Isabelle plays with a hexagonal sheet of paper. With each move she rotates the hexagon by the same angle in the same direction. The illustration shows the sheet at the start and after the first move. After how many moves does the sheet look the same as it did at the beginning?
Show answer
Answer: A — 6
Show hints
Hint 1 of 2
The single dotted wedge acts as a marker — track where it lands after one move.
Still stuck? Show hint 2 →
Hint 2 of 2
Find the rotation angle of one move, then see how many moves complete a full turn back to the start.
Show solution
Approach: track the marker wedge under repeated equal rotations
The colouring has no rotational symmetry, so the sheet only looks identical after a whole turn brings every wedge home.
Comparing the start and the first move, the unique dotted wedge has shifted by one position — a 60° rotation.
A 60° step needs \(360^\circ \div 60^\circ = 6\) moves to return to the original picture, which is (A).
In Kangaroo Town the unit of distance is the ‘hop’ instead of the kilometre. Kangaroo Klaus needs 12 minutes to cover one hop. How many hops can he cover in 12 hours?
Show answer
Answer: A — 60
Show hints
Hint 1 of 2
First find how many minutes are in 12 hours.
Still stuck? Show hint 2 →
Hint 2 of 2
Each hop costs 12 minutes, so divide the total minutes by 12.
Show solution
Approach: convert to a common time unit then divide
12 hours = 12 × 60 = 720 minutes.
Each hop takes 12 minutes, so he covers \(720 \div 12 = 60\) hops, which is (A).
The picture shows the menu of a burger restaurant. The rain has washed away some of the numbers. The burgers are listed by price in increasing order, the cheapest being the “veggie” burger. What is the smallest possible price of the “deluxe” burger?
Show answer
Answer: B — 6.80
Show hints
Hint 1 of 2
Only the cents are readable; the whole-euro parts are hidden, but the prices increase down the list.
Still stuck? Show hint 2 →
Hint 2 of 2
Step down the menu choosing the smallest whole-euro part that keeps each price above the one before it.
Show solution
Approach: build the cheapest increasing price chain
Five circles, each with an area of 8 cm², overlap to form the figure shown. Each overlapping region has an area of 1 cm². What is the total area of the figure, in cm²?
Show answer
Answer: B — 36
Show hints
Hint 1 of 2
Add the five circle areas, then fix the double counting where they overlap.
Still stuck? Show hint 2 →
Hint 2 of 2
Each of the four overlaps was counted twice, so subtract it once.
Show solution
Approach: inclusion–exclusion on overlapping areas
Five circles give 5 × 8 = 40 cm² if simply added.
The four overlap regions (1 cm² each) were each counted twice, so subtract 4 cm².
Total figure area = \(40 - 4 = 36\) cm², which is (B).
When Paul sets the number 0000 on his bicycle lock, he sees 8888 at the point marked with an x. To open the lock he must turn the rings so that 2815 shows at the x mark. What number is then next to the arrows?
Show answer
Answer: A — 4037
Show hints
Hint 1 of 2
Compare the two visible windows: at the arrows you see 0, two rows up at the x you see 8.
Still stuck? Show hint 2 →
Hint 2 of 2
Each ring is offset by a fixed amount; find the digit at the arrows for each target digit at the x.
Show solution
Approach: use the fixed offset between the x-window and the arrow-window
When the arrows show 0 the x shows 8, so on every ring the arrow digit is 2 more than the x digit (the rows run …,8,9,0,… downward).
Setting 2,8,1,5 at the x makes the arrows read 2+2, 8+2, 1+2, 5+2 (mod 10) = 4, 0, 3, 7.
A mouse wants to reach a piece of cheese. From each square it can only move to the square to its right or to the square directly below it. How many different paths lead the mouse to a piece of cheese?
Show answer
Answer: C — 8
Show hints
Hint 1 of 2
Instead of drawing every route, write a number in each square: how many ways the mouse can reach that square.
Still stuck? Show hint 2 →
Hint 2 of 2
A square's number is the sum of the numbers in the square just left of it and the square just above it, since those are the only ways in.
Show solution
Approach: fill in path counts square by square
Write 1 in the start square; every other square gets the total of the square to its left and the square above it (the only squares that can flow into it).
Filling the staircase this way, the counts grow 1, then 2 and 3, and so on down toward the cheese.
The cheese squares collect a total of 8 paths, which is (C).
In a 60 m hurdles race there are 5 hurdles. The first hurdle is 12 m after the start, and the distance between any two consecutive hurdles is 8 m. How far is the last hurdle from the finish line?
Show answer
Answer: E — 16 m
Show hints
Hint 1 of 2
Find the position of the last (5th) hurdle measured from the start.
Still stuck? Show hint 2 →
Hint 2 of 2
Then subtract that from the 60 m finish line.
Show solution
Approach: locate the last hurdle, then measure to the finish
Hurdle 1 is at 12 m and each later hurdle is 8 m further: 12, 20, 28, 36, 44.
The 5th hurdle is at 12 + 4×8 = 44 m.
Distance to the finish = \(60 - 44 = 16\) m, which is (E).
Emma wants to write a number in each circle (see diagram) so that each number equals the sum of the numbers in the two adjacent circles. She has already written two numbers. Which number will she write in the grey circle?
Show answer
Answer: D — −3
Show hints
Hint 1 of 2
Every circle equals the sum of the two circles touching it, and the 1 and 2 are not next to each other.
Still stuck? Show hint 2 →
Hint 2 of 2
Start with the empty circle between the 1 and the 2 — it must equal \(1+2\) — then keep applying the rule around the ring.
Show solution
Approach: propagate the neighbour-sum rule around the ring
The top circle sits between the 1 and the 2, so it equals \(1+2=3\).
Now use 'each circle = sum of its two neighbours' going around: the right circle is \(2-3=-1\), the left circle is \(1-3=-2\), and the grey bottom circle (between them) is \(-1+(-2)=-3\).
Sanja has two bowls of numbered balls. The left bowl holds seven balls numbered 1, 2, 6, 7, 10, 11 and 12, with arithmetic mean 7.0. The right bowl holds five balls numbered 3, 4, 5, 8 and 9, with arithmetic mean 5.8. Sanja wants to increase the arithmetic mean of both bowls. Which ball must she move from the left bowl to the right bowl to do this?
Show answer
Answer: A — 6
Show hints
Hint 1 of 2
Moving a ball must push BOTH averages up — think about what value keeps each average rising.
Still stuck? Show hint 2 →
Hint 2 of 2
The moved ball must be below the left average (so the left mean rises) yet above the right average (so the right mean rises).
Show solution
Approach: bound the ball value from both averages
Left sum is 7×7 = 49, right sum is 5.8×5 = 29.
To raise the left mean the removed ball must be below 7; to raise the right mean it must be above 5.8.
The only left-bowl ball between 5.8 and 7 is 6, which is (A).
Theo is on a treadmill and sees two stopwatches. The left one shows the time elapsed since he started his workout; the right one shows the time remaining until the end. At the moment shown the left reads 14:58 and the right reads 21:32. At some point both stopwatches will display the same time. What will they display then?
Show answer
Answer: D — 18:15
Show hints
Hint 1 of 2
One clock counts up and the other counts down at the same speed, so their sum stays constant.
Still stuck? Show hint 2 →
Hint 2 of 2
They show the same time exactly halfway through that constant total.
Show solution
Approach: the two readings always sum to the full workout time
Elapsed + remaining is constant: 14:58 + 21:32 = 36:30 (the whole workout).
They are equal at half of that total: \(36{:}30 \div 2 = 18{:}15\), which is (D).
In a room there are 10 more people who always tell the truth than there are people who always lie. Everyone in the room was asked, “Are you telling the truth?” and all 20 people answered yes. How many liars are in the room?
Show answer
Answer: B — 5
Show hints
Hint 1 of 2
Both truth-tellers and liars answer 'yes' to 'Are you telling the truth?' — so the answers tell you nothing new.
Still stuck? Show hint 2 →
Hint 2 of 2
Just use that truth-tellers outnumber liars by 10 and there are 20 people.
Show solution
Approach: solve the count from the two given totals
A truth-teller says 'yes' truthfully; a liar also says 'yes' (lying), so all 20 say yes regardless.
If liars = L, truth-tellers = L + 10, and (L+10) + L = 20.
So \(2L = 10\), giving \(L = 5\) liars, which is (B).
A bee, a mouse, a beetle and a cat want to take a group photo, so they line up next to each other. The cat is not allowed to stand next to the mouse. In how many different ways can the animals line up?
Show answer
Answer: B — 12
Show hints
Hint 1 of 2
Count all line-ups, then remove the bad ones where the cat and mouse are side by side.
Still stuck? Show hint 2 →
Hint 2 of 2
Treat the cat-and-mouse pair as a single block to count the forbidden arrangements.
Show solution
Approach: total minus the cat-next-to-mouse cases
Four animals can line up in 4! = 24 ways.
Glue cat+mouse together: 3 items in 3! = 6 orders, times 2 for the internal order = 12 forbidden line-ups.
Allowed line-ups = \(24 - 12 = 12\), which is (B).
The two bookworms Linki and Rechti eat their way through a row of books. Linki starts from the left and Rechti from the right, both at the same time. Linki eats through a book cover in 3 days and through all the pages of a book in 2 days. Rechti eats through a book cover in 1 day and through all the pages of a book in 2 days. In which book (see illustration) do the two meet?
Show answer
Answer: B — B
Show hints
Hint 1 of 2
Give each worm a 'days to chew through' cost for a cover and for a set of pages, then send them toward each other.
Still stuck? Show hint 2 →
Hint 2 of 2
Add up the days each worm spends, layer by layer, until their totals show them reaching the same book.
Show solution
Approach: accumulate the days each worm needs until they meet
Linki (from the left) needs 3 days per cover and 2 days per book of pages; Rechti (from the right) needs 1 day per cover and 2 days per book of pages.
After 13 days Linki has chewed through book A entirely and reached the pages of book B \((3+2+3+3+2)\); in those same 13 days Rechti has chewed through E, D and C and into book B \((1+2+1+1+2+1+1+2+1+1)\).
Emus, snakes and kangaroos live together on an Australian farm. Emus have two legs and no tail. Kangaroos have four legs and a tail. Snakes have no legs but a tail. Every animal has two eyes. Altogether they have 18 eyes, 7 tails and 24 legs. How many kangaroos live on the farm?
Show answer
Answer: E — 5
Show hints
Hint 1 of 2
Every animal has 2 eyes, so the eye count gives the total number of animals.
Still stuck? Show hint 2 →
Hint 2 of 2
Use tails to split snakes+kangaroos from emus, then legs to isolate the kangaroos.
Show solution
Approach: set up and solve from eyes, tails and legs
18 eyes mean 9 animals; 7 tails are carried by kangaroos and snakes, so 2 emus.
Legs: emus give 2×2 = 4, so kangaroos give 24 − 4 = 20 legs.
Each kangaroo has 4 legs, so there are \(20 \div 4 = 5\) kangaroos, which is (E).
Elke draws quarter circles on a sheet of paper measuring 12 cm × 9 cm, with the centres at the four corners. She shades the resulting region in the middle of the figure (not drawn to scale). How long is the distance marked with the question mark?
Show answer
Answer: B — 6 cm
Show hints
Hint 1 of 2
Each quarter circle reaches exactly its radius along the side it starts from, so name the radii and mark where each arc meets the bottom edge.
Still stuck? Show hint 2 →
Hint 2 of 2
Where two arcs touch, their radii must fit together along that side — use that to pin the radii, then read the marked length off the 12 cm bottom.
Show solution
Approach: use the corner-circle radii along the sides
Each arc reaches its own radius along the edges from its corner; where two arcs meet on the 9 cm right side, those two radii add up to 9.
Those same radii mark off points along the 12 cm bottom edge, and the question-mark length is what is left between the bottom-left corner and the bottom-right arc.
Working the radii through the 12 cm width leaves the marked distance equal to 6 cm, which is (B).
In the six-digit number PAPAYA, different letters stand for different digits and equal letters stand for equal digits. It is also given that Y = P + P = A + A + A. What is the value of P × A × P × A × Y × A?
Show answer
Answer: A — 432
Show hints
Hint 1 of 2
The clues say Y = 2P and Y = 3A — so 2P = 3A with single digits.
Still stuck? Show hint 2 →
Hint 2 of 2
Find digits where twice P equals three times A, then read off Y.
Show solution
Approach: solve the digit equations, then multiply
Y = P + P = 2P and Y = A + A + A = 3A, so 2P = 3A.
The simplest distinct digits are A = 2, P = 3, giving Y = 6.
Then \(P\times A\times P\times A\times Y\times A = 3\cdot2\cdot3\cdot2\cdot6\cdot2 = 432\), which is (A).
Manuela takes a total of 17 shots at goal over two soccer training sessions. In the first session she scores with 60% of her shots, and in the second she scores with 75% of her shots. How many goals does she score in the second session?
Show answer
Answer: D — 9
Show hints
Hint 1 of 2
Both hit-counts must be whole numbers: 60% of the first session and 75% of the second.
Still stuck? Show hint 2 →
Hint 2 of 2
Split 17 shots so that 60% of the first part and 75% of the second part are both integers.
Show solution
Approach: force whole-number hits to fix the split
60% needs the first count to be a multiple of 5; 75% needs the second to be a multiple of 4.
With shots adding to 17, the split is 5 and 12 (5 is a multiple of 5, 12 of 4).
Second-session hits = 75% of 12 = 9, which is (D).
Louise places three rectangular pictures as shown in the figure. What is the size of the angle α?
Show answer
Answer: B — 70°
Show hints
Hint 1 of 2
Each picture is a rectangle, so every corner of it is a right angle (90°) — that is the key fact to lean on.
Still stuck? Show hint 2 →
Hint 2 of 2
Use a right-angle corner together with the marked 62° to find a small leftover angle, then combine it with the 42°.
Show solution
Approach: angle-chase using the rectangles' right angles
Because each picture is a rectangle, the corner sitting at the 62° mark is a right angle, so beyond the 62° there is \(90^\circ-62^\circ=28^\circ\) left over.
That 28° lines up next to the 42° gap, so \(\alpha=42^\circ+28^\circ\).
Anurag lives 1 km from his school and sets off at the same time every day. Walking, he travels at 4 km/h; cycling, he travels at 15 km/h. When he walks, he arrives 5 minutes before school starts. How many minutes before school starts does he arrive if he cycles?
Show answer
Answer: E — 16
Show hints
Hint 1 of 2
Find how long the 1 km walk takes and how long the 1 km ride takes.
Still stuck? Show hint 2 →
Hint 2 of 2
Cycling saves the time difference, so add that saving to the 5 minutes he already had to spare.
Show solution
Approach: compare travel times and the fixed start moment
Walking 1 km at 4 km/h takes 15 minutes; cycling 1 km at 15 km/h takes 4 minutes.
He leaves at the same moment, so cycling gets him there 15 − 4 = 11 minutes earlier than walking.
Walking he is 5 minutes early, so cycling he is \(5 + 11 = 16\) minutes early, which is (E).
In rectangle ABCD, the points E and F lie on side DC (see diagram) so that ∠EBA = ∠DFA = 45° and AB + EF = 20 cm. How long is side BC?
Show answer
Answer: D — 10 cm
Show hints
Hint 1 of 2
A 45° line rises exactly as much as it runs, so each slanted line shifts sideways by the rectangle's height as it climbs from the bottom to the top.
Still stuck? Show hint 2 →
Hint 2 of 2
Write where E and F land on the top edge in terms of the width AB and the height BC, then plug into AB + EF = 20.
Show solution
Approach: use the 45° lines to locate E and F on the top side
The 45° line from B reaches the top after moving in by the height \(BC\), and likewise the 45° line from A; so each of E and F sits a distance \(BC\) horizontally inside an end of the top edge.
Measuring along the top edge then gives \(EF=2\,BC-AB\), so \(AB+EF=2\,BC\); the two slants effectively double the height.
So \(2\,BC=20\), giving \(BC=\) 10 cm, which is (D).
If the height of a cuboid is reduced by 3 cm, its surface area decreases by 60 cm² and the result is a cube. What is the volume of the original cuboid, in cm³?
Show answer
Answer: D — 200
Show hints
Hint 1 of 2
When you trim 3 cm off the height you remove a band around the side — that lost area is the side perimeter times 3.
Still stuck? Show hint 2 →
Hint 2 of 2
After trimming it's a cube, so the base is a square; find its side from the lost surface area.
Show solution
Approach: relate the surface-area loss to the base perimeter
Cutting the height by 3 cm removes a strip of lateral surface = (base perimeter)×3 = 60, so the base perimeter is 20 and each square side is 5.
The result is a cube of side 5, so the original height was 5 + 3 = 8.
Original volume = \(5\times5\times8 = 200\) cm³, which is (D).
In quadrilateral ABCD, the points N and K are marked on sides BC and AD so that BN = 2·NC and AK = KD. The areas of triangles ABN and CKD are shown in the figure. What is the area of quadrilateral ABCD?
Show answer
Answer: A — 13
Show hints
Hint 1 of 2
A point that splits a side in a ratio splits a triangle's area in the same ratio (same height).
Still stuck? Show hint 2 →
Hint 2 of 2
Stretch triangle ABN up to ABC using BN:NC, and double CKD up to ACD using the midpoint K.
Show solution
Approach: scale each known triangle to a piece of the quadrilateral
BN = 2·NC means BN:BC = 2:3, so triangle ABC has area 6·(3/2) = 9.
K is the midpoint of AD, so triangle ACD = 2·(area CKD) = 2·2 = 4.
Splitting ABCD by diagonal AC: area = \(9 + 4 = 13\), which is (A).
Martin wants to fill the cells in the diagram so that each cell contains either a cross or a circle, with no row, column or diagonal containing four consecutive identical symbols. What will the grey column contain in the completed diagram?
Show answer
Answer: B — 2 circles and 4 crosses
Show hints
Hint 1 of 2
No row, column or diagonal may hold four of the same symbol in a row — that rule forces almost every empty cell once you start from the ones already filled.
Still stuck? Show hint 2 →
Hint 2 of 2
Work cell by cell from the given symbols; whenever three matching symbols line up, the fourth must be the opposite one.
Show solution
Approach: propagate the no-four-in-a-row constraint
Start from the already-filled cells and apply the rule: whenever a line would otherwise get four equal symbols in a row, the next cell is forced to be the other symbol.
Chasing these forced choices through the grid fills the grey column uniquely.
The grey column ends up with 2 circles and 4 crosses, which is (B).
The letters p, q, r, s and t stand for five consecutive positive integers, but not necessarily in that order. It is given that p + q = 69 and s + t = 72. Which number does r stand for?
Show answer
Answer: C — 34
Show hints
Hint 1 of 2
Five consecutive integers add up to 5 times the middle one.
Still stuck? Show hint 2 →
Hint 2 of 2
Add p+q and s+t, then compare with the total of all five to isolate r.
Show solution
Approach: use the total of the five consecutive integers
Adding the two given sums, \((p+q)+(s+t)=69+72=141\), accounts for every letter except \(r\).
The five consecutive integers must be 33, 34, 35, 36, 37 (their two largest add to 72 and two others add to 69), and together they total 175.
Annie, Bibi, Clara and Doris each live on a different floor of a four-storey building, and other people live there too. 25 people live above Annie, 5 people live below Bibi, 17 people live below Clara, and 22 people live above Doris. How many people live in the building in total?
Show answer
Answer: A — 27
Show hints
Hint 1 of 2
For any resident, (people above) + (themselves) + (people below) is the whole building, so each clue tells you both an 'above' and a 'below' count once you call the total \(N\).
Still stuck? Show hint 2 →
Hint 2 of 2
Annie has the most people above her, so she is the lowest of the four friends; rank the four and make their above/below counts line up.
Show solution
Approach: rank the four friends and reconcile the counts
Annie has 25 people above her — the most of the four — so she is the lowest friend, and the most people above means the fewest below; in fact only Clara (17 below) and the others can sit above her.
Ranking everyone in one line, Annie is 26th from the top, Doris 23rd, Bibi has 5 below and Clara 17 below; fitting these together with one person below Annie forces the total.
The counts only agree when the building holds 27 people, which is (A).
