🦘 Math Kangaroo Grade All Felix 1-2 Ecolier 3-4 Benjamin 5-6 Kadett 7-8 Junior 9-10 Student 11-12 ⇄ switch contest
Mock Test

Mock Test (random)

30 problems — one per position, pulled from random authored years. Hints and solutions are locked until you submit. Retake as often as you want — every attempt is saved to your test history (if you're logged in).

30 problems Reshuffle
Paper:

Pick a paper and hit Start to take it as a timed test — or browse past papers untimed to read the original questions, figures and answers at your own pace.

← Back Tip: real Math Kangaroo is about 75 minutes — set a timer if you want the full pressure.
Problem 1 · 2016 Math Kangaroo Easy
Spatial & Visual Reasoning symmetry

Which of the following road signs has the most axes of symmetry?

Figure for Math Kangaroo 2016 Problem 1
Show answer
Answer: C — The no-entry sign.
Show hints
Hint 1 of 3
Imagine folding each sign along a straight line so the two halves land exactly on top of each other.
Still stuck? Show hint 2 →
Hint 2 of 3
Try both a left-right fold and a top-bottom fold on every sign, then count how many folds work.
Still stuck? Show hint 3 →
Hint 3 of 3
A plain horizontal bar inside a circle matches itself for both folds.
Show solution
Approach: fold each sign and count the lines that match
  1. An axis of symmetry is a fold line where one half lands perfectly on the other half.
  2. The arrow signs match only one fold (or none, once an arrowhead points a direction), and the car shape matches just its up-down fold.
  3. The no-entry sign (a horizontal bar in a circle) matches a left-right fold AND a top-bottom fold, so it has 2 folds.
  4. Two is the most of any sign, so the answer is the no-entry sign, choice (C).
Mark: · log in to save
Problem 2 · 2016 Math Kangaroo Easy
Arithmetic & Operations division

Mike cuts a pizza into four equally big pieces. Then he cuts each piece into three equally big pieces. Into how many equally big pieces did Mike cut the pizza?

Show answer
Answer: E — 12
Show hints
Hint 1 of 2
Picture the four pieces side by side, then split each one.
Still stuck? Show hint 2 →
Hint 2 of 2
Each of the four pieces turns into three pieces.
Show solution
Approach: count the pieces in equal groups
  1. After the first cuts there are 4 pieces.
  2. Each of those 4 pieces is split into 3, so you get 4 groups of 3.
  3. That is \(4 \times 3 = 12\) pieces, choice (E).
Mark: · log in to save
Problem 3 · 2010 Math Kangaroo Easy
Arithmetic & Operations total-then-divide

A fly has 6 legs and a spider has 8. Together, 2 flies and 3 spiders have as many legs as 10 birds and …

Show answer
Answer: C — 4 cats
Show hints
Hint 1 of 2
First count all the legs on the left side of the comparison.
Still stuck? Show hint 2 →
Hint 2 of 2
Subtract the birds' legs, then see how many 4-legged animals fill the gap.
Show solution
Approach: count legs, then divide the leftover
  1. 2 flies = 2×6 = 12 legs; 3 spiders = 3×8 = 24 legs; together 36 legs.
  2. 10 birds have 10×2 = 20 legs.
  3. The remaining 36−20 = 16 legs come from cats: 16÷4 = 4 cats.
Mark: · log in to save
Problem 4 · 2014 Math Kangaroo Easy
Algebra & Patterns substitution

In three differently sized baskets there are 48 balls in total. Together the smallest and the biggest basket hold twice as many balls as the middle one. The smallest basket holds half as many balls as the middle one. How many balls are there in the biggest basket?

Show answer
Answer: C — 24
Show hints
Hint 1 of 2
Let the middle basket be your unit and write the others in terms of it.
Still stuck? Show hint 2 →
Hint 2 of 2
Three quantities add to 48 — turn the words into one equation in the middle amount.
Show solution
Approach: express all baskets through the middle one
  1. Let the middle basket hold m balls. The smallest holds m/2.
  2. Smallest + biggest = 2m, so biggest = 2m − m/2 = 3m/2.
  3. Total: m/2 + m + 3m/2 = 3m = 48, so m = 16.
  4. Biggest = 3m/2 = 24 balls.
Mark: · log in to save
Problem 5 · 2016 Math Kangaroo Easy
Spatial & Visual Reasoning reflectiontransformations
Figure for Math Kangaroo 2016 Problem 5
Show answer
Answer: B — Picture B
Show hints
Hint 1 of 2
A flip downward reflects the picture top-to-bottom; a flip to the right reflects it left-to-right.
Still stuck? Show hint 2 →
Hint 2 of 2
Apply the two reflections in turn to the original diagram.
Show solution
Approach: apply two reflections to the figure
  1. Flipping the card downward reflects the design across a horizontal line (top and bottom swap).
  2. Flipping it to the right then reflects across a vertical line (left and right swap).
  3. Doing both turns the original into the picture shown in (B).
Mark: · log in to save
Problem 6 · 2023 Math Kangaroo Medium
Spatial & Visual Reasoning reflectionsymmetry

Christoph folds a see-through piece of foil along the dashed line. What can he then see? (Choose from pictures A–E.)

Figure for Math Kangaroo 2023 Problem 6
Show answer
Answer: A
Show hints
Hint 1 of 2
Folding along the dashed line flips the figure like a mirror.
Still stuck? Show hint 2 →
Hint 2 of 2
Reflect each digit across the fold line and read the result.
Show solution
Approach: reflect the pattern across the fold line
  1. Folding the transparent foil mirrors the drawing over the dashed line.
  2. Each mark lands on its mirror image, turning the shapes into readable digits.
  3. The reflected result reads the pattern shown in option A.
Mark: · log in to save
Problem 7 · 2022 Math Kangaroo Medium
Spatial & Visual Reasoning symmetry

Otto fastens his licence plate to the car upside down, but it doesn’t matter because the plate looks exactly the same that way. Which of these plates could be Otto’s?

Show answer
Answer: B — 60 SOS 09
Show hints
Hint 1 of 2
Turning the plate upside down rotates it 180 degrees; which digits still read as valid digits after that?
Still stuck? Show hint 2 →
Hint 2 of 2
Only 0, 1, 8 (and the pair 6/9 swapping) survive a 180 turn; the whole string must read the same.
Show solution
Approach: test each plate for 180-degree symmetry
  1. Rotating 180 degrees, 0 stays 0, 1 stays 1, 8 stays 8, while 6 and 9 swap.
  2. The string must read identically after flipping and reversing order.
  3. Only 60 SOS 09 reads the same upside down, so the answer is B.
Mark: · log in to save
Problem 8 · 2025 Math Kangaroo Easy
Logic & Word Problems work-backwardsum-constraint

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?

Figure for Math Kangaroo 2025 Problem 8
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
  1. Prices rise: veggie 3.70 < classic _.30 < hot bacon _.60 < cheesy _.50 < double _.10 < deluxe _.80.
  2. Smallest classic above 3.70 is 4.30; then hot bacon 4.60; cheesy must beat 4.60 so 5.50; double beats 5.50 so 6.10.
  3. Deluxe must beat 6.10 and end in .80, so the least it can be is 6.80, which is (B).
Mark: · log in to save
Problem 9 · 2024 Math Kangaroo Medium
Number Theory factorization

Noah starts with the number 1 and multiplies it with either 6 or 10. He then multiplies the result again by either 6 or 10. He repeats this process several times. Which of the following numbers can he not obtain in this way?

Show answer
Answer: B — \(2^{90}\cdot 3^{30}\cdot 5^{80}\)
Show hints
Hint 1 of 2
Each multiplication by 6 or 10 always adds one factor of 2, plus either a 3 or a 5.
Still stuck? Show hint 2 →
Hint 2 of 2
So in the final number the power of 2 must equal the powers of 3 and 5 added together — check which option breaks this.
Show solution
Approach: count factors of 2 against 3 and 5
  1. 6 = 2·3 and 10 = 2·5, so each step adds exactly one 2 and either one 3 or one 5.
  2. After several steps the power of 2 equals (power of 3) + (power of 5).
  3. Option B has 2⁹⁰·3³⁰·5⁸⁰, but 30 + 80 = 110 ≠ 90.
  4. So B cannot be obtained.
Mark: · log in to save
Problem 10 · 2019 Math Kangaroo Medium
Logic & Word Problems caseworkwork-backward

Every day the three kangaroos Alex, Bob and Carl go for a walk. If Alex does not wear a hat, then Bob wears a hat. If Bob does not wear a hat, then Carl wears a hat. Today Carl does not wear a hat. Which kangaroos can we be sure are wearing a hat today?

Show answer
Answer: E — only Bob
Show hints
Hint 1 of 2
Carl wears no hat. Use the rule 'if Bob has no hat then Carl wears one' in reverse.
Still stuck? Show hint 2 →
Hint 2 of 2
Since Carl has no hat, Bob must have one; then check whether Alex is forced.
Show solution
Approach: contrapositive reasoning from Carl
  1. Rule: if Bob has no hat, Carl wears one. Carl has no hat, so Bob must have a hat.
  2. Rule: if Alex has no hat, Bob wears one — already satisfied, so Alex is not forced either way.
  3. Thus for certain only Bob is wearing a hat.
  4. Answer (E) only Bob.
Mark: · log in to save
Problem 11 · 2017 Math Kangaroo Medium
Fractions, Decimals & Percents fraction-to-decimal

Ant Annie starts at the left end of the stick and crawls 23 of the length of the stick. Ladybird Bob starts at the right end of the stick and crawls 34 of the length of the stick. Which fraction of the length of the stick are they then apart from each other?

Figure for Math Kangaroo 2017 Problem 11
Show answer
Answer: D512
Show hints
Hint 1 of 2
Put both bugs' positions on the same scale, measured from the left end.
Still stuck? Show hint 2 →
Hint 2 of 2
Subtract the two positions to get the gap.
Show solution
Approach: locate both on [0,1] and subtract
  1. Annie is at 2/3 from the left. Bob, 3/4 from the right, is at 1 − 3/4 = 1/4 from the left.
  2. They have passed each other; the gap is 2/3 − 1/4 = 8/12 − 3/12 = 5/12.
Mark: · log in to save
Problem 12 · 2015 Math Kangaroo Medium
Spatial & Visual Reasoning cube-viewscareful-counting

Every one of these six building blocks consists of 5 little cubes. The little cubes are either white or grey. Cubes of equal colour don’t touch each other. How many little white cubes are there in total?

Figure for Math Kangaroo 2015 Problem 12
Show answer
Answer: C — 12
Show hints
Hint 1 of 2
In each block of 5 cubes the colours alternate, so they go grey-white-grey-white-grey.
Still stuck? Show hint 2 →
Hint 2 of 2
Count the white cubes in one block, then multiply by the number of blocks.
Show solution
Approach: count white per block, then scale to all blocks
  1. Because same-colour cubes never touch, each block of 5 alternates colour, giving 3 of one colour and 2 of the other.
  2. Each block ends up with 2 white cubes.
  3. With 6 blocks that is 6 x 2 = 12 white cubes, choice C.
Mark: · log in to save
Problem 13 · 2014 Math Kangaroo Medium
Spatial & Visual Reasoning tiling-tessellationcomposition

Erwin has the four paper pieces shown. He has to cover a special shape exactly with these four pieces. In which drawing can he do this, when the one piece is placed as shown? (Choose the matching picture.)

Figure for Math Kangaroo 2014 Problem 13
Show answer
Answer: C
Show hints
Hint 1 of 3
The piece that is already placed covers part of the shape, so look at the empty gap that is left.
Still stuck? Show hint 2 →
Hint 2 of 3
Ask whether the other three pieces can fill that gap with no holes and no sticking out.
Still stuck? Show hint 3 →
Hint 3 of 3
Try each drawing and keep the only one that the pieces fit perfectly.
Show solution
Approach: see which outline the leftover three pieces fill exactly
  1. Once the shown piece is set down, the empty space that remains has a fixed shape.
  2. Imagine sliding the other three pieces in like a little jigsaw, covering every square with no gaps and no overlaps.
  3. Only one of the drawings lets all four pieces fit exactly.
  4. That drawing is C.
Mark: · log in to save
Problem 14 · 2017 Math Kangaroo Stretch
Arithmetic & Operations off-by-onecareful-counting

A kangaroo always does ten jumps within a minute. Then he has a three minute break. How many minutes does it need in order to do 50 jumps?

Show answer
Answer: D — 17
Show hints
Hint 1 of 2
50 jumps means 5 separate one-minute jumping sessions; count the breaks BETWEEN them.
Still stuck? Show hint 2 →
Hint 2 of 2
There is no break after the last session, so there is one fewer break than sessions.
Show solution
Approach: add jumping minutes and the breaks between sessions
  1. 10 jumps per minute, so 50 jumps need 5 one-minute jumping sessions.
  2. Between the 5 sessions there are 4 breaks of 3 minutes each (no break after the last).
  3. Total time = 5 jumping minutes + 4 breaks of 3 minutes = 5 + 12 = 17 minutes.
  4. So it needs 17 minutes.
Mark: · log in to save
Problem 15 · 2013 Math Kangaroo Hard
Algebra & Patterns substitution

The sides of rectangle ABCD are parallel to the coordinate axes. The rectangle lies below the x-axis and to the right of the y-axis, as shown. For each of the points A, B, C, D the quotient (y-coordinate) : (x-coordinate) is calculated. For which point will you obtain the smallest quotient?

Figure for Math Kangaroo 2013 Problem 15
Show answer
Answer: DD
Show hints
Hint 1 of 2
For every corner the quotient is (negative y) over (negative x), so it is positive; you want the smallest positive value.
Still stuck? Show hint 2 →
Hint 2 of 2
Smallest quotient = smallest |y| paired with largest |x| — find that corner.
Show solution
Approach: compare y/x at each corner
  1. In the diagram the rectangle has both coordinates negative, so each quotient y : x is positive.
  2. The smallest quotient comes from the corner closest to the x-axis (smallest |y|) and farthest from the y-axis (largest |x|).
  3. That is the top-left corner, point D.
Mark: · log in to save
Problem 16 · 2009 Math Kangaroo Hard
Algebra & Patterns sum-constraintsubstitution

The diagram shows an object with 6 triangular faces. On each corner there is a number (two are shown). The sum of the numbers on the corners of each face is the same. What is the sum of all 5 numbers?

Figure for Math Kangaroo 2009 Problem 16
Show answer
Answer: C — 17
Show hints
Hint 1 of 2
Equal face-sums force just two values: the two tips are equal and the three middle corners are equal.
Still stuck? Show hint 2 →
Hint 2 of 2
Use the two shown numbers to tell which value is a tip and which is a middle corner.
Show solution
Approach: exploit equal face sums to find the two values
  1. The solid is a triangular bipyramid: two tips plus three middle corners. Equal face sums force all three middle corners equal and both tips equal.
  2. The shown 1 is a tip and the shown 5 is a middle corner, so the numbers are 1, 1 (tips) and 5, 5, 5 (middle).
  3. Their total is 1+1+5+5+5 = 17.
Mark: · log in to save
Problem 17 · 2014 Math Kangaroo Stretch
Number Theory divisibility

Lea plays with her marbles, placing them in small groups on the table. In groups of three, two marbles are left over. In groups of five, again two are left over. How many more marbles does Lea need so that she can place them in groups of three and in groups of five with none left over?

Show answer
Answer: E — 13
Show hints
Hint 1 of 2
Leaving 2 over for both groups of three and groups of five means leaving 2 over for groups of fifteen.
Still stuck? Show hint 2 →
Hint 2 of 2
She wants a multiple of 15; find the next multiple of 15 above her current count and see how many more marbles that needs.
Show solution
Approach: use that 2 left over mod 3 and mod 5 means 2 left over mod 15
  1. Leaving 2 over in groups of three and in groups of five means leaving 2 over in groups of fifteen.
  2. So her smallest possible count is 2; to split evenly into both 3s and 5s she needs a multiple of 15.
  3. The next multiple of 15 after 2 is 15, which is 15 − 2 = 13 more marbles.
  4. She needs 13 more marbles.
Mark: · log in to save
Problem 18 · 2022 Math Kangaroo Hard
Logic & Word Problems work-backward

The bus stops in the villages A, B, C and D lie along a road in this order, and neighbouring villages are 10 km apart. There are 10 children in village A, 20 in B, 30 in C and 40 in D. Every child takes the bus to school. A new school will be built where the total number of kilometres travelled by all the children is as small as possible. Where will the new school be built?

Show answer
Answer: D — in C
Show hints
Hint 1 of 3
The best spot has about as many children on one side of the school as on the other side.
Still stuck? Show hint 2 →
Hint 2 of 3
Start at one end and add up children until you reach more than half of all of them.
Still stuck? Show hint 3 →
Hint 3 of 3
The village where you pass the halfway count is the best place for the school.
Show solution
Approach: walk from one end and stop where you pass half the children
  1. There are 10 + 20 + 30 + 40 = 100 children in all, so half of them is 50.
  2. Counting from A: A has 10, then A and B have 30, then A, B and C have 60 - we pass 50 right at C.
  3. Since just as many children sit on each side once we reach C, building the school in C makes the total travel smallest.
  4. So the answer is D.
  5. Check by trying neighboursMoving the school 10 km from C toward D saves 40 children 10 km each (400 km) but costs the other 60 children 10 km each (600 km), a net loss; moving it toward B is worse too, so C truly is best.
Mark: · log in to save
Problem 19 · 2012 Math Kangaroo Stretch
Number Theory place-value

Gregory made two 3-digit numbers from the digits 1, 2, 3, 4, 5, 6. Each digit was used only once. Afterwards he added the two numbers together. What is the largest answer he could have got?

Show answer
Answer: D — 1173
Show hints
Hint 1 of 2
To make a sum large, put the biggest digits where they count the most.
Still stuck? Show hint 2 →
Hint 2 of 2
Give the two hundreds places the largest digits, then the tens, then the units.
Show solution
Approach: place the largest digits in the highest places
  1. Use 6 and 5 in the hundreds places, 4 and 3 in the tens, 2 and 1 in the units.
  2. The two numbers add to (600 + 500) + (40 + 30) + (2 + 1).
  3. That is 1100 + 70 + 3 = 1173.
  4. The largest possible sum is 1173.
Mark: · log in to save
Problem 20 · 2016 Math Kangaroo Stretch
Algebra & Patterns substitution

In an enclosure there are 2016 kangaroos. Each of them is either red or grey, and there is at least one red and at least one grey kangaroo amongst them. For each kangaroo K we calculate the fraction obtained, if you take the number of kangaroos of the other colour divided by the kangaroos of the own colour (including K itself). Determine the sum of these 2016 fractions.

Show answer
Answer: A — 2016
Show hints
Hint 1 of 2
Group the kangaroos by colour: say r red and g grey, with r + g = 2016.
Still stuck? Show hint 2 →
Hint 2 of 2
Add the fractions colour by colour and watch the counts cancel.
Show solution
Approach: sum by colour
  1. Each red kangaroo contributes g/r and there are r of them, totalling g.
  2. Each grey kangaroo contributes r/g and there are g of them, totalling r.
  3. The grand total is g + r = 2016.
Mark: · log in to save
Problem 21 · 2022 Math Kangaroo Stretch
Algebra & Patterns sum-constraintcasework

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?

Figure for Math Kangaroo 2022 Problem 21
Show answer
Answer: B — 1
Show hints
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.
Still stuck? Show hint 2 →
Hint 2 of 2
Comparing the four equal block-sums forces a tidy rule on the four corner cells of the grid.
Show solution
Approach: equal block-sums link the four corners
  1. 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.
  2. The given corners are 2, 4 and 3; the missing corner pairs with 4 across the diagonal, so (missing) + 4 = 2 + 3 = 5.
  3. Therefore the fourth corner is 5 - 4 = 1, so the answer is B.
Mark: · log in to save
Problem 22 · 2016 Math Kangaroo Stretch
Logic & Word Problems magic-squaresum-constraint

Kirsten has written numbers into 5 of the 10 circles. She wants to write numbers into the remaining circles so that the sum of the three numbers along every side of the pentagon is always the same. Which number does she have to write into the circle marked X?

Figure for Math Kangaroo 2016 Problem 22
Show answer
Answer: D — 13
Show hints
Hint 1 of 3
Every side of the pentagon uses three circles and they all add to the same total.
Still stuck? Show hint 2 →
Hint 2 of 3
Find a side that already has two numbers filled in to pin down that common total.
Still stuck? Show hint 3 →
Hint 3 of 3
Then walk around the pentagon, filling each missing circle from the side total until you reach X.
Show solution
Approach: find the common side-total, then fill circles one at a time
  1. Call the common total of each side \(S\); a side that already shows two numbers tells you \(S\) once you know the third.
  2. Using the five given numbers and that fixed total \(S\), fill the empty circles one side at a time, each missing circle being \(S\) minus the two known circles on its side.
  3. Carrying this around to the marked circle gives \(X = 13\), choice (D).
Mark: · log in to save
Problem 23 · 2013 Math Kangaroo Stretch
Ratios, Rates & Proportions proportioncareful-counting

Peter has bought a rug that is 36 dm wide and 60 dm long. The rug is covered in squares that each contain either a sun or a moon, as shown in the picture. There are exactly nine squares along the width of the rug. The full length of the rug cannot be seen. How many moons would you see if you could see the entire rug?

Figure for Math Kangaroo 2013 Problem 23
Show answer
Answer: B — 67
Show hints
Hint 1 of 3
If 9 squares fit across the 36 dm width, work out how wide one square is.
Still stuck? Show hint 2 →
Hint 2 of 3
Use that square size to find how many squares fit along the 60 dm length.
Still stuck? Show hint 3 →
Hint 3 of 3
Multiply rows by columns for the total squares, then the sun/moon pattern splits them almost in half.
Show solution
Approach: find the grid size, then count one colour
  1. 36 dm ÷ 9 = 4 dm per square, so the length 60 dm holds 60 ÷ 4 = 15 squares.
  2. The rug is 9 × 15 = 135 squares in a checkerboard of suns and moons.
  3. The moons fall on the smaller of the two colour groups, giving 67 moons.
Mark: · log in to save
Problem 24 · 2018 Math Kangaroo Stretch
Logic & Word Problems balance-scalelogic

Each of the four balls weighs either 10, 20, 30 or 40 grams. The two balances are shown. Which ball weighs 30 grams?

Figure for Math Kangaroo 2018 Problem 24
Show answer
Answer: C — C
Show hints
Hint 1 of 3
The four weights 10, 20, 30, 40 add up to 100 grams altogether — keep that total handy.
Still stuck? Show hint 2 →
Hint 2 of 3
The second scale is level, so the two balls on one side weigh exactly as much as the single ball C on the other side.
Still stuck? Show hint 3 →
Hint 3 of 3
Try out which single weight can be split into two of the other weights, and check it against the tilted first scale.
Show solution
Approach: use the level scale to say C equals two other balls added, then test which weight that can be
  1. The second scale balances, so B and D together weigh the same as C alone — that means C is made by adding two of the other weights.
  2. Among 10, 20, 30 and 40, the only weight that is the sum of two of the others is 30 = 10 + 20, so C must be 30 grams (with B and D being 10 and 20, leaving A as 40).
  3. Check the first scale: A and B together (40 + 20 = 60) are heavier than C and D (30 + 10 = 40), and that side does tip down, so everything fits — the 30 gram ball is C.
Mark: · log in to save
Problem 25 · 2016 Math Kangaroo Stretch
Algebra & Patterns substitutiondifference-of-squares

The equations \(x^2 + ax + b = 0\) and \(x^2 + bx + a = 0\) both have real solutions. It is known that the sum of the squares of the solutions of the first equation is equal to the sum of the squares of the solutions of the second equation, and that \(a \ne b\). Then \(a + b\) equals

Show answer
Answer: B — -2
Show hints
Hint 1 of 3
By Vieta, the sum of the squares of the roots is \((\text{sum})^2 - 2(\text{product})\).
Still stuck? Show hint 2 →
Hint 2 of 3
Set the two sums-of-squares equal, then factor the resulting symmetric equation.
Still stuck? Show hint 3 →
Hint 3 of 3
The condition \(a \ne b\) lets you cancel one factor.
Show solution
Approach: Vieta plus factoring
  1. For \(x^2+ax+b\) the roots have sum \(-a\) and product \(b\), so their squares sum to \(a^2 - 2b\); for \(x^2+bx+a\) it is \(b^2 - 2a\).
  2. Setting \(a^2 - 2b = b^2 - 2a\) gives \(a^2 - b^2 + 2a - 2b = 0\), i.e. \((a-b)(a+b+2) = 0\).
  3. Since \(a \ne b\), the other factor vanishes: \(a + b = -2\), answer B.
Mark: · log in to save
Problem 26 · 2019 Math Kangaroo Stretch
Number Theory place-valuecareful-counting

The integers from 1 to 99 are written down in ascending order without a gap. This sequence of digits is divided up into triples (groups of three):

\(123456789101112\cdots979899 \longrightarrow (123)(456)(789)(101)(112)\cdots(979)(899).\)

Which of the following triples is not obtained?

Show answer
Answer: B — (444)
Show hints
Hint 1 of 2
Write 1234567891011… and chop into groups of three; track where each number lands.
Still stuck? Show hint 2 →
Hint 2 of 2
Check each candidate triple against the actual grouping — one never appears.
Show solution
Approach: generate the triples and check membership
  1. Concatenate 1 to 99 and split into threes; this gives a fixed list of triples.
  2. Scanning it, (123), (464), (646) and (888) all occur as groups.
  3. The triple (444) never lines up, so the answer is (444).
Mark: · log in to save
Problem 27 · 2025 Math Kangaroo Stretch
Geometry & Measurement area-fractionratio

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?

Figure for Math Kangaroo 2025 Problem 27
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
  1. BN = 2·NC means BN:BC = 2:3, so triangle ABC has area 6·(3/2) = 9.
  2. K is the midpoint of AD, so triangle ACD = 2·(area CKD) = 2·2 = 4.
  3. Splitting ABCD by diagonal AC: area = \(9 + 4 = 13\), which is (A).
Mark: · log in to save
Problem 28 · 2013 Math Kangaroo Stretch
Ratios, Rates & Proportions distance-speed-timeevaluate-formula

A car starts at point A and drives on a straight road at 50 km/h. Every hour after that, another car leaves point A with a speed 1 km/h faster than the one before. The last car leaves A 50 hours after the first car and drives at 100 km/h. What is the speed of the car that is leading 100 hours after the start of the first car?

Show answer
Answer: C — 75 km/h
Show hints
Hint 1 of 2
The car leaving at hour k has speed 50 + k and, by 100 hours, has driven for 100 − k hours.
Still stuck? Show hint 2 →
Hint 2 of 2
Maximise the distance (50 + k)(100 − k) over k from 0 to 50.
Show solution
Approach: maximise distance travelled at t = 100
  1. The car leaving at hour k (speed 50 + k) has travelled (50 + k)(100 − k) by t = 100.
  2. This product is largest at k = 25, giving 75 × 75 = 5625.
  3. That leading car's speed is 75 km/h.
Mark: · log in to save
Problem 29 · 2022 Math Kangaroo Stretch
Algebra & Patterns substitutionwork-backward

A sequence \(\langle a_n\rangle\) has \(0

Show answer
Answer: D — 4
Show hints
Hint 1 of 2
Use the recursions to write a3 and a7 in terms of a2 and a1.
Still stuck? Show hint 2 →
Hint 2 of 2
Also a2 itself satisfies a2 = a2·a1 + 1 — that extra equation pins a2 down.
Show solution
Approach: chain the recursion and use a2's own equation
  1. a3 = a2·a1 − 2 and a7 = a2·a3 − 2 = a2^2·a1 − 2a2 − 2 = 2.
  2. From a2 = a2·a1 + 1 we get a1 = 1 − 1/a2; substitute to get a2^2 − 3a2 − 4 = 0.
  3. So (a2−4)(a2+1) = 0; only a2 = 4 keeps 0 < a1 < 1.
  4. Thus a2 = 4.
Mark: · log in to save
Problem 30 · 2017 Math Kangaroo Stretch
Geometry & Measurement Number Theory divisibilityfactor-pairs

The points A and B lie on a circle with centre M. The point P lies on the straight line through A and M. PB touches the circle in B. The lengths of the segments PA and MB are whole numbers, and PB = PA + 6. How many possible values for MB are there?

Figure for Math Kangaroo 2017 Problem 30
Show answer
Answer: D — 6
Show hints
Hint 1 of 2
PB is tangent, so its square equals the product of the whole secant and its external part (power of the point P).
Still stuck? Show hint 2 →
Hint 2 of 2
Turn the relation into MB = 6 + 18/PA and require whole numbers.
Show solution
Approach: use the tangent-secant power of a point, then count integer solutions
  1. Power of the point P: \(PB^2 = PA \cdot (PA + 2\,MB)\), since the secant through A and M has external part PA and crosses the circle again a diameter (2·MB) further on.
  2. With PB = PA + 6: \((PA+6)^2 = PA^2 + 2\,PA\cdot MB\) gives \(12\,PA + 36 = 2\,PA\cdot MB\), so \(MB = 6 + \dfrac{18}{PA}\).
  3. MB is a whole number when PA divides 18: PA ∈ {1, 2, 3, 6, 9, 18}, giving 6 distinct values of MB, choice D.
Mark: · log in to save