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.A club has 8 members. In how many ways can a group of 3 be chosen to attend a conference, if the three places are all the same?[4]
A56
B336
C24
D112
E28
2.A three-digit code uses only the digits 1, 2 and 3, and digits may repeat. How many such codes contain at least one 3?[5]
A8
B9
C12
D19
E27
3.Four beads - one red, one green, one blue, one yellow - are threaded on a circular bracelet. Two bracelets are the same if one can be rotated into the other. How many different bracelets are there?[5]
A3
B4
C6
D12
E24
4.A drawer holds 10 red socks, 10 blue socks and 10 green socks, all mixed up in the dark. How many socks must you take out to be sure of having a matching pair?[5]
A2
B3
C4
D11
E31
5.Five beads of five different colours are threaded on a circular bracelet. Two bracelets count as the same if one can be turned or flipped into the other. How many different bracelets are there?[5]
A12
B20
C24
D60
E120
6.Ten different whole numbers, each bigger than zero, add up to 100. What is the largest that the biggest of them can be?[5]
A45
B50
C55
D91
E100
7.A cube 3 units on each side is painted all over and then cut into unit cubes. How many of them have exactly two painted faces?[5]
A6
B8
C12
D24
E27
8.An ordinary six-sided die is rolled three times. What is the probability that at least one roll shows a six?[5]
A91/216
B1/2
C125/216
D3/216
E1/6
9.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
10.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 2 · 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 — 56KA-0163Selections of a few objects
BI counted ordered selections, so I counted the same three people once for every order they could stand in.
CI multiplied the 8 members by the 3 places, which counts something quite different from a selection.
DI divided the 336 ordered selections by 3 instead of by 3 factorial.
EI chose 2 people rather than 3.
2.D — 19KA-0018Complementary counting
AI counted the codes that avoid 3 entirely and gave that as my answer without subtracting.
BI counted the codes with a 3 in the first position only and forgot the other two positions.
CI counted the codes with exactly one 3 and forgot the ones with two or three of them.
EI counted every possible code and forgot to remove the ones with no 3 at all.
3.C — 6KA-0025Counting up to symmetry
AI also treated bracelets that are mirror images as the same, but only rotations were allowed.
BI divided the number of beads by itself rather than dividing the number of arrangements by the number of rotations.
DI divided the 24 arrangements by 2 instead of by the 4 rotations of a circle of four beads.
EI counted every arrangement in a line and forgot that rotating a bracelet gives the same bracelet.
4.C — 4KA-0040The pigeonhole principle
AI answered with the lucky case, where the first two socks happen to match.
BI used the number of colours as my answer without adding one for the sock that must repeat.
DI worked from the number of socks of each colour instead of the number of colours.
EI took the whole drawer, guaranteeing a pair but far more socks than are needed.
5.A — 12KA-0071Overcount, then correct
BI divided the 120 arrangements by 6 rather than by the 10 movements that leave a bracelet looking the same.
CI allowed for turning the bracelet but forgot it can also be flipped over.
DI halved the 120 for flipping but forgot that turning also gives the same bracelet.
EI counted every arrangement in a line, as if the bracelet had a fixed first bead.
6.C — 55KA-0078Extremal / worst-case counting
AI found the smallest possible total of the other nine and gave that instead of what is left.
BI assumed the biggest number could be at most half the total.
DI made the other nine numbers all equal to 1, forgetting that they have to be different from each other.
EI ignored the other nine numbers entirely, as if the biggest could take the whole total.
7.C — 12KA-0090Counting cubes in a stack, including hidden ones
AI counted the middle cube of each face, which has exactly one painted face rather than two.
BI counted the corner cubes, which have three painted faces.
DI counted every cube that is painted at all except the corners, without separating one face from two.
EI gave the total number of small cubes rather than the ones with exactly two painted faces.
8.A — 91/216KA-0168Complementary counting
BI added 1/6 three times, which counts the overlapping cases more than once and would exceed 1 for seven rolls.
CI worked out the probability of NO six and forgot to subtract it from 1.
DI found the probability of three sixes instead of at least one.
EI gave the probability for a single roll and ignored that there are three.
9.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.
10.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.