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 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
2.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
3.On a 4 by 4 board, how many aligned squares of any size are there, counting 1 by 1, 2 by 2, 3 by 3 and 4 by 4?[5]
A16
B20
C26
D30
E36
4.Five children sit in a row. Two of them are twins who insist on sitting next to each other. In how many orders can the five sit?[5]
A12
B24
C48
D60
E120
5.A box holds 4 red, 5 blue and 6 green balls, mixed up. How many must be taken out without looking to be sure of having 3 of the same colour?[5]
A3
B6
C7
D9
E13
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.How many squares of any size can be found on a 3 by 3 board?[5]
A9
B10
C12
D13
E14
8.A code uses the letters A, B and C, each exactly once. How many codes do not start with A?[5]
A2
B3
C4
D6
E9
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 1 · 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.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.
2.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.
3.D — 30KA-0047Counting by position (sliding window)
AI counted only the sixteen smallest squares and stopped there.
BI counted the small squares and the single big one, and forgot every size in between.
CI counted the 2 by 2 squares as four rather than nine, splitting the board into blocks.
EI used 4 by 4 positions for every size instead of shrinking the range as the square grows.
4.C — 48KA-0070Counting with restrictions
AI treated the twins as one child and forgot they can also swap places with each other.
BI glued the twins together into a single block but then forgot that the block can be arranged with the others in more ways than I counted.
DI halved the 120 total, assuming the twins are together in exactly half the arrangements.
EI counted every arrangement of five children and forgot the twins' condition entirely.
5.C — 7KA-0077The pigeonhole principle
AI answered with the luckiest case, where the first three balls happen to match.
BI found the worst case of two of each colour and forgot to take one more.
DI multiplied the three colours by the three balls I need, rather than thinking about the worst case.
EI worked from the number of balls of each colour rather than from the number of colours.
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.E — 14KA-0133Counting shapes hidden inside a figure
AI counted the nine small squares and stopped, missing every larger one.
BI counted the small squares and the whole board, forgetting the middle size entirely.
CI found the 2 by 2 squares by splitting the board into blocks, getting two instead of four.
DI slid the 2 by 2 square across and down but missed one of its four positions.
8.C — 4KA-0140The multiplication principle
AI counted the codes that do start with A rather than the ones that do not.
BI counted the letters available for the first position instead of counting whole codes.
DI counted every arrangement of the three letters and forgot the restriction.
EI allowed letters to repeat, which the words each exactly once rule out.
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.