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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.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
3.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
4.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
5.One of nine identical-looking coins is slightly heavier. Using only a balance, what is the smallest number of weighings that is certain to find it?[5]
A1
B2
C3
D4
E8
6.The numbers 1 to 10 stand in a row. A move swaps two neighbours. After exactly 45 moves, can the row be back in its starting order?[5]
AYes, always
BYes, if the swaps are chosen well
CNo, because 45 is not a multiple of 10
DNo, because the number of moves is odd
EIt depends which numbers are swapped
7.The numbers 1 to 8 are on a board. A move rubs out two of them and writes their difference, larger minus smaller. After seven moves one number is left. Can it be 1?[5]
AYes, and there are many ways
BYes, but only one way
CNo, the last number is always even
DNo, the last number is always 0
EIt depends on the order of the moves
8.A class of 12 scored an average of 8. Another class of 18 scored an average of 13. What is the average for all 30 students together?[5]
A10.5
B11
C11.5
D12
E21
9.What is the value of (1 - 1/2) x (1 - 1/3) x (1 - 1/4) x ... x (1 - 1/10)?[5]
A1/10
B1/9
C1/2
D9/10
E1
10.You are 20 minutes into a 75-minute paper of 30 questions, still on question 9, and four minutes in with no progress. What is the best move?[5]
AKeep going — four minutes are already invested
BAnswer it with your best guess, mark it, and move on
CSkip it, leave it blank, and come back at the end
DGo back and re-check questions 1 to 8
EJump to question 30 and work backwards
11.Squares of dots grow: 1 dot, then 4, then 9, then 16. How many dots are in the seventh square?[5]
A25
B36
C42
D49
E64
12.What is the last digit of 3 multiplied by itself 2026 times, that is, of 3 to the power 2026?[5]
A3
B9
C7
D1
E6
13.One tap fills a tank in 6 hours. A second tap fills the same tank in 4 hours. Both taps are opened together on an empty tank. How long does it take to fill?[5]
A2 hours 24 minutes
B5 hours
C10 hours
D2 hours
E1 hour 12 minutes
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
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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 — 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.
3.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.
4.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.
5.B — 2KA-0094Weighing and balance puzzles
AI assumed one weighing could separate nine possibilities, but it has only three outcomes.
CI split the coins into halves each time, which wastes the balance's third outcome.
DI weighed the coins one against another in pairs rather than in groups.
EI compared each coin with a known good one in turn, which always works but is nowhere near the fewest.
6.D — No, because the number of moves is oddKA-0097Parity arguments
AI assumed enough moves can undo anything, without asking what each move preserves.
BI tried a few sequences that nearly worked and assumed a better choice would finish the job.
CI reached for the number of items rather than for what a single swap actually changes.
EI thought the choice of swaps could change the outcome, when every swap has the same effect.
7.C — No, the last number is always evenKA-0098Invariants and monovariants
AI found sequences ending in small numbers and assumed 1 was among the reachable ones.
BI assumed a hard-looking question must have exactly one answer.
DI found one sequence ending in 0 and assumed every sequence must end there.
EI thought the order could change the outcome, when the quantity that decides it never changes.
8.B — 11KA-0112Mixtures and weighted averages
AI averaged the two averages, ignoring that the classes are different sizes.
CI weighted the averages but used the wrong class size against each one.
DI leaned towards the larger class's average without computing the totals.
EI added the two averages together instead of combining them.
9.A — 1/10KA-0125Spot the trick vs. decide to grind
BI cancelled the chain but stopped one factor early, leaving a 9 on the bottom.
CI worked out the first bracket and assumed the rest made no difference.
DI gave the last bracket on its own instead of the whole product.
EI assumed everything cancels completely, leaving nothing behind.
10.B — Answer it with your best guess, mark it, and move onKA-0126Time triage: now, later, or guess
AI let the time already spent decide, but that time is gone whatever I do next.
CI moved on but left a blank, which scores nothing if I never get back to it.
DI spent time re-reading work I had no reason to doubt, while 21 unseen questions waited.
EI assumed the hardest questions are the best use of time, when the unseen easy ones are worth the same each.
11.D — 49KA-0138Growth patterns and figurate numbers
AI continued the pattern to the fifth square instead of the seventh.
BI stopped one square short, giving the sixth instead of the seventh.
CI added 6 to the sixth square's 36, treating the differences as constant.
EI continued one square too far and gave the eighth.
12.B — 9KA-0158Last-digit behavior of products and powers
AI used the first entry of the cycle because 2026 is even, without working out where in the cycle it lands.
CI counted the cycle starting from zero, so a remainder of 2 put me one place too far along.
DI assumed a power that large must end in 1.
EI multiplied 3 by 2 and used the last digit of that.
13.A — 2 hours 24 minutesKA-0159Combined work rates
BI averaged the two times, but two taps together must be faster than either one alone.
CI added the two times, which would be right for filling two tanks one after the other.
DI halved the smaller of the two times, guessing that two taps must simply be twice as fast as one.
EI found 2 hours 24 minutes correctly and then halved it a second time.
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.