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Problem 1 · 2016 Math Kangaroo
Easy
Spatial & Visual Reasoningsymmetry
Which of the following road signs has the most axes of symmetry?
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Answer: C — The no-entry sign.
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Hint 1 of 3
Imagine folding each sign along a straight line so the two halves land exactly on top of each other.
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Hint 2 of 3
Try both a left-right fold and a top-bottom fold on every sign, then count how many folds work.
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Hint 3 of 3
A plain horizontal bar inside a circle matches itself for both folds.
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Approach: fold each sign and count the lines that match
An axis of symmetry is a fold line where one half lands perfectly on the other half.
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.
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.
Two is the most of any sign, so the answer is the no-entry sign, choice (C).
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?
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Answer: E — 12
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Hint 1 of 2
Picture the four pieces side by side, then split each one.
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Hint 2 of 2
Each of the four pieces turns into three pieces.
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Approach: count the pieces in equal groups
After the first cuts there are 4 pieces.
Each of those 4 pieces is split into 3, so you get 4 groups of 3.
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?
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Answer: C — 24
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Hint 1 of 2
Let the middle basket be your unit and write the others in terms of it.
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Hint 2 of 2
Three quantities add to 48 — turn the words into one equation in the middle amount.
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Approach: express all baskets through the middle one
Let the middle basket hold m balls. The smallest holds m/2.
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?
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Answer: B — 60 SOS 09
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Hint 1 of 2
Turning the plate upside down rotates it 180 degrees; which digits still read as valid digits after that?
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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.
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Approach: test each plate for 180-degree symmetry
Rotating 180 degrees, 0 stays 0, 1 stays 1, 8 stays 8, while 6 and 9 swap.
The string must read identically after flipping and reversing order.
Only 60 SOS 09 reads the same upside down, so the answer is B.
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?
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Answer: B — 6.80
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Hint 1 of 2
Only the cents are readable; the whole-euro parts are hidden, but the prices increase down the list.
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Hint 2 of 2
Step down the menu choosing the smallest whole-euro part that keeps each price above the one before it.
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Approach: build the cheapest increasing price chain
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?
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Answer: B — \(2^{90}\cdot 3^{30}\cdot 5^{80}\)
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Hint 1 of 2
Each multiplication by 6 or 10 always adds one factor of 2, plus either a 3 or a 5.
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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.
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Approach: count factors of 2 against 3 and 5
6 = 2·3 and 10 = 2·5, so each step adds exactly one 2 and either one 3 or one 5.
After several steps the power of 2 equals (power of 3) + (power of 5).
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?
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Answer: E — only Bob
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Hint 1 of 2
Carl wears no hat. Use the rule 'if Bob has no hat then Carl wears one' in reverse.
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Hint 2 of 2
Since Carl has no hat, Bob must have one; then check whether Alex is forced.
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Approach: contrapositive reasoning from Carl
Rule: if Bob has no hat, Carl wears one. Carl has no hat, so Bob must have a hat.
Rule: if Alex has no hat, Bob wears one — already satisfied, so Alex is not forced either way.
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?
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Answer: D — 512
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Hint 1 of 2
Put both bugs' positions on the same scale, measured from the left end.
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Hint 2 of 2
Subtract the two positions to get the gap.
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Approach: locate both on [0,1] and subtract
Annie is at 2/3 from the left. Bob, 3/4 from the right, is at 1 − 3/4 = 1/4 from the left.
They have passed each other; the gap is 2/3 − 1/4 = 8/12 − 3/12 = 5/12.
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?
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Answer: C — 12
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Hint 1 of 2
In each block of 5 cubes the colours alternate, so they go grey-white-grey-white-grey.
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Hint 2 of 2
Count the white cubes in one block, then multiply by the number of blocks.
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Approach: count white per block, then scale to all blocks
Because same-colour cubes never touch, each block of 5 alternates colour, giving 3 of one colour and 2 of the other.
Each block ends up with 2 white cubes.
With 6 blocks that is 6 x 2 = 12 white cubes, choice C.
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.)
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Answer: C
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Hint 1 of 3
The piece that is already placed covers part of the shape, so look at the empty gap that is left.
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Hint 2 of 3
Ask whether the other three pieces can fill that gap with no holes and no sticking out.
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Hint 3 of 3
Try each drawing and keep the only one that the pieces fit perfectly.
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Approach: see which outline the leftover three pieces fill exactly
Once the shown piece is set down, the empty space that remains has a fixed shape.
Imagine sliding the other three pieces in like a little jigsaw, covering every square with no gaps and no overlaps.
Only one of the drawings lets all four pieces fit exactly.
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?
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Answer: D — D
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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.
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Hint 2 of 2
Smallest quotient = smallest |y| paired with largest |x| — find that corner.
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Approach: compare y/x at each corner
In the diagram the rectangle has both coordinates negative, so each quotient y : x is positive.
The smallest quotient comes from the corner closest to the x-axis (smallest |y|) and farthest from the y-axis (largest |x|).
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?
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Answer: C — 17
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Hint 1 of 2
Equal face-sums force just two values: the two tips are equal and the three middle corners are equal.
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Hint 2 of 2
Use the two shown numbers to tell which value is a tip and which is a middle corner.
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Approach: exploit equal face sums to find the two values
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.
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).
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?
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Answer: E — 13
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Hint 1 of 2
Leaving 2 over for both groups of three and groups of five means leaving 2 over for groups of fifteen.
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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.
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Approach: use that 2 left over mod 3 and mod 5 means 2 left over mod 15
Leaving 2 over in groups of three and in groups of five means leaving 2 over in groups of fifteen.
So her smallest possible count is 2; to split evenly into both 3s and 5s she needs a multiple of 15.
The next multiple of 15 after 2 is 15, which is 15 − 2 = 13 more marbles.
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?
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Answer: D — in C
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Hint 1 of 3
The best spot has about as many children on one side of the school as on the other side.
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Hint 2 of 3
Start at one end and add up children until you reach more than half of all of them.
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Hint 3 of 3
The village where you pass the halfway count is the best place for the school.
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Approach: walk from one end and stop where you pass half the children
There are 10 + 20 + 30 + 40 = 100 children in all, so half of them is 50.
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.
Since just as many children sit on each side once we reach C, building the school in C makes the total travel smallest.
So the answer is D.
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.
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?
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Answer: D — 1173
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Hint 1 of 2
To make a sum large, put the biggest digits where they count the most.
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Hint 2 of 2
Give the two hundreds places the largest digits, then the tens, then the units.
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Approach: place the largest digits in the highest places
Use 6 and 5 in the hundreds places, 4 and 3 in the tens, 2 and 1 in the units.
The two numbers add to (600 + 500) + (40 + 30) + (2 + 1).
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.
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Answer: A — 2016
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Hint 1 of 2
Group the kangaroos by colour: say r red and g grey, with r + g = 2016.
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Hint 2 of 2
Add the fractions colour by colour and watch the counts cancel.
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Approach: sum by colour
Each red kangaroo contributes g/r and there are r of them, totalling g.
Each grey kangaroo contributes r/g and there are g of them, totalling r.
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.
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?
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Answer: D — 13
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Hint 1 of 3
Every side of the pentagon uses three circles and they all add to the same total.
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Hint 2 of 3
Find a side that already has two numbers filled in to pin down that common total.
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Hint 3 of 3
Then walk around the pentagon, filling each missing circle from the side total until you reach X.
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Approach: find the common side-total, then fill circles one at a time
Call the common total of each side \(S\); a side that already shows two numbers tells you \(S\) once you know the third.
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.
Carrying this around to the marked circle gives \(X = 13\), choice (D).
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?
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Answer: B — 67
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Hint 1 of 3
If 9 squares fit across the 36 dm width, work out how wide one square is.
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Hint 2 of 3
Use that square size to find how many squares fit along the 60 dm length.
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Hint 3 of 3
Multiply rows by columns for the total squares, then the sun/moon pattern splits them almost in half.
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Approach: find the grid size, then count one colour
36 dm ÷ 9 = 4 dm per square, so the length 60 dm holds 60 ÷ 4 = 15 squares.
The rug is 9 × 15 = 135 squares in a checkerboard of suns and moons.
The moons fall on the smaller of the two colour groups, giving 67 moons.
Each of the four balls weighs either 10, 20, 30 or 40 grams. The two balances are shown. Which ball weighs 30 grams?
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Answer: C — C
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Hint 1 of 3
The four weights 10, 20, 30, 40 add up to 100 grams altogether — keep that total handy.
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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.
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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.
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Approach: use the level scale to say C equals two other balls added, then test which weight that can be
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.
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).
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.
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
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Answer: B — -2
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Hint 1 of 3
By Vieta, the sum of the squares of the roots is \((\text{sum})^2 - 2(\text{product})\).
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Hint 2 of 3
Set the two sums-of-squares equal, then factor the resulting symmetric equation.
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Hint 3 of 3
The condition \(a \ne b\) lets you cancel one factor.
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Approach: Vieta plus factoring
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 \(b^2 - 2a\).
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?
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Answer: A — 13
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Hint 1 of 2
A point that splits a side in a ratio splits a triangle's area in the same ratio (same height).
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Hint 2 of 2
Stretch triangle ABN up to ABC using BN:NC, and double CKD up to ACD using the midpoint K.
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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).
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?
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Answer: C — 75 km/h
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Hint 1 of 2
The car leaving at hour k has speed 50 + k and, by 100 hours, has driven for 100 − k hours.
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Hint 2 of 2
Maximise the distance (50 + k)(100 − k) over k from 0 to 50.
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Approach: maximise distance travelled at t = 100
The car leaving at hour k (speed 50 + k) has travelled (50 + k)(100 − k) by t = 100.
This product is largest at k = 25, giving 75 × 75 = 5625.
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?
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Answer: D — 6
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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).
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Hint 2 of 2
Turn the relation into MB = 6 + 18/PA and require whole numbers.
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Approach: use the tangent-secant power of a point, then count integer solutions
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.
With PB = PA + 6: \((PA+6)^2 = PA^2 + 2\,PA\cdot MB\) gives \(12\,PA + 36 = 2\,PA\cdot MB\), so \(MB = 6 + \dfrac{18}{PA}\).
MB is a whole number when PA divides 18: PA ∈ {1, 2, 3, 6, 9, 18}, giving 6 distinct values of MB, choice D.
