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1.Five children queue up. Ana is somewhere ahead of Bo. Cal is behind Bo but ahead of Dee. Eve is last. Who is second in the queue?[4]
AAna
BBo
CCal
DDee
Eit cannot be decided
2.Two apples balance three pears. One apple weighs 90 grams. How many grams does one pear weigh?[4]
A45
B60
C90
D135
E180
3.A number is doubled, then 6 is added, then the result is halved. The answer is 11. What was the original number?[4]
A4
B8
C11
D16
E22
4.Ana says "Bo is lying." Bo says "Cal is lying." Cal says "Ana and Bo are both lying." How many of the three are telling the truth?[5]
A0
B1
C2
D3
Eit cannot be decided
5.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
6.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
7.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
8.Can a 10 by 10 board be covered exactly by T-shaped tiles of four squares each, with no gaps and no overlaps?[5]
AYes, and it is straightforward
BYes, but the arrangement is fiddly
CNo, because 100 is not a multiple of 4
DNo, because the two colours cannot balance
EIt depends how the tiles are turned
9.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
10.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
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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.B — BoKA-0092Ordering and ranking from clues
AI gave the child who is first rather than the one who is second.
CI placed Cal directly after Ana, reading ahead of Dee as meaning immediately ahead.
DI built the order backwards from the last position instead of forwards from the first.
EI assumed the clues left several orders open without checking that they chain into a single one.
2.B — 60KA-0093Weighing and balance puzzles
AI halved the apple's weight, as if one apple balanced two pears.
CI assumed a pear must weigh the same as an apple because the two sides balance.
DI multiplied the apple's weight by three halves the wrong way round, making the pear heavier.
EI gave the total weight of the two apples rather than the weight of one pear.
3.B — 8KA-0100Working backwards
AI undid the operations in the order they were written instead of in reverse order.
CI gave the final answer back, assuming the three steps cancelled each other out.
DI doubled the 11 and subtracted the 6 but forgot the final halving that undoes the doubling.
EI undid only the halving and stopped there without touching the other two steps.
4.B — 1KA-0017Truth-tellers and liars
AI assumed everyone could be lying at once without checking that Cal's statement would then be true.
CI found one consistent truth-teller and added another without testing whether both could hold together.
DI assumed everyone was telling the truth without noticing that Ana's statement then contradicts Bo's.
EI gave up after one assumption led to a contradiction, instead of trying the other assumption.
5.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.
6.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.
7.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.
8.D — No, because the two colours cannot balanceKA-0101Coloring arguments
AI checked that 100 divides by 4 and treated that as proof that a covering exists.
BI assumed a covering must exist somewhere and that I simply had not found it yet.
CI gave a reason that is not even true, since 100 really is a multiple of 4.
EI thought orientation could rescue it, but every turn of a T covers the same mixture of colours.
9.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.
10.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.