Pick what you want, print the page. The answer key comes with it, on its own sheet, and it tells you the mistake behind every wrong choice — not just which letter is right.
This exact URL always produces this exact sheet — bookmark it and the answer key will still match next term. Print with your browser; the controls and the site navigation are left off the page.
1.What is the remainder when 7 to the power 100 is divided by 5?[3]
A1
B2
C3
D4
E0
2.What is the size of one interior angle of a regular twelve-sided polygon?[3]
A150 degrees
B30 degrees
C165 degrees
D1800 degrees
E144 degrees
3.A sequence is defined by a(1) = 2 and a(n+1) = 2 a(n) - 1 for every n. What is a(6)?[3]
A33
B31
C64
D17
E63
4.A committee of 4 must be formed from 5 boys and 4 girls, and it must contain exactly 2 boys and 2 girls. How many different committees are possible?[4]
A60
B126
C20
D240
E40
5.Each of the four edges of a square is painted either black or white. Two paintings count as the same if one can be rotated onto the other. How many genuinely different paintings are there?[4]
A6
B16
C4
D8
E5
6.A driver travels from one town to another at an average of 60 km/h and returns along the same road at an average of 40 km/h. What is the average speed for the whole journey?[4]
A48 km/h
B50 km/h
C52 km/h
D24 km/h
E100 km/h
7.What is the smallest number n such that ANY collection of n whole numbers must contain two of them whose difference is divisible by 7?[5]
A8
B7
C14
D15
E4
8.An 8 by 8 chessboard has two opposite corner squares removed, leaving 62 squares. Each domino covers exactly two squares that share an edge. Can the 62 squares be covered exactly by 31 dominoes?[5]
ANo, because the two removed corners share a colour, so 32 squares of one colour remain and only 30 of the other
BYes, because 62 is even and 31 dominoes cover exactly 62 squares
CNo, because 62 is not divisible by 4
DYes, but only if the dominoes may be placed diagonally
ENo, because the board is no longer rectangular
9.How many diagonals does a convex polygon with 12 sides have? A diagonal joins two vertices that are not already joined by a side.[5]
A54
B66
C108
D120
E42
Kangaroo Atlas · https://kangaroo-atlas.vercel.app · seed 3 · CC BY-NC-SA 4.0 · Independent project, not affiliated with Math Kangaroo in USA, NFP.
Answer key — Math Kangaroo practice
Each wrong choice carries the thinking that produces it. When a student picks C, this is what they were doing.
1.A — 1KA-0169Remainders and modular cycles
BI reduced 7 to 2 correctly but then used a remainder of 1 when dividing 100 by 4.
CI read the cycle backwards and landed on the entry before the right one.
DI used 100 divided by 4 giving 25 and read off the 25th entry as if the cycle had length 25.
EI assumed a power that large must be divisible by 5.
2.A — 150 degreesKA-0170Angles on lines, in triangles, in polygons
BI found the exterior angle and answered with that instead of the interior one.
CI used 360 divided by 12 subtracted from 195, or otherwise slipped a step in the arithmetic.
DI gave the total of all twelve interior angles rather than one of them.
EI used a ten-sided polygon by mistake.
3.A — 33KA-0171Recursive rules
BI used the rule 2a(n) - 1 but started the sequence at 1 instead of 2.
CI doubled six times and ignored the minus one entirely.
DI stopped at a(5), one term early.
EI computed 2^6 - 1, applying the minus one only once at the very end.
4.A — 60KA-0172Selections of a few objects
BI chose any 4 from all 9 people and ignored the two-and-two requirement.
CI added the two counts instead of multiplying them.
DI treated the choices as ordered, counting the same committee several times.
EI used 5 x 4 x 2 or a similar shortcut rather than counting each selection properly.
5.A — 6KA-0173Counting up to symmetry
BI counted every colouring of the four edges and forgot that rotations make some of them identical.
CI divided 16 by 4, but that only works when no painting is left unchanged by a rotation, and some are.
DI halved 16, treating only the 180 degree turn as a symmetry.
EI listed by how many edges are black but forgot that two black edges can be adjacent or opposite.
6.A — 48 km/hKA-0174Rate, time, distance
BI averaged the two speeds, which would only be right if equal TIME were spent at each, not equal distance.
CI weighted the two speeds by their own values instead of by the time each leg took.
DI halved the correct answer, dividing by the total distance twice.
EI added the two speeds together, which gives a number faster than either leg and so cannot be an average of them.
7.A — 8KA-0175The pigeonhole principle
BI used the number of possible remainders without adding one for the pair that must collide.
CI doubled 7, thinking I needed two full sets of remainders.
DI doubled 7 and added one, applying the pigeonhole idea to the wrong number of boxes.
EI guessed from small cases without identifying what the boxes actually are.
8.A — No, because the two removed corners share a colour, so 32 squares of one colour remain and only 30 of the otherKA-0176Coloring arguments
BI checked that the counts match but a matching count does not make a covering possible.
CI invented a divisibility condition; dominoes cover two squares, so only divisibility by 2 could matter.
DI changed the rules rather than testing them; the question says dominoes cover squares sharing an edge.
EI appealed to the shape, but plenty of non-rectangular regions can be tiled by dominoes.
9.A — 54KA-0177Overcount, then correct
BI counted every line joining two vertices and forgot to remove the 12 sides.
CI counted each diagonal from both of its endpoints and forgot to halve.
DI used 12 x 10 without halving, double counting every diagonal.
EI subtracted 24 rather than 12, removing each side twice.