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Problem 1 · 2019 Math Kangaroo
Easy
Spatial & Visual Reasoningspatial-reasoning
The higher someone stands on the podium, the better the ranking. Which number got third place?
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Answer: E — 5
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Hint 1 of 2
Higher on the podium means a better rank, so first place is at the very top.
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Hint 2 of 2
Find which step is the third-highest and read off the number standing there.
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Approach: read the podium by height
The taller the step, the better the place: the highest step is 1st.
Order the steps from highest to lowest and take the third one down.
The child standing on that third-highest step wears number 5.
The base of a triangle is extended by 50% and its height is reduced by one third. What is the ratio of the area of the new triangle to the area of the original triangle?
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Answer: B — 1:1
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Hint 1 of 2
A triangle's area is proportional to base times height.
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Hint 2 of 2
Multiply the two change factors together and see what happens to the product.
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Approach: multiply the two scale factors
Extending the base by 50% makes it \(\tfrac{3}{2}\) of the old base; reducing the height by one third makes it \(\tfrac{2}{3}\) of the old height.
Area scales by \(\tfrac{3}{2}\times\tfrac{2}{3}=1\), so the area is unchanged.
Sam paints the 9 small squares in the shape either white, grey or black. What is the minimum number he must paint over so that no two squares sharing a side have the same colour?
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Answer: A — 2
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Hint 1 of 2
Find every pair of side-touching squares that currently share a colour; those are the trouble spots.
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Hint 2 of 2
Try to fix several clashes at once by changing a single well-chosen square, and count the fewest squares you must repaint.
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Approach: spot the colour clashes and repaint the fewest squares to break them all
Scan the grid for neighbours that share a colour: the two grey squares touch, and the black squares touch.
Changing two carefully chosen squares is enough to separate every same-colour pair, while one change still leaves a clash somewhere.
Jan cannot draw very accurately, but he tried to make a roadmap of his village. The relative positions of the houses and the street crossings are all correct, but although three of the roads are actually straight, Qurwik street is not. Who lives on Qurwik street?
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Answer: C — Carol
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Hint 1 of 2
Three of the four roads are straight; the one that bends is Qurwik street.
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Hint 2 of 2
Find the resident whose connecting road is the curved one, not a straight segment.
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Approach: identify the single curved road in the map
The crossings and house positions are correct, and only one road is drawn curved instead of straight.
That curved road is Qurwik street, and the house it serves belongs to Carol.
The pie chart shows the number of inhabitants in the five zones of a city. The central zone has the same population as the north, west and east zones combined, and the south zone has half as many inhabitants as the west zone. What is the difference, in percentage points, between the inhabitants of the north and east zones?
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Answer: D — 13%
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Hint 1 of 2
The central slice (47%) equals north+west+east together.
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Hint 2 of 2
Use 'south is half of west' to pin down which slices are which.
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Approach: match the percentage slices to the zones using the constraints
The non-central slices are 11, 6, 24, 12 (summing to 53), and central 47 = north+west+east, so south = 53 − 47 = 6.
South is half of west, so west = 12.
North + east = 47 − 12 = 35, which forces the pair to be 11 and 24.
18 cubes are coloured white, grey or black and stacked into the block shown. The figures show the white part and the black part on their own. Which of the following is the grey part?
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Answer: E
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Hint 1 of 2
The whole block is white cubes plus black cubes plus grey cubes, with nothing missing.
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Hint 2 of 2
The grey part is just the leftover cubes — the exact shape that fills the holes the white and black pieces leave behind.
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Approach: the grey piece is the leftover after taking out the white and black
Imagine lifting the white cubes and the black cubes out of the big block.
Whatever cubes are still sitting there make up the grey part — it exactly fills the gaps the white and black left.
Matching that leftover shape to the pictures gives option E.
Among 10 different given positive integers, exactly five are divisible by 5 and exactly seven are divisible by 7. Let M be the largest of these numbers. What is the smallest possible value of M?
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Answer: E — a different value
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Hint 1 of 2
At least how many numbers must be divisible by both 5 and 7?
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Hint 2 of 2
Inclusion–exclusion forces ≥2 multiples of 35; the second-smallest is 70, so M is at least 70.
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Approach: inclusion–exclusion, then minimise the maximum
5 + 7 − (multiples of 35) ≤ 10, so at least 2 numbers are multiples of 35.
The two smallest multiples of 35 are 35 and 70, so 70 must appear and is the largest.
M = 70, which is not among A–D, so the answer is a different value.
Five swimmers from a school are training for a relay race. The five participants swim the same distance, one after the other, without stopping. The coach stops the intermediate time after each swimmer. The first swimmer takes 2 minutes and 8 seconds. The stopwatches show the total time after the first, second, third, fourth and fifth swimmer (see picture). Which swimmer swam the distance the fastest?
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Answer: D — the fourth
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Hint 1 of 2
Each stopwatch shows the total time after that many swimmers, so subtract to get each leg.
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Hint 2 of 2
All swam the same distance, so the fastest is the one with the shortest individual leg.
Show solution
Approach: difference of consecutive total times
The watches show running totals: 2:08, 4:07, 6:10, 8:05, 10:03.
Subtract each total from the one before to get each swimmer’s own time: 128 s, 119 s, 123 s, 115 s, 118 s.
Same distance means fastest = shortest time, and 115 s is the smallest.
The year 2022 has three equal digits. This is the third time that Tortoise Eva has experienced a year where the same digit appears three times. What is the minimum age that Tortoise Eva can be this year?
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Answer: C — 23
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Hint 1 of 2
List recent years whose four digits include the same digit three times.
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Hint 2 of 2
She has lived through three such years; to make her as young as possible, pick the latest possible earlier two.
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Approach: find the three latest triple-digit years up to 2022
2022 has three 2s. The two latest earlier such years are 2000 (three 0s) and 1999 (three 9s).
If 2022 is her third, she was alive in 1999, so she was born by 1999.
One of the two sides of a rectangle has length 6 cm. In the rectangle circles are drawn next to each other in such a way that their centres form an equilateral triangle. What is the shortest distance between the two grey circles (in cm)?
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Answer: C — \(2\sqrt{3} - 2\)
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Hint 1 of 2
Three equal circles span the 6 cm side, so each has diameter 2 and radius 1.
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Hint 2 of 2
Equal touching circles have centres 2 apart; stack the equilateral rows to find the two grey centres.
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Approach: locate the grey centres, then subtract the two radii
Three touching circles fill the 6 cm width, so each diameter is 2 and each radius is 1.
Centres of touching circles are 2 apart and form equilateral triangles, so going down two rows the centres drop by 2 × √3 = 2√3 vertically.
The two grey circles' centres are 2√3 apart; subtract the two radii (1 + 1 = 2) to get the gap.
In a tournament FC Barcelona scored a total of three goals, and conceded one goal. In the tournament the team had won one game, lost one game and drawn one game. What was the score in the game that FC Barcelona won?
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Answer: B — 3:0
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Hint 1 of 2
The single goal they conceded must have come in the game they lost.
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Hint 2 of 2
That fixes the lost and drawn scores, leaving the goals for the win.
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Approach: place the one conceded goal, then balance the totals
They conceded only 1 goal all tournament, and a loss needs at least one conceded, so the loss was 0:1.
With no goals left to concede, the draw must be 0:0.
All 3 scored goals fall in the win, so the won game was 3:0.
In the (x, y)-plane the coordinate axes are drawn as usual. The point A(1, −10), which lies on the parabola \(y=ax^{2}+bx+c\), was marked. Then the coordinate axes and most of the parabola were erased, leaving the sketch on the right. Which of the following statements could be false?
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Answer: E — \(c<0\)
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Hint 1 of 2
Use the point A(1, −10): substituting x = 1 gives a + b + c = −10.
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Hint 2 of 2
That fixes some statements as always true; look for the one that the picture does not force.
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Approach: substitute the known point and test each statement
A(1,−10) gives a+b+c = −10 < 0, so that statement is always true; the upward shape forces a > 0.
But c is the value at x = 0, which the trimmed picture does not pin down — it may be positive.
Maria writes the numbers 1, 2, 3, 4, 5, 6 and 7, each exactly once, into the number wall. Each upper box equals the sum of the two boxes just below it. The bottom-left box already holds 6. Which number must she write in the box with the star?
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Answer: D — 4
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Hint 1 of 2
The box sitting on top of the 6 is 6 plus its neighbour, and it can be at most 7.
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Hint 2 of 2
That forces the neighbour, then keep building upward with 1 to 7 each used once.
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Approach: use that no box can exceed 7 to fix the numbers, then build upward
Every box is the sum of the two below it, and no number is bigger than 7.
The box above the 6 is 6 + (its right neighbour), so that neighbour must be 1, giving 6 + 1 = 7.
The leftover numbers 2, 3, 5 must fill the rest; placing the bottom row as 6, 1, 3, 2 gives the next row 7, 4, 5 - and that uses 1 to 7 exactly once.
The starred middle-top box is 1 + 3 = 4, option D.
There are three great circles on a sphere that intersect each other at right angles. Starting at point S, a little bug moves along the great circles in the direction indicated. At each crossing it turns alternately to the right or to the left. How many quarter circles does it crawl along until it is back at point S?
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Answer: A — 6
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Hint 1 of 2
The three perpendicular great circles meet at six points (the axis tips); each arc between them is a quarter circle.
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Hint 2 of 2
Track the turns: with the right/left alternation the path closes after surprisingly few quarter-arcs.
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Approach: trace the alternating-turn path between the six crossing points
Three mutually perpendicular great circles cross at six points (like the six face-centres of a cube); each arc from one crossing to the next neighbouring crossing is a quarter circle.
From S the bug reaches a crossing after one quarter arc, turns, takes the next quarter arc, and so on, alternating right and left.
Tracing this alternating walk closes the loop after visiting six such crossings, so it covers six quarter circles before returning to S.
Jenny writes numbers in a \(3 \times 3\) table so that the four numbers in every \(2 \times 2\) square have the same sum. Three corner cells are already filled in (see diagram). Which number does she write in the fourth corner?
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Answer: B — 1
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Hint 1 of 2
The four overlapping 2x2 squares all share the same sum; compare two of them that overlap in a row or column.
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Hint 2 of 2
Comparing the four equal block-sums forces a tidy rule on the four corner cells of the grid.
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Approach: equal block-sums link the four corners
Comparing the four equal 2x2 sums cancels the shared middle cells and forces the two diagonal corner-pairs to have equal sums: one corner plus its opposite equals the other two corners added.
The given corners are 2, 4 and 3; the missing corner pairs with 4 across the diagonal, so (missing) + 4 = 2 + 3 = 5.
Therefore the fourth corner is 5 - 4 = 1, so the answer is B.
Counting & ProbabilityLogic & Word Problemscareful-countingcasework
Snow White organises a chess tournament for the seven dwarfs, lasting several days. Every dwarf has to play every other dwarf exactly once. On Monday Grumpy plays 1 game, Sneezy plays 2, Sleepy 3, Bashful 4, Happy 5 and Doc 6 games. How many games does Dopey, the 7th dwarf, play on Monday?
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Answer: C — 3
Show hints
Hint 1 of 3
Each dwarf can play at most the other six, and the given counts 1, 2, 3, 4, 5, 6 are all different.
Still stuck? Show hint 2 →
Hint 2 of 3
Start from the extremes: the one who played 6 games played everybody, while the one who played only 1 must have played that same dwarf.
Still stuck? Show hint 3 →
Hint 3 of 3
Keep pairing the highest remaining count with the lowest, peeling off the players whose schedule is now complete.
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Approach: pair the largest count with the smallest, working inward
Doc played 6 games, so he played everyone — including Grumpy, whose single game must therefore be against Doc only.
Happy played 5: everyone except Grumpy (Grumpy is already finished), so Happy played the other five, including Dopey; Sneezy's 2 games are then Doc and Happy.
Bashful played 4: he must be Doc, Happy, Sleepy and Dopey; Sleepy's 3 are Doc, Happy and Bashful.
So Dopey was played by Doc, Happy and Bashful only — that is 3 games, option C.
Johannes wrote the numbers 6, 7 and 8 in the circles as shown. He wants to write the numbers 1, 2, 3, 4 and 5 in the remaining circles so that the sum of the numbers along each side of the square is 13. What will be the sum of the numbers in the grey circles?
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Answer: E — 16
Show hints
Hint 1 of 3
The grey circles are the four corners, and the white circles are the four middles.
Still stuck? Show hint 2 →
Hint 2 of 3
Add up the four sides together: each corner sits on two sides, so it gets counted twice, while each middle is counted once.
Still stuck? Show hint 3 →
Hint 3 of 3
You also know all eight numbers (the corners plus the middles) are 1 to 8.
Show solution
Approach: add all four sides at once
Add the four side-totals together: 13 + 13 + 13 + 13 = 52. In that big sum each corner is counted twice (it touches two sides) and each middle is counted once.
All eight numbers 1, 2, 3, 4, 5, 6, 7, 8 add up to 36, so the corners and middles together make 36.
Take the 52 away from the 36 idea: 52 is one extra copy of every corner above the full 36, so the grey corners add to 52 − 36 = 16, answer E.
A cube is cut in three directions as shown, to produce eight cuboids (each cut is parallel to one of the faces of the cube). What is the ratio of the total surface area of the eight cuboids to the surface area of the original cube?
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Answer: D — 2 : 1
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Hint 1 of 2
Each straight cut makes two new faces, each equal to the cross-section it slices.
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Hint 2 of 2
Add the new face area from the three cuts to the original surface and compare.
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Approach: count the new surface from each cut
Let the cube's surface be 6 (in face units). Each middle cut adds two faces of area 1, i.e. +2 per cut.
Three cuts add 6, giving total 6 + 6 = 12.
The ratio of new total to original is 12 : 6 = 2 : 1.
Three big boxes P, Q and R are stored in a warehouse. The upper picture on the right shows their placement seen from above. The boxes are so heavy that they can only be rotated 90° around a vertical edge, as indicated in the pictures below. Now the boxes should be rotated to stand against the wall in a certain order. Which arrangement is possible? (Choice E: all four arrangements are possible.)
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Answer: B
Show hints
Hint 1 of 3
Each allowed move is a 90° turn about a vertical edge, which both slides a box and rotates its top label a quarter turn.
Still stuck? Show hint 2 →
Hint 2 of 3
So a box's final orientation is tied to how far it travelled: track position and label-rotation together.
Still stuck? Show hint 3 →
Hint 3 of 3
Test each pictured line-up and reject any whose letters point the wrong way for the moves needed to reach it.
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Approach: couple each box's final position to its forced label orientation
Tipping a box 90° about a vertical edge moves it one step and turns its label a quarter turn, so position and orientation change together — they cannot be set independently.
Walking the boxes to the wall one move at a time, each box's letter must end pointing the way that number of quarter-turns dictates.
Comparing the four pictured line-ups, only one has every letter oriented consistently with the moves that put the boxes there.
David wants to place the twelve numbers from 1 to 12 in a circle so that two adjacent numbers always differ by 2 or 3. Which numbers are therefore adjacent?
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Answer: D — 6 and 8
Show hints
Hint 1 of 2
List which numbers each value is allowed to sit next to (differ by 2 or 3).
Still stuck? Show hint 2 →
Hint 2 of 2
1 and 2 have very few options — those force several neighbours; build out from there.
Show solution
Approach: forced neighbours then complete the loop
Each number's neighbours must differ from it by 2 or 3, and each has exactly two neighbours in the circle.
1 can only touch 3 and 4; 2 can only touch 4 and 5; chaining these and continuing yields the loop 6–3–1–4–2–5–7–10–12–9–11–8 back to 6.
In that loop 8 sits next to 6, so 6 and 8 are adjacent.
A number of oranges, peaches, apples and bananas are placed in a row. What is the minimum number of fruits needed so that each kind of fruit lies next to each other kind at least once somewhere in the row?
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Answer: C — 8
Show hints
Hint 1 of 2
There are four fruit types, so 6 unordered pairs must each appear side by side somewhere.
Still stuck? Show hint 2 →
Hint 2 of 2
Think of fruits as dots and 'were neighbours' as edges of K₄; you want a near-Eulerian walk.
Show solution
Approach: cover all six pairs with the fewest neighbour-links
With 4 types there are 6 pairs to realise as adjacencies; a row of n fruits has only n−1 adjacencies.
Each type would need to touch the other three, but all four 'dots' have odd degree 3 in the pair-graph K₄.
Repeating one link fixes the parity, giving 7 needed adjacencies, hence 8 fruits.
A row such as orange-peach-apple-orange-banana-peach-apple-banana shows 8 works, so the answer is 8.
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?
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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).
A piece of string is folded as shown in the diagram by folding it in the middle, then folding it in the middle again and finally folding it in the middle once more. Then this folded piece of string is cut so that several pieces emerge. Amongst the resulting pieces there are some with length 4 m and some with length 9 m. Which of the following lengths cannot be the total length of the original piece of string? (In the picture, “Schnitt” marks where the cut is made.)
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Answer: C — 72 m
Show hints
Hint 1 of 3
Folding in half three times stacks the string into eight equal layers before the single cut.
Still stuck? Show hint 2 →
Hint 2 of 3
Call the two parts the cut makes in one folded layer \(a\) and \(b\); then the whole string has length \(8(a+b)\).
Still stuck? Show hint 3 →
Hint 3 of 3
Work out the lengths of the pieces in terms of \(a\) and \(b\), then test each total to see whether both a 4 m and a 9 m piece can appear.
Show solution
Approach: lengths of the unfolded pieces
Three folds make 8 stacked layers, so the folded packet has length \(s=a+b\) where the cut lands distance \(a\) from the folded edge, and the whole string is \(8s=8(a+b)\).
Unfolding, the cut points split the string into pieces of just three lengths: \(a\) (the two ends), \(2a\), and \(2b\); so both a 4 m piece and a 9 m piece must appear among \(a,2a,2b\).
A total of 52, 68 or 88 m can be split this way with a 4 m and a 9 m piece, but for a total of 72 m we get \(s=9\), and then \(a,2a,2b\) can never give both 4 and 9 at once.
In a game of dominoes the tiles always have to be placed so that the touching halves of two adjacent domino tiles show the same number of dots. Paul has six domino tiles in front of him (see diagram). In several steps he tries to arrange them in a correct order. In each step he is allowed either to swap any two domino tiles or to turn one domino tile 180° around. What is the minimum number of steps he needs to arrange the domino tiles correctly?
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Answer: C — 3
Show hints
Hint 1 of 2
Treat each domino value as a connection and look for a chain that uses all six with matching ends.
Still stuck? Show hint 2 →
Hint 2 of 2
Count the fewest swaps and 180° turns needed to turn the given row into such a chain.
Show solution
Approach: find a matching chain, count the moves to reach it
The six dominoes are 4-6, 3-1, 4-2, 3-4, 6-1, 2-6; one valid chain is 4-6, 6-1, 1-3, 3-4, 4-2, 2-6.
Starting from the given order, this is reachable by turning one tile and swapping tiles — three moves in total.
No arrangement is reachable in fewer, so the minimum is 3.
