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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.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.Seven cups all stand upside down. In one move you must turn over exactly two cups. Can all seven ever stand the right way up?[5]
AYes, in 4 moves
BYes, in 7 moves
CYes, but it takes many moves
DIt depends which two cups you pick
ENo, it is impossible
4.What is the units digit of 7 multiplied by itself 2026 times, that is 7 to the power 2026?[5]
Ait cannot be found without a calculator
B1
C3
D7
E9
5.A bag holds 5 black and 6 white stones. You repeatedly remove two stones: if they match you put a black one in, if they differ you put a white one in. What colour is the last stone?[5]
Ablack
Bwhite
Cit depends on the order of the moves
Dthe bag never gets down to one stone
Ewhite if the first two stones match
6.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
7.A circle of radius 5 is drawn inside a square of side 10, touching all four sides. What fraction of the square is outside the circle, to the nearest tenth?[5]
Text description of the figure
A square with a circle drawn inside it, the circle touching the midpoint of each of the four sides. The region inside the square but outside the circle is shaded.
A0.1
B0.2
C0.3
D0.5
E0.8
8.How many three-digit whole numbers contain at least one digit 7?[5]
A243
B252
C271
D280
E648
9.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
10.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
11.How many squares of any size can be found on a 3 by 3 board?[5]
A9
B10
C12
D13
E14
12.A square sheet is folded in half, then in half again. Two holes are punched through all the layers. How many holes are there when it is unfolded?[5]
A2
B4
C6
D8
E16
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.The numbers 1 to 10 are written on a board. You repeatedly rub out any two of them and write down their positive difference instead, until a single number is left. What can be said about that final number?[5]
AIt is always odd
BIt is always even
CIt can be either odd or even
DIt is always zero
EIt is always 1
15.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
Math Kangaroo Star · 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.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.E — No, it is impossibleKA-0028Parity arguments
AI found a sequence that turned over most of the cups and assumed the last one could be fixed somehow.
BI matched the number of moves to the number of cups without checking whether the target is reachable at all.
CI assumed that with enough moves any arrangement can be reached.
DI thought the choice of which cups to flip could change whether the target is reachable.
4.E — 9KA-0037Last-digit behavior of products and powers
AI assumed a number this large has no reachable last digit and gave up on the cycle.
BI found the cycle 7, 9, 3, 1 but used a remainder of 0 when the remainder is actually 2.
CI found the cycle but counted its positions starting at zero, shifting my answer along by one.
DI assumed the units digit of any power of 7 is always 7.
5.A — blackKA-0043Invariants and monovariants
BI noticed there are more white stones than black ones and guessed the majority colour would survive.
CI tried a few orders, saw different-looking positions along the way, and assumed the ending must vary too.
DI did not notice that every move takes two stones out and puts one back, so the count falls by exactly one each time.
EI assumed the first move settles the outcome instead of looking for a quantity that no move can change.
6.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.
7.B — 0.2KA-0052Area by subtraction (shaded regions)
AI estimated by eye from the four corner pieces without comparing them to the whole square.
CI used a diameter of 5 instead of a radius of 5 when working out the circle's area.
DI assumed the circle covers half the square because it touches all four sides.
EI found the fraction covered by the circle and gave that instead of the fraction outside it.
8.B — 252KA-0075Complementary counting
AI counted the numbers made entirely of digits other than 7 in every position, including a leading zero.
CI counted the numbers with a 7 in each position separately and forgot that some were counted twice.
DI added three lots of 90 and one extra hundred, double counting the 700s.
EI counted the numbers with no 7 at all and gave that instead of subtracting it.
9.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.
10.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.
11.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.
12.D — 8KA-0134Paper folding and hole punching
AI forgot that the punch goes through every layer, not just the top one.
BI undid only one of the two folds before counting.
CI doubled twice for the folds but added the second hole instead of doubling for it too.
EI counted three folds instead of two, doubling one time too many.
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 — It is always oddKA-0167Parity arguments
BI checked the parity of the count of numbers rather than of their sum.
CI tried two examples, got different answers, and concluded nothing is forced.
DI assumed differences must shrink all the way to nothing.
EI found one sequence of moves ending at 1 and assumed every sequence must.
15.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.