Work through the levels in order. Type your answer and press Check — open the worked solution if you get stuck. Answers like -6, 3/4 or 0.25 are typed exactly as shown. A ⭐ Bonus Challenge at the end is extra credit — skip it if the rest was hard work.
30 questions · official format, shuffled order
1 What is the value of 2 − 4 + 6 − 8 + 10 − 12 + … + 98 − 100?
Show worked solution
The 50 terms make 25 pairs, each equal to −2, so the total is 25 × (−2) = −50. (−25 counts each pair as −1; 50 has the sign the wrong way round.)
2 A 5 cm by 3 cm rectangle is cut from the corner of a 12 cm by 8 cm rectangle, leaving the shape below. All its corners are right angles. What is its perimeter?
Show worked solution
The 5 cm inner edge replaces 5 cm of the top, and the 3 cm inner edge replaces 3 cm of the right side, so the perimeter is still 2 × (12 + 8) = 40 cm. (32 cm leaves out both inner edges; 35 cm and 37 cm leave out one of them.)
3 The mean of five numbers is 12. One of the numbers is removed, and the mean of the other four is 10. Which number was removed?
Show worked solution
The five add to 5 × 12 = 60 and the four add to 4 × 10 = 40, so the removed number is 60 − 40 = 20. (2 is the drop in the mean, 12 − 10; 22 would leave a total of 38.)
4 Which of these numbers is the largest?
Show worked solution
250 = (25)10 = 3210, 330 = 2710, 520 = 2510, and 1010 and 610 are already tenth powers. The largest base is 32, so the answer is 250. (330 looks bigger because 3 > 2, but its tenth-power base is only 27.)
5 The length of a rectangle is increased by 20% and its width is decreased by 20%. What happens to its area?
Show worked solution
1.2 × 0.8 = 0.96, so the area decreases by 4%. ("Stays the same" adds +20% and −20% instead of multiplying.)
6 How many squares have all four of their corners on dots of this 4 by 4 grid? (Tilted squares count.)
Show worked solution
Upright: 9 + 4 + 1 = 14. Tilted inside a 2 by 2 box: 4 boxes × 1 = 4. Tilted inside the 3 by 3 box: 2. Total 14 + 4 + 2 = 20. (14 forgets the tilted squares; 18 misses the two in the 3 by 3 box.)
7 In triangle ABC, AB = AC and ∠BAC = 40°. The line BD bisects ∠ABC and meets AC at D. What is ∠BDC?
Show worked solution
The base angles are (180° − 40°) ÷ 2 = 70°, so ∠ABD = 35°. ∠BDC is the exterior angle of triangle ABD: 40° + 35° = 75°. (105° is ∠ADB, the angle on the other side of D; 70° is a base angle.)
8 How many of the positive divisors of 360 are odd?
Show worked solution
Odd divisors are the divisors of 32 × 5 = 45: there are (2 + 1)(1 + 1) = 6 of them (1, 3, 5, 9, 15, 45). (24 is the count of all divisors; 12 halves it, which is wrong because the factors of 2 are not split evenly.)
9 A cyclist rides the first half of a journey at 60 km/h and the second half, the same distance, at 40 km/h. What is her average speed for the whole journey?
Show worked solution
Each half 120 km: 2 h at 60 km/h and 3 h at 40 km/h, so 240 km in 5 h = 48 km/h. (50 km/h averages the speeds, but she spends longer at the slower speed.)
10 The grid is made of 1 cm squares. What is the area of the shaded quadrilateral?
Show worked solution
The box is 6 × 5 = 30 cm2. The corner triangles are 4, 3, 4 and 3 cm2, which is 14 cm2. So the shape is 30 − 14 = 16 cm2. (14 cm2 is the corner triangles, not the shape; 30 cm2 is the whole box.)
11 n is the smallest positive whole number for which 12n is a perfect cube. What is n?
Show worked solution
12 = 22 × 3 needs one more 2 and two more 3s: n = 2 × 32 = 18, and 12 × 18 = 216 = 63. (12 gives 144 = 122, a square, not a cube; 6 gives 72, which still lacks a 3.)
12 x is a number with x + 1/x = 5. What is x2 + 1/x2?
Show worked solution
(x + 1/x)2 = x2 + 2 + 1/x2 = 25, so x2 + 1/x2 = 23. (25 forgets the middle term 2; 27 adds it instead of subtracting.)
13 How many three-digit numbers have their digits strictly increasing from left to right (for example 147 or 589)?
Show worked solution
Each choice of 3 digits from 1–9 gives exactly one such number: 9 × 8 × 7 ÷ 6 = 84. 0 cannot be used: it would have to come first. (120 chooses from all ten digits, including 0.)
14 A right-angled triangle has shorter sides 5 cm and 12 cm. What is the length h of the perpendicular from the right angle to the longest side?
Show worked solution
The longest side is 13 cm. Area = ½ × 5 × 12 = 30 = ½ × 13 × h, so h = 60/13 cm. (30/13 cm forgets the ½ on one side; 13/2 cm is half the hypotenuse.)
15 Two fair six-sided dice are rolled. What is the probability that the total is a prime number?
Show worked solution
2: 1 way, 3: 2, 5: 4, 7: 6, 11: 2, a total of 15 out of 36 = 5/12. (7/18 = 14/36 misses one of the ways, usually the single way to make 2.)
16 What is the sum of the digits of 1020 − 20?
Show worked solution
105 − 20 = 99 980 has three 9s and an 8. In the same way 1020 − 20 is eighteen 9s, then 8, then 0: 18 × 9 + 8 = 170. (180 is twenty 9s; 171 is nineteen 9s — both miscount how many 9s are left.)
17 Six teams play a tournament in which every team plays every other team once. A win scores 3 points, a draw 1 point to each team and a loss 0. At the end the six teams have 40 points altogether. How many games were drawn?
Show worked solution
There are 6 × 5 ÷ 2 = 15 games. With no draws there would be 15 × 3 = 45 points, and each draw hands out 1 point fewer. 45 − 40 = 5 draws. (Dividing 40 by 3 or treating a draw as 1 point in total gives the smaller options.)
18 Squares ABCD (side 10 cm) and BEFG (side 6 cm) sit side by side as shown. What is the area of the shaded triangle DBF?
Show worked solution
The rectangle is 16 × 10 = 160. Around the triangle: ADB = ½ × 10 × 10 = 50, BEF = ½ × 6 × 6 = 18, and the top-right triangle D, F and the corner above E = ½ × 16 × 4 = 32. So 160 − 50 − 18 − 32 = 60 cm2. (50 cm2 is triangle ABD, half the big square; 30 cm2 is half the right answer.)
19 Students arrive at a hall. If they sit 4 to a bench, 3 students have nowhere to sit. If they sit 5 to a bench, every student has a seat, two benches are left empty and every other bench is full. How many students are there?
Show worked solution
4b + 3 = 5(b − 2) gives b = 13, so there are 4 × 13 + 3 = 55 students (and 5 × 11 = 55 ✓). (60 and 65 are multiples of 5 but 60 − 3 and 65 − 3 are not multiples of 4; 35 = 5 × 7 would need 9 benches, but 4 × 9 + 3 = 39.)
20 How many of the whole numbers from 1 to 100 can be written as a2 − b2, where a and b are whole numbers (0 is allowed)?
Show worked solution
Every odd number works (2k + 1 = (k + 1)2 − k2) and every multiple of 4 works (4k = (k + 1)2 − (k − 1)2). A number that is 2 more than a multiple of 4 never works, because (a − b)(a + b) is then odd × odd or even × even. There are 25 of those, so 100 − 25 = 75. (50 counts only the odd numbers; 25 is the count that fail.)
21 A right-angled triangle has sides 3, 4 and 5. A semicircle has its diameter along the side of length 4, with one end at the right angle, and it touches the side of length 5. What is its radius?
Show worked solution
The centre O is r from the side of length 3 (that side is perpendicular to the diameter) and r from the side of length 5 (the semicircle touches it). Splitting the triangle at O: ½ × 3 × r + ½ × 5 × r = 6, so 4r = 6 and r = 1.5. (1 is the radius of the inscribed circle; 1.2 is half the height to the hypotenuse.)
22 What is the remainder when 1! + 2! + 3! + … + 50! is divided by 100? (n! = 1 × 2 × 3 × … × n.)
Show worked solution
10! and every later term contain 2 × 5 × 10, so they end in 00. The last two digits of 1 + 2 + 6 + 24 + 120 + 720 + 5040 + 40 320 + 362 880 are 1 + 2 + 6 + 24 + 20 + 20 + 40 + 20 + 80 = 213 → 13. The remainder is 13. (33 stops at 4!; 53 and 73 stop at 5! and 6!.)
23 Moving only right or up along the streets, how many routes are there from A to B if the middle crossing is closed?
Show worked solution
All routes: choose which 4 of the 8 moves go right, 70. Routes through the middle: 6 ways to reach it × 6 ways on from it = 36. So 70 − 36 = 34. (36 is the routes through the closed crossing; 70 ignores the closure.)
24 How many ordered pairs (x, y) of positive whole numbers satisfy 1/x + 1/y = 1/6?
Show worked solution
6y + 6x = xy, so xy − 6x − 6y + 36 = 36, that is (x − 6)(y − 6) = 36. Both brackets must be positive (if both were negative, x and y would be below 6 and 1/x + 1/y would exceed 1/6). 36 has 9 positive divisors, giving 9 ordered pairs, from (7, 42) to (42, 7). (5 counts unordered pairs; 18 doubles the 9.)
25 The numbers 1, 2, 3, …, 10 are written on a board. A move is: rub out any two numbers a and b and write a + b − 1 in their place. After 9 moves only one number is left. What is it?
Show worked solution
The total starts at 55 and each move lowers it by exactly 1, whichever numbers are chosen. After 9 moves the one number left is 55 − 9 = 46. (45 counts 10 moves; 54 subtracts only once.)
26 How many four-digit numbers (1000 to 9999) have digits that add up to 10?
Show worked solution
First digit 1 + a, so a + b + c + d = 9 with each at least 0. Stars and bars: 9 stars and 3 bars in 12 places, 12 × 11 × 10 ÷ 6 = 220 ways. Every digit is then at most 9 except one case: a = 9, which makes the first digit 10. Remove it: 220 − 1 = 219.
27 What is the smallest positive whole number that has exactly 18 positive divisors?
Show worked solution
18 = 18, 9 × 2, 6 × 3 or 3 × 3 × 2. The smallest number for each: 217; 28 × 3 = 768; 25 × 32 = 288; 22 × 32 × 5 = 180. The smallest is 180.
28 ABCD is a square of side 40 cm. P is a point inside it with PA = PB, and PA equal to the distance from P to the side CD. Find PA in centimetres.
Show worked solution
PA = PB puts P 20 cm from A along AB and y cm above it, so PA2 = 202 + y2. The distance to CD is 40 − y. Then 400 + y2 = (40 − y)2 = 1600 − 80y + y2, so 80y = 1200 and y = 15. PA = 40 − 15 = 25 cm (check: 202 + 152 = 625 = 252).
29 For every whole number n ≥ 1, the sum of the first n terms of a sequence is 3n2 + 2n. What is the 80th term?
Show worked solution
Term n = S(n) − S(n − 1) = 3n2 + 2n − 3(n − 1)2 − 2(n − 1) = 6n − 1. (Check: S(1) = 5 = 6 × 1 − 1.) So term 80 = 480 − 1 = 479.
30 Some numbers are chosen from 1, 2, 3, …, 30 so that no chosen number is double another chosen number. What is the largest number of numbers that can be chosen?
Show worked solution
Every number is an odd number times a power of 2, so 1–30 splits into chains: 1-2-4-8-16 (best 3), 3-6-12-24 (2), 5-10-20 (2), 7-14-28 (2), 9-18, 11-22, 13-26, 15-30 (1 each), and 17, 19, 21, 23, 25, 27, 29 alone (7). In a chain of length k you can take at most every other number, so at most ⌈k/2⌉. Total 3 + 2 + 2 + 2 + 4 + 7 = 20, reached by taking, for example, 1, 4, 16, 3, 12, 5, 20, 7, 28, 9, 11, 13, 15 and 17–29 odd.