1
arithmetic
easy
Mr Lim paid a total of $98 400 for his new car. He made the first payment of $30 000. For the remaining amount, he paid a fixed amount monthly for the next 60 months. How much did he pay each month?
Answer: 1140
Remaining amount = 98400 − 30000 = $68400.
Monthly payment = 68400 ÷ 60 = $1140.
2
fractions
medium
Edward bought some apples. 3/5 of the apples were rotten and were thrown away. He sold 1/4 of the remaining apples and had 60 apples left. How many apples did Edward have at first?
Answer: 200
After throwing away rotten apples, remaining = 2/5 of total.
After selling 1/4 of the remaining, 3/4 of (2/5 of total) = 60.
(3/10) of total = 60, so total = 200.
3
geometry
hard
In the figure below, ABC and BCD are isosceles triangles (tick marks show AB = AC, and CD = BD). ∠BAC = 46° and ∠ACD = 150°. Find ∠BDC.
Answer: 14
Triangle ABC is isosceles with AB = AC, so ∠ABC = ∠ACB = (180° − 46°)/2 = 67°.
∠BCD = ∠ACD − ∠ACB = 150° − 67° = 83°.
Triangle BCD is isosceles with CD = BD, so its base angles ∠DBC = ∠DCB = 83°.
∠BDC = 180° − 83° − 83° = 14°.
4
arithmetic
medium
Every day, David saves $1.40 while his brother saves $2.20. David started saving 20 days earlier before his brother started saving. When David has saved $63, how much has his brother saved?
Answer: 55
David has been saving for 63 ÷ 1.40 = 45 days.
His brother has been saving for 45 − 20 = 25 days.
Brother's savings = 25 × $2.20 = $55.
5
measurement
easy
Ali had a roll of wire measuring 1 m long. He cut 9 pieces of wire, each measuring 8 cm, from the roll of wire. The remaining wire was used to form a square. Find the area of the square. Give your answer in cm².
Answer: 49
Remaining wire = 100 − 9×8 = 100 − 72 = 28 cm (the square's perimeter).
Side of square = 28 ÷ 4 = 7 cm.
Area = 7 × 7 = 49 cm².
6
fractions
medium
Mrs Lim baked 265 cakes and pies at first. After she sold 25 cakes and 2/7 of the pies, she had an equal number of cakes and pies left. How many cakes did she have at first?
Answer: 125
Let cakes = c, pies = p, c + p = 265.
c − 25 = (5/7)p (remaining cakes = remaining pies).
Substituting c = 265 − p: 240 − p = (5/7)p → 240 = (12/7)p → p = 140.
c = 265 − 140 = 125.
7
number-patterns
medium
Study the number pattern below. Based on the pattern, what is the difference in value between the first number and the last number in Row 30?
Answer: 58
Row n has (2n−1) numbers, and the last number in Row n is n² (since 1+3+5+...+(2n−1) = n²).
Row 30 ends at 30² = 900, and Row 29 ends at 29² = 841, so Row 30 starts at 842.
Difference = 900 − 842 = 58.
8
averages
medium
A group of 11 students saved an average of $385. When two of the students lost all their savings, the average savings of the remaining students became $288. What was the total amount of money lost by the two students?
Answer: 1643
Total of 11 students = 11 × 385 = $4235.
Total of the other 9 students = 9 × 288 = $2592.
Amount lost by the two students = 4235 − 2592 = $1643.
9
percentage
easy
Mr Lim sold a total of 60 gift vouchers. 40% of the gift vouchers sold were $5 in value each and the rest were $10 in value each. How much money did Mr Lim collect from the sale of the $10 gift vouchers?
Answer: 360
40% of 60 = 24 vouchers at $5; the rest, 60 − 24 = 36 vouchers, at $10.
Money from $10 vouchers = 36 × $10 = $360.
10
geometry
medium
Jerome glued a 2-cm cube and a 5-cm cube together as shown below (the small cube sits on top of the large cube). He then painted the whole solid blue. What was the total area of all the painted faces of the solid? Give your answer in cm².
Answer: 166
Surface area of 5-cm cube = 6×5² = 150 cm²; of 2-cm cube = 6×2² = 24 cm².
Where the cubes meet, a 2×2 = 4 cm² patch is unpainted on each cube.
Total painted area = (150 − 4) + (24 − 4) = 146 + 20 = 166 cm².
11
ratio
medium
At a fruit stall, the ratio of the number of apples to the number of oranges was 9 : 5. 1/2 of the apples and 1/5 of the oranges were rotten. There were 253 rotten apples and oranges. How many apples and oranges were there altogether at first?
Answer: 644
Let apples = 9k, oranges = 5k. Rotten = 9k/2 + k = 11k/2 = 253 → k = 46.
Apples = 414, oranges = 230. Total = 414 + 230 = 644.
12
geometry
hard
Square ABCD overlaps Triangle EFG to form Square ERCS as shown below. AB = EF = EG and R and S are the midpoints of EF and EG respectively. The area of Square ERCS is 81 cm². Find the area of the whole figure. Give your answer in cm².
Answer: 405
Square ERCS has area 81, so its side ER = 9. Since R is the midpoint of EF, EF = 2×9 = 18, so the square ABCD's side AB = EF = 18, giving area 18² = 324.
Triangle EFG is right-angled at E (marked) with EF = EG = 18, so its area = (1/2)×18×18 = 162.
Whole figure area = square + triangle − overlap = 324 + 162 − 81 = 405 cm².
13
averages
medium
Mrs Lim measured and recorded the height of a group of 5 children: Vinesh 1.62 m, Sue 1.58 m, Lili 1.55 m, Raja ? m, Ken 1.53 m. The average height of the 5 children was 157 cm. One more child, Jack, joined the group and the average height of the six children became 159 cm. What was the average height of Jack and Raja? Give your answer in cm.
Answer: 163
Sum of 5 heights = 5×157 = 785 cm. Known four = 162+158+155+153 = 628 cm. Raja = 785−628 = 157 cm.
Sum of 6 heights = 6×159 = 954 cm. Jack = 954−785 = 169 cm.
Average of Jack and Raja = (169+157)/2 = 163 cm.
14
algebra
medium
Kenny and James have the same number of coins. Kenny has 20-cent coins and James has 50-cent coins. James has $35.70 more than Kenny. How many coins do Kenny and James have altogether?
Answer: 238
Let each have n coins. 0.50n − 0.20n = 35.70 → 0.30n = 35.70 → n = 119.
Total coins = 119 + 119 = 238.
15
ratio
easy
Azahar Money Changer offers an exchange rate of 0.75 US dollars for 1 Singapore dollar. At this rate, how many Singapore dollars will Mrs Lim get when she gives the money changer 675 US dollars?
Answer: 900
1 SGD = 0.75 USD, so 1 USD = 1/0.75 SGD.
675 USD = 675 ÷ 0.75 = S$900.
16
algebra
medium
In January, the total expenditure of Alice, Bernice and Carol was $1510. In February, Alice increased her expenditure by $120, Bernice doubled her expenditure and Carol reduced her expenditure by $130. In the end, their expenditure became the same. What was Alice's expenditure in January?
Answer: 480
Let the common February value be X. Alice(Jan) = X−120, Bernice(Jan) = X/2, Carol(Jan) = X+130.
(X−120) + X/2 + (X+130) = 1510 → 2.5X + 10 = 1510 → X = 600.
Alice's January expenditure = 600 − 120 = $480.
17
volume
hard
Sam cut out Cube Y from a larger wooden Cuboid X along the dotted line as shown below. The original volume of the Cuboid X was 6 times the volume of Cube Y. Then, Sam cut out as many 2-cm cubes as possible from the remaining Cuboid X. What was the volume of Cuboid X that was left uncut? Give your answer in cm³.
Answer: 491
Cube Y has side 7 cm, volume 343 cm³. Original Cuboid X volume = 6×343 = 2058 cm³.
Since X has a 7×7 cross-section (matching Y's face), its length = 2058 ÷ 49 = 42 cm.
After removing Y (7 cm from one end), the remaining cuboid is 35 × 7 × 7 = 1715 cm³.
Fitting 2-cm cubes: floor(35/2)=17, floor(7/2)=3, floor(7/2)=3 → 17×3×3 = 153 cubes, using 153×8 = 1224 cm³.
Volume left uncut = 1715 − 1224 = 491 cm³.
18
fractions
hard
A book costs 3 times as much as a file. John spent 5/9 of his money on some books. He spent 3/4 of the remainder on 9 files. How many books did John buy?
Answer: 5
Let total money = M, file price = f. Spent on files = 9f = (3/4)×(4/9)M = M/3 → f = M/27.
Spent on books = (5/9)M = (number of books)×3f = (number of books)×(M/9).
Number of books = (5/9)M ÷ (M/9) = 5.
19
percentage
medium
Sharon was looking for a new laptop. At Shop W, usual price $2830 with 20% discount (no GST). At Shop X, usual price $2500 with 18% discount, plus 8% GST on the discounted price. Sharon bought the laptop at the lower price. How much did she pay for the laptop?
Answer: 2214
Shop W: 2830 × 0.80 = $2264.
Shop X: 2500 × 0.82 = $2050, then ×1.08 = $2214.
Lower price is Shop X's $2214.
20
algebra
medium
Beatrice sold some files at $7.50 each and 18 bookmarks at $1.50 each. Jen sold the same number of files at $7.80 each and 18 bookmarks at $1.30 each. They collected the same amount of money. How many files did each girl sell?
Answer: 12
7.50n + 18×1.50 = 7.80n + 18×1.30
7.50n + 27 = 7.80n + 23.4 → 3.6 = 0.30n → n = 12.
21
geometry
hard
In the figure below, PQRS is a rhombus and PQTU is a trapezium (right angles at T and U). PT is a straight line, crossing side RS at O. ∠QRS = 124°, ∠ROP = 82° and ∠STU = 18°. Find the sum of ∠SPT and ∠RST.
Answer: 78
Rhombus: ∠QRS = 124° so ∠SRT = 180−124 = 56° (Q,R,T colinear). In triangle ORT: ∠ROT = 180−82 = 98°, ∠ORT = 56°, so ∠OTR = 26°.
Trapezium PQTU is right-angled at both T and U, so ∠QTU = 90°; ∠PTU = 90−26 = 64°; ∠PTS = 64−18 = 46° (S lies between P and U).
Triangle OPS: ∠POS = 98° (vertical to ∠ROT), ∠OSP = 180−124 = 56° (rhombus angle at S), so ∠SPT = ∠OPS = 180−98−56 = 26°.
Triangle OST: ∠SOT = 82° (vertical to ∠ROP), ∠OTS = 46°, so ∠RST = ∠OST = 180−82−46 = 52°.
Sum = 26 + 52 = 78°.
22
averages
medium
A group of students spent an average of 70 minutes each day for online learning. Then, 3 students who each spent 30 minutes each day joined the group. In the end, the average time spent online learning for the whole group of students became 55 minutes. How many students were there in total in the end?
Answer: 8
Let original group size = n. (70n + 3×30) / (n+3) = 55 → 70n + 90 = 55n + 165 → 15n = 75 → n = 5.
Final total = 5 + 3 = 8 students.
23
measurement
medium
Green and red flags are tied to a rope in the pattern as shown below (alternating green, red, starting and ending with green). The widths of the green flags are 14 cm and the red flags are 11 cm. The first and the last flag is tied 12 cm from each end of the rope. The gap between each flag is 6 cm. There is a total of 97 flags tied to the rope. What is the length of the rope in centimetres?
Answer: 1814
97 flags alternating starting and ending with green → 49 green, 48 red.
Flag widths: 49×14 + 48×11 = 686 + 528 = 1214 cm.
Gaps between the 97 flags: 96×6 = 576 cm.
End gaps: 2×12 = 24 cm.
Total = 1214 + 576 + 24 = 1814 cm.
24
fractions
medium
Joshua spent 1/8 of his monthly salary and an additional $50 on food. He spent 2/5 of the remainder and an additional $70 on household expenses. He was left with $2630. How much was Joshua's monthly salary?
Answer: 5200
Let salary = S. After food: (7/8)S − 50.
After household: (3/5)×[(7/8)S − 50] − 70 = 2630.
(21/40)S − 30 − 70 = 2630 → (21/40)S = 2730 → S = 5200.
25
volume
medium
Rectangular Tank A measuring 20 cm by 7 cm by 23 cm is filled with water to the brim. Rectangular Tank B measuring 8 cm by 11 cm by 13 cm is filled with some water. After 1/5 of the water from Tank A is poured into Tank B, Tank B is 5/8 filled with water. How much water was in Tank B at first? Give your answer in cm³.
Answer: 71
Tank A volume = 20×7×23 = 3220 cm³. 1/5 poured = 644 cm³.
Tank B capacity = 8×11×13 = 1144 cm³. 5/8 filled = 715 cm³.
Water in Tank B at first = 715 − 644 = 71 cm³.
26
percentage
medium
The table below shows the rates for a cruise to Penang. Mr Potter, his wife and 3 children aged 8, 10 and 15 planned to go on a cruise. Mr Potter chose the Ocean View package and received an early bird discount of 15% off his total cost, then paid additional 8% GST on the discounted price. How much did he have to pay in all? Round your answer to the nearest dollar.
Answer: 2290
The 15-year-old is not below 12, so pays adult rate. Adults: Mr Potter, wife, 15-year-old = 3×$645 = $1935. Children (8,10): 2×$280 = $560.
Total = 1935+560 = $2495. After 15% discount: 2495×0.85 = $2120.75. After 8% GST: 2120.75×1.08 = $2290.41 ≈ $2290.
27
percentage
hard
1000 concert tickets were on sale over a week at $25 each. The graph below shows the number of tickets left unsold at the end of each day (Mon 900, Tue 700, Wed 550, Thu 450, Fri 250, Sat 150). On Sunday, the concert organiser held a special promotion to sell all the remaining tickets: for every 5 tickets bought, the 6th ticket was given free. Sunshine School bought all the remaining tickets with this special promotion. How much did the concert organiser collect from the sale of the remaining tickets on Sunday?
Answer: 3125
Unsold at end of Saturday = 150 tickets, all sold on Sunday.
150 ÷ 6 = 25 groups exactly (each group: pay for 5, get 1 free).
Collected = 25 × 5 × $25 = $3125.
28
percentage
medium
During a sale, a discount of 10% was given to every 12 chocolate bars bought. Each chocolate bar cost $8. Dan paid a total of $331.20 for some chocolate bars. How many chocolate bars did he buy?
Answer: 45
Every full set of 12 costs 12×8×0.9 = $86.40 (discounted); extra bars beyond full sets cost $8 each.
3 sets = $259.20; remainder = 331.20−259.20 = $72 = 9 bars at $8.
Total bars = 3×12 + 9 = 45.
29
algebra
hard
Eggs were only sold in trays of 30 eggs at a shop. Mr Lee bought some such trays of eggs and re-packed them into carton boxes of 12 eggs. He needed 6 more eggs to have exact carton boxes of 12 eggs. He also needed 26 more carton boxes than the number of trays he had bought. How many eggs did Mr Lee buy from the shop?
Answer: 510
Let trays = T, eggs bought = 30T. (30T+6)/12 = T+26 → 30T+6 = 12T+312 → 18T = 306 → T = 17.
Eggs bought = 30×17 = 510. (Check: (510+6)/12 = 43 = 17+26 ✓)
30
ratio
medium
The ratio of the number of female members to the number of male members in a chess club was 4 : 3. When 38 male members left the club and 28 new female members joined the club, the number of male members to the number of female members became 1 : 2. How many male members were there in the chess club in the end?
Answer: 118
Let female = 4k, male = 3k. (3k−38)/(4k+28) = 1/2 → 6k−76 = 4k+28 → 2k = 104 → k = 52.
Male members in the end = 3×52 − 38 = 118.