When the 5 pieces shown are fitted together correctly, the result is a rectangle with a calculation written on it. What is the answer to this calculation?
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Answer: A — −100
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Hint 1 of 2
Fit the jigsaw pieces into a rectangle so the symbols line up into a single calculation.
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Hint 2 of 2
Once assembled it reads a short arithmetic expression — just evaluate it.
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Approach: assemble the pieces into the expression and compute
The five pieces fit together to spell out a calculation using the digits 2, 0, 2, 1 and a minus sign.
A student correctly added the two two-digit numbers on the left of the board (AB + CD) and got the answer 137. What answer will he get if he adds the two four-digit numbers on the right of the board (ADCB + CBAD)?
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Answer: B — 13 837
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Hint 1 of 2
Write the right-hand sum ADCB + CBAD in terms of the totals A+C and B+D.
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Hint 2 of 2
The left-hand fact AB + CD = 137 already pins down 10(A+C) + (B+D).
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Approach: express the big sum using A+C and B+D
From AB + CD = 137 we get 10(A+C) + (B+D) = 137, so A+C = 13 and B+D = 7.
Byron is 5 cm taller than Aaron, but 10 cm shorter than Caron. Darren is 10 cm taller than Caron, but 5 cm shorter than Erin. Which of the following statements is true?
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Answer: E — Aaron is 30 cm shorter than Erin
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Hint 1 of 2
Anchor everyone to Aaron's height and step through the comparisons one at a time.
A rectangular chocolate bar is made of equal squares. Neil breaks off two complete strips of squares and eats the 12 squares he obtains. Later, Jack breaks off one complete strip of squares from the same bar and eats the 9 squares he obtains. How many squares of chocolate are left in the bar?
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Answer: D — 45
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Hint 1 of 2
Neil's two equal strips total 12, so a strip in that direction holds 6 — that fixes one side of the bar.
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Hint 2 of 2
Jack's strip runs the other way; remember Neil already removed two rows before Jack broke his strip.
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Approach: recover the bar's dimensions from the strip sizes
Neil's two equal strips give 12 squares, so each strip holds 6: one side of the bar is 6.
Jack's strip runs the other way and holds 9, but Neil had already removed 2 squares from that direction, so the full bar was 6 by (9+2) = 11, i.e. 66 squares.
The area of the large square is 16 cm² and the area of each small (corner) square is 1 cm². What is the total area of the central flower, in cm²?
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Answer: C — 4
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Hint 1 of 2
The big square is 4×4 since its area is 16, and it is split into a 4×4 grid of unit squares.
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Hint 2 of 2
Find the flower's area by combining its petal pieces using the symmetry of the figure.
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Approach: decompose the unit-grid square
Area 16 means the big square is 4 by 4, so the unit corner squares have side 1 and the centre is the middle of the grid.
The flower has 8 yellow petals; the 4 'straight' petals each reach from the centre to a side, and the 4 'diagonal' petals reach toward the corner unit squares.
By symmetry the 8 petals together tile half of the inner 2×2 region around the centre plus matching slivers, and the pieces total 4 cm².
Costa is building a new fence in his garden. He uses 25 planks of wood, each of which are 30 cm long. He arranges these planks so that there is the same slight overlap between any two adjacent planks. The total length of Costa's new fence is 6.9 metres. What is the length, in centimetres, of the overlap between any pair of adjacent planks?
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Answer: B — 2.5
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Hint 1 of 2
With 25 planks there are 24 overlaps, all equal.
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Hint 2 of 2
Compare the total plank length to the actual fence length to find the overlap total.
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Approach: account for total length lost to overlaps
Laid end to end the 25 planks would span 25·30 = 750 cm.
The fence is only 690 cm, so 750 − 690 = 60 cm is lost to overlapping.
There are 24 equal overlaps, so each is 60 ÷ 24 = 2.5 cm.
Five identical right-angled triangles can be arranged so that their larger acute angles touch to form the star shown in the diagram. It is also possible to form a different star by arranging more of these triangles so that their smaller acute angles touch. How many triangles are needed to form the second star?
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Answer: D — 20
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Hint 1 of 2
Five large acute angles meeting at a point fill 360°, so each large acute angle is 72°.
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Hint 2 of 2
The small acute angle is 90° minus the large one; see how many fill a full turn.
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Approach: use the angles meeting at the star's centre
For the first star, 5 equal larger acute angles surround the centre: 360° ÷ 5 = 72°.
The triangle is right-angled, so the smaller acute angle is 90° − 72° = 18°.
For the second star, smaller angles meet at the centre: 360° ÷ 18° = 20 triangles.
Five squares are positioned as shown. The small square indicated has area 1. What is the value of h?
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Answer: C — 4 m
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Hint 1 of 2
The marked small square has area 1, so its side is 1; use it as the unit of length.
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Hint 2 of 2
Work along the staircase of squares to express h in those units.
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Approach: measure with the unit square
The marked small square has area 1, so its side is 1; use that as the unit of length along the figure.
Each larger square's side is set by stacking on the one beside it, so the side lengths grow by fixed steps measured in those units.
The arrow h reaches across the top from the small square's structure to the far edge of the big right-hand square, and summing those side-steps gives h = 4 m.
There are 20 questions in a quiz. Each correct answer scores 7 points, each wrong answer scores −4 points, and each question left blank scores 0 points. Eric took the quiz and scored 100 points. How many questions did he leave blank?
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Answer: B — 1
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Hint 1 of 2
Let the numbers of correct, wrong and blank answers add to 20, with score 7c − 4w = 100.
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Hint 2 of 2
Search for whole-number solutions and read off the blanks.
Show solution
Approach: solve the score equation in whole numbers
With c correct and w wrong: 7c − 4w = 100 and c + w ≤ 20.
c = 16, w = 3 gives 112 − 12 = 100, and that is the only fit.
A rectangular strip of paper of dimensions 4 cm × 13 cm is folded as shown in the diagram. 2 rectangles are formed with areas P and Q where P = 2Q. What is the value of x?
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Answer: C — 6 cm
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Hint 1 of 2
The 45° fold makes the overlap a right-isosceles triangle, linking the two rectangle sizes.
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Hint 2 of 2
Use P = 2Q together with the strip's fixed width and length to pin down x.
Show solution
Approach: relate the two rectangle areas through the fold
The strip has width 4, so both rectangles are 4 wide: P = 4x and Q = 4y, where x and y are their lengths.
The 45° fold turns the strip square across its width, so the diagonal overlap uses a 4-by-4 square, and the three lengths fit the strip: x + 4 + y = 13.
P = 2Q gives x = 2y; with x + y = 9 this makes y = 3 and x = 6.
Logic & Word ProblemsRatios, Rates & Proportionsratiocasework
A box of fruit contains twice as many apples as pears. Christy and Lily divided them up so that Christy had twice as many pieces of fruit as Lily. Which one of the following statements is always true?
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Answer: E — Christy took as many pears as Lily got apples.
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Hint 1 of 2
Let there be p pears and 2p apples, total 3p; Christy ends with 2p pieces and Lily with p.
Still stuck? Show hint 2 →
Hint 2 of 2
Test each statement against every possible split — only one holds in all cases.
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Approach: check each claim over all valid splits
Pears = p, apples = 2p, total 3p; Christy has 2p pieces, Lily has p.
If Christy takes a apples she takes 2p−a pears; Lily then gets the remaining 2p−a apples.
So Christy's pears always equal Lily's apples; the other options can fail.
Logic & Word ProblemsAlgebra & Patternssubstitution
Three villages are connected by paths as shown. From Downend to Uphill, the detour via Middleton is 1 km longer than the direct path. From Downend to Middleton, the detour via Uphill is 5 km longer than the direct path. From Uphill to Middleton, the detour via Downend is 7 km longer than the direct path. How long is the shortest of the three direct paths between the villages?
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Answer: C — 3 km
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Hint 1 of 2
Name the three direct distances and turn each 'detour is k longer' fact into an equation.
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Hint 2 of 2
Add all three equations to get the total of the distances quickly.
Show solution
Approach: set up and add the detour equations
Let the direct paths be DU = a, DM = b, UM = c. The detours give b+c = a+1, a+c = b+5, a+b = c+7.
Adding all three: 2(a+b+c) = (a+b+c) + 13, so a+b+c = 13.
Then a = 6, b = 4, c = 3; the shortest direct path is 3 km.
In a particular fraction the numerator and denominator are both positive. The numerator of this fraction is increased by 40%. By what percentage should its denominator be decreased so that the new fraction is double the original fraction?
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Answer: C — 30%
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Hint 1 of 2
Increasing the top by 40% multiplies the fraction by 1.4; you want the new fraction to be twice the old.
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Hint 2 of 2
Find the multiplier the denominator needs, then convert it to a percent decrease.
Show solution
Approach: balance the multipliers on top and bottom
Multiplying the numerator by 1.4 while dividing the denominator by (1 − p) should double the fraction: 1.4 / (1 − p) = 2.
Logic & Word ProblemsSpatial & Visual Reasoningcareful-countingcasework
A triangular pyramid is built with 20 cannon balls, as shown. Each cannon ball is labelled with one of A, B, C, D or E. There are 4 cannon balls with each type of label. The picture shows the labels on the cannon balls on 3 of the faces of the pyramid. What is the label on the hidden cannon ball in the middle of the fourth face?
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Answer: D — D
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Hint 1 of 2
Each label A–E is used exactly four times across the 20 balls.
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Hint 2 of 2
Tally how many of each label already appear on the three shown faces (counting shared edge balls once); the centre ball must be the label still short of four.
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Approach: count each label and find the one not yet at full quota
There are 4 balls of each label. Tally the labels visible on the three shown faces, counting shared edge/corner balls once.
One label falls one short of its quota of 4; that missing ball is the hidden centre of the fourth face.
Logic & Word ProblemsCounting & Probabilitysum-constraintcomplementary-counting
A box contains only green, red, blue and yellow counters. There is always at least one green counter amongst any 27 counters chosen from the box; always at least one red counter amongst any 25 counters chosen; always at least one blue amongst any 22 counters chosen and always at least one yellow amongst any 17 counters chosen. What is the largest number of counters that could be in the box?
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Answer: B — 29
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Hint 1 of 2
'Any 27 chosen contain a green' means you can never pick 27 with no green — so the non-green counters number at most 26.
Still stuck? Show hint 2 →
Hint 2 of 2
Write the same kind of bound for each colour and add them up.
Number TheoryLogic & Word Problemsdivisibilitycasework
2021 coloured kangaroos are arranged in a row and are numbered from 1 to 2021. Each kangaroo is coloured either red, grey or blue. Amongst any three consecutive kangaroos, there are always kangaroos of all three colours. Bruce guesses the colours of five kangaroos. These are his guesses: Kangaroo 2 is grey; Kangaroo 20 is blue; Kangaroo 202 is red; Kangaroo 1002 is blue; Kangaroo 2021 is grey. Only one of his guesses is wrong. What is the number of the kangaroo whose colour he guessed incorrectly?
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Answer: B — 20
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Hint 1 of 2
'Any three in a row use all three colours' forces the colouring to repeat with period 3.
Still stuck? Show hint 2 →
Hint 2 of 2
So a kangaroo's colour depends only on its position modulo 3; compare the guesses at equal residues.
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Approach: use the period-3 structure
With every three consecutive kangaroos all different, the colour pattern repeats every 3 positions.
Positions 2, 20 and 2021 are all 2 (mod 3), so they must share one colour.
Guesses say k2 = grey, k20 = blue, k2021 = grey; the lone disagreement (k20) must be the wrong one.
A 3×4×5 cuboid consists of 60 identical small cubes. A termite eats its way along the diagonal from P to Q. This diagonal does not intersect the edges of any small cube inside the cuboid. How many of the small cubes does it pass through on its journey?
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Answer: C — 10
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Hint 1 of 2
The diagonal of an a×b×c grid crosses a + b + c interior 'sheets', but crossings at shared edges count once.
Still stuck? Show hint 2 →
Hint 2 of 2
Use a + b + c minus the pairwise gcds plus the triple gcd.
Show solution
Approach: count unit cubes a space-diagonal passes through
For a 3×4×5 box the count is a+b+c − gcd(a,b) − gcd(b,c) − gcd(a,c) + gcd(a,b,c).
Logic & Word ProblemsCounting & Probabilitycaseworkcareful-counting
In a town there are 21 knights who always tell the truth and 2000 knaves who always lie. A wizard divided 2020 of these 2021 people into 1010 pairs. Every person in a pair described the other person as either a knight or a knave. As a result, 2000 people were called knights and 20 people were called knaves. How many pairs of two knaves were there?
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Answer: D — 995
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Hint 1 of 3
Work out what each type of pair (two knights, two knaves, mixed) makes the partners say.
Still stuck? Show hint 2 →
Hint 2 of 3
Only mixed pairs produce 'knave' answers, two each — that pins down the number of mixed pairs.
Still stuck? Show hint 3 →
Hint 3 of 3
Then use the 21 knights to back out the other pair types.
Show solution
Approach: classify pairs by the labels they generate
In a same-type pair both say 'knight'; in a mixed pair both say 'knave'.
The 20 'knave' calls come 2 per mixed pair, so there are 10 mixed pairs (using 10 knights and 10 knaves).
With one knight left out, the other 10 knights form 5 knight-knight pairs; the remaining 1990 knaves form 995 knave-knave pairs.
Logic & Word ProblemsCounting & Probabilitycasework
In a tournament each of the 6 teams plays one match against every other team. In each round of matches, 3 take place simultaneously. A TV station has already decided which match it will broadcast for each round, as shown in the table. In which round will team D play against team F?
Round
1
2
3
4
5
Broadcast match
A–B
C–D
A–E
E–F
A–C
Show answer
Answer: A — 1
Show hints
Hint 1 of 2
The five rounds form a schedule where each round is three disjoint matches covering all six teams.
Still stuck? Show hint 2 →
Hint 2 of 2
Each team plays once per round, so a team's partners across the rounds are all different; reconstruct from the broadcasts.
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Approach: reconstruct the round-by-round pairing
A's broadcast matches are B (R1), E (R3) and C (R5), so A must meet D and F in rounds 2 and 4; since R2 already shows C–D, A plays F in R2 and D in R4.
Round 4 shows A–D and E–F, so its third match is B–C; round 2's leftover pair is B–E.
Now D still needs B, E, F and E still needs C, D: E–C must fall in round 1 (E is busy in the others), forcing E–D into round 5.
That leaves round 1 as A–B, C–E and the last pair D–F, so D plays F in round 1 — answer A.
The diagram shows a quadrilateral divided into 4 smaller quadrilaterals with a common vertex K. The other labelled points divide the sides of the large quadrilateral into three equal parts. The numbers indicate the areas of the corresponding small quadrilaterals. What is the area of the shaded quadrilateral?
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Answer: C — 6
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Hint 1 of 2
Because the side-points trisect the big quadrilateral's sides, triangles sharing the vertex K have areas in fixed ratios.
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Hint 2 of 2
Set up the proportional relations among the four corner quadrilaterals and solve for the shaded one.
Show solution
Approach: use the equal-trisection area ratios
The labelled points trisect each side of the big quadrilateral, so triangles that share the vertex K and sit on equal side-thirds have equal areas.
Splitting each of the four small quadrilaterals through K into two triangles and matching the equal-base pairs ties the three known areas 8, 10 and 18 to the shaded one.
Solving those balance relations gives the shaded quadrilateral an area of 6.
