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Worksheet generator
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.
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1.Eight people meet and everyone shakes hands with everyone else exactly once. How many handshakes take place?[3]
A16
B28
C36
D56
E64
2.A rectangle is divided by two vertical lines into three parts. How many rectangles of any size are in the picture?[3]
Text description of the figure
A wide rectangle divided by two vertical lines into three smaller rectangles side by side.
A3
B4
C6
D7
E9
3.How many different two-digit numbers can be written using only the digits 1, 2 and 3, if a digit may be used twice?[3]
A3
B6
C9
D12
E27
4.How many three-digit numbers can be written using only the digits 1 and 2, with repeats allowed?[3]
A2
B3
C4
D6
E8
5.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
6.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
7.How many two-digit numbers have digits that add up to 8?[5]
A4
B7
C8
D9
E16
8.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
9.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
10.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
11.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
12.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
13.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
14.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
15.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
Math Kangaroo Star · https://kangaroo-atlas.vercel.app · seed 8 · 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.B — 28KA-0035Handshakes and pairs
AI doubled the number of people, as if each person made two handshakes in total.
CI let each person shake hands with all eight people including themselves before halving.
DI multiplied 8 by 7 and forgot that each handshake was counted from both sides.
EI multiplied 8 by 8, counting every ordered pair including a person with themselves.
2.C — 6KA-0054Counting shapes hidden inside a figure
AI counted only the three small rectangles drawn as separate cells and stopped there.
BI counted the three small rectangles and the whole one, and forgot the two made of two cells.
DI counted a rectangle made of the left and right cells, even though they are not next to each other.
EI assumed any pair of the four vertical lines makes a rectangle without checking each one.
3.C — 9KA-0072Systematic listing
AI counted only the numbers with two identical digits, like 11, 22 and 33.
BI required the two digits to be different, even though repeats are allowed.
DI allowed a fourth digit that was not in the list.
EI counted three-digit numbers instead of two-digit ones.
4.E — 8KA-0143Systematic listing
AI counted the two available digits rather than the numbers they can build.
BI counted the three positions instead of the numbers.
CI used only two of the three positions when multiplying.
DI multiplied 2 by 3, mixing the number of digits with the number of positions.
5.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.
6.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.
7.C — 8KA-0012Casework
AI counted each pair of digits once instead of counting both orders, such as 17 and 71.
BI listed the pairs starting from 1 and 7 and forgot the number 80, where the second digit is zero.
DI included 08 as a two-digit number, but a two-digit number cannot start with zero.
EI counted both orders of every pair and then counted the pairs that reverse to themselves twice as well.
8.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.
9.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.
10.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.
11.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.
12.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.
13.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.
14.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.
15.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.