: E
The smallest positive number divisible by 2, 5, and 7 is their LCM, which is 70 (since 2, 5, 7 are pairwise coprime, LCM = 2×5×7 = 70). Working backwards, the original number is 70 + 10 = 80. Independently verified: 80 − 10 = 70, and 70 is indeed divisible by 2, 5, and 7.
10
geometry-counting-rectangles
hard
How many rectangles of any size are there in the figure given?
A
7
B
6
C
8
D
9
E
10
Show answer
Answer: D
Counting by size: there are 5 smallest (1-part) rectangles, 1 two-part rectangle, 1 three-part rectangle, 1 four-part rectangle, and 1 five-part rectangle (the whole figure), giving 5+1+1+1+1 = 9. Independently re-derived by examining the figure directly: the figure is a square with a full-width strip across the top, and the remaining area below split into a left column and two narrower columns on the right (one of which is further split horizontally near the bottom) — systematically enumerating every rectangular region bounded by the drawn grid lines confirms 9 distinct rectangles.
11
triangular-numbers
medium
A pile of cards has numbers from 1 to 100 written on them. The first child gets a card with the number 1, the second child gets two cards with numbers 2 and 3, the third child gets 3 cards with numbers 4, 5, and 6, and so on. Which child will get a card with the number 50 written on it?
A
10th
B
12th
C
15th
D
20th
E
22nd
Show answer
Answer: A
The k-th child receives k cards, and the last (largest) card number that child receives is the triangular number 1+2+⋯+k = k(k+1)/2. Child 9's last card is 45 (9×10/2), and child 10's last card is 55 (10×11/2). Since 46 ≤ 50 ≤ 55, the card numbered 50 is held by the 10th child. Independently verified: the 10th child receives cards numbered 46 through 55, which does include 50.
12
combinatorics-coin-tosses
medium
Grace and Freddy tossed a coin three times each. If Grace got a head in the first toss and Freddy got a tail in the last toss, how many different ways are there that they get the same outcome?
A
4
B
2
C
6
D
8
E
3
Show answer
Answer: B
For Grace and Freddy's 3-toss outcomes to be identical, the shared sequence must start with H (Grace's requirement) and end with T (Freddy's requirement), i.e. it has the form H _ T where the middle toss can be either H or T. This gives exactly 2 possible shared outcomes: HHT and HTT. Independently verified by listing all 8 possible 3-toss sequences (HHH, HHT, HTH, HTT, TTH, THT, THH, TTT) and checking which start with H and end with T — only HHT and HTT qualify.
13
logic-ordering
medium
Bret has fewer marbles than Ruby but more than Teresa. Patrick has fewer marbles than Suzy and fewer marbles than Bret. Teresa has fewer marbles than Suzy but she does not have the least number of marbles. Who has the least number of marbles?
A
Teresa
B
Bret
C
Patrick
D
Ruby
E
Suzy
Show answer
Answer: C
From Teresa < Bret < Ruby, Ruby is not the least, and Bret is not the least (Teresa is below him). From Patrick < Suzy, Suzy is not the least. Teresa is explicitly stated not to be the least. By elimination, only Patrick remains, so Patrick has the least number of marbles. Independently verified: Patrick < Bret and Patrick < Suzy, and since Teresa (who is less than Bret) is stated not to be the least, someone must be below Teresa — the only candidate consistent with all constraints is Patrick, confirming he is least.
14
sequences
medium
In the sequence below, each letter represents a number. The sum of any three neighbouring numbers is 19. What is the value of the letter D?
A, 7, 9, B, C, D
A
10
B
9
C
7
D
3
E
5
Show answer
Answer: B
Since A+7+9=19, A=3. The three numbers 3, 7, 9 (summing to 19) then repeat every three positions, so B=A=3, C=7, and D=9. Independently verified by checking every neighbouring triple: 3+7+9=19, 7+9+3=19, 9+3+7=19 — all equal 19, confirming D=9.
15
ratio-word-problem
medium
Todd, Yasmine and Dorothy distributed a box of 35 chocolates in such a way that Todd got the same number of chocolates as Dorothy but twice as much as Yasmine. How many chocolates did Dorothy get?
A
14
B
7
C
21
D
28
E
20
Show answer
Answer: A
Let Yasmine's share be 1 unit; then Todd and Dorothy each have 2 units. Total: 2+2+1 = 5 units = 35, so 1 unit = 7. Dorothy's share is 2 units = 14. Independently verified: Todd=14, Yasmine=7, Dorothy=14, and 14+7+14 = 35.
Section B
Sec B Integers — 10 Questions
(+5)
16
time-word-problem
easy
Carry took a nap in the afternoon for 35 minutes. If she woke up at 4.00 pm, what time did she sleep? Write your answer in 4-digit number format, e.g., if the time is 1.50 am or 1.50 pm, write 0150.
Show answer
Answer: 0325
35 minutes before 4:00 pm: since there are 60 minutes in an hour, 60−35 = 25 minutes past 3:00 pm, i.e. 3:25 pm. Written in the required 4-digit format, this is 0325. Independently verified: 3:25 pm + 35 minutes = 4:00 pm.
17
rate-word-problem
medium
Paul takes 4 minutes and 30 seconds to jog 1 km. If he continues jogging at the same constant pace, how long will he take to jog 10 km? Write your answer in minutes.
Show answer
Answer: 0045
Paul's pace is 4 minutes 30 seconds (= 4.5 minutes) per km. For 10 km at the same constant pace: 4.5 × 10 = 45 minutes. Independently verified via the doubling approach used in the source: 2 km takes 9 minutes, and 10 km = 5×2 km takes 9×5 = 45 minutes.
18
fractions-word-problem
medium
Veena was watching a movie on Netflix. She finished watching one-third of the movie and took a break. If she would have watched for 12 minutes more, she would have finished watching half of the movie. How long is the movie (in minutes)?
Show answer
Answer: 0072
The extra 12 minutes bridges the gap between 1/3 and 1/2 of the movie: 1/2 − 1/3 = 3/6 − 2/6 = 1/6 of the movie = 12 minutes. So the whole movie (6/6) is 12 × 6 = 72 minutes. Independently verified: 1/3 of 72 = 24 minutes, 1/2 of 72 = 36 minutes, and 36 − 24 = 12, matching the given extra time.
19
consecutive-numbers
easy
If the sum of three even consecutive numbers is 78, find the largest number.
Show answer
Answer: 0028
Let the three consecutive even numbers be n, n+2, n+4. Their sum is 3n+6 = 78, so n = 24, and the numbers are 24, 26, 28. The largest is 28. Independently verified: 24+26+28 = 78.
20
digit-puzzle
hard
A four-digit number lock was installed at Tim's door. He said that the tens digit is the sum of the hundreds and thousands digits. Also, the ones digit is one-third of the tens digit. If the sum of all the digits is 21, find the digit at the tens place.
Show answer
Answer: 0009
Let the tens digit be T, so T = (hundreds digit + thousands digit), and the ones digit is T/3. The sum of all four digits is: ones + tens + hundreds + thousands = (T/3) + T + T = (T/3) + 2T = 21 (since hundreds+thousands = T). Multiplying by 3: T + 6T = 63, so 7T = 63, T = 9. Independently verified: with T=9, ones = 3, and hundreds+thousands = 9 (e.g. 3 and 6, or any valid split), giving digit sum 3+9+9 = 21 ✓, matching the source's own tabulated case (ones=3, tens=9, hundreds+thousands=9, sum=21).
21
ratio-word-problem
medium
Olivia gave 18 cookies to Lara. Now, Lara has twice as many cookies as Olivia. If the total number of cookies they had at first was 105, how many cookies did Olivia have at first?
Show answer
Answer: 0053
The total number of cookies (105) stays the same before and after the exchange. After the exchange, let Olivia have 1 unit and Lara have 2 units: 1+2 = 3 units = 105, so 1 unit = 35. Olivia has 35 cookies after giving away 18, so she had 35+18 = 53 cookies at first. Independently verified: before, Olivia had 53 and Lara had 105−53 = 52; after Olivia gives 18, Olivia has 35 and Lara has 70, and 70 = 2×35 ✓.
22
arithmetic-sum
medium
Numbers 11, 12, 13, 14, 15 and 16 were printed on 6 cards. Two players picked three cards each. If the sum of numbers of Player 1's cards was 42 and Player 2 has three cards with consecutive numbers, what is the sum of the smallest and the greatest of Player 2's cards?
Show answer
Answer: 0026
The sum of all six numbers is 11+12+13+14+15+16 = 81. Since Player 1's cards sum to 42, Player 2's cards sum to 81−42 = 39. Three consecutive numbers summing to 39: n + (n+1) + (n+2) = 39 ⟹ 3n+3 = 39 ⟹ n = 12, so Player 2 has 12, 13, 14. The sum of the smallest and greatest is 12+14 = 26. Independently verified: 12+13+14 = 39, and the remaining cards 11, 15, 16 (Player 1's) sum to 42 ✓.
23
arithmetic-sequences
medium
Each term of a sequence of numbers was obtained by subtracting the same number from the previous term. The 1st term is 257 and the 5th term is 209. What is the 15th term of this sequence?
Show answer
Answer: 0089
From the 1st to the 5th term is 4 steps, and the total decrease is 257−209 = 48, so each step subtracts 48÷4 = 12. From the 5th to the 15th term is 10 more steps, a further decrease of 10×12 = 120. So the 15th term is 209−120 = 89. Independently verified: the common difference is −12, so term n = 257 − 12(n−1); term 15 = 257 − 12(14) = 257 − 168 = 89.
24
arithmetic-series
easy
What is the sum of the first 10 multiples of 6?
Show answer
Answer: 0330
The first 10 multiples of 6 are 6×1, 6×2, ..., 6×10, so their sum is 6×(1+2+⋯+10) = 6×55 = 330. Independently verified: 1+2+⋯+10 = 10×11/2 = 55, and 6×55 = 330.
25
number-pattern
medium
Find the value of P.
[Four boxes are shown, each with a top number and three bottom numbers: box 1 has top 18 and bottom 4, 5, 2; box 2 has top 17 and bottom 7, 3, 4; box 3 has top 20 and bottom 5, 5, 5; box 4 has top 38 and bottom P, 5, 7.]
Show answer
Answer: 0009
The pattern in each box is: top = (1st bottom number) × (2nd bottom number) − (3rd bottom number). Checking: 4×5−2=18 ✓, 7×3−4=17 ✓, 5×5−5=20 ✓. Applying the same rule to the fourth box: P×5−7 = 38, so 5P = 45, P = 9. Independently verified: 9×5−7 = 45−7 = 38, matching the box's top number.