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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.Toma spent 4 euros, then spent half of what was left, and now has 6 euros. How much did she start with?[3]
A10
B14
C16
D20
E24
2.Ana, Bo and Cal each keep a different pet: a cat, a dog and a fish. Ana does not keep the cat. Bo keeps neither the cat nor the fish. Who keeps the cat?[4]
AAna
BBo
CCal
Dit cannot be decided
Eeither Ana or Cal
3.Cards numbered 1 to 10 lie face up. You take any six of them. Must two of your cards add up to 11?[4]
AYes, always
BYes, but only if you take the card numbered 1
CNo, six cards can be chosen that avoid it
DOnly if the six numbers are consecutive
EIt cannot be decided without knowing the cards
4.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
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.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
7.One hundred apples are packed into twelve boxes. What is the largest number n for which you can always be sure that some box holds at least n apples?[5]
A8
B9
C10
D12
E100
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.Four children stand in a line. Ana is not first. Bo stands directly behind Ana. Cal is last. Who is first?[5]
AAna
BBo
CCal
DDee
Eit cannot be decided
10.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
Math Kangaroo Star · https://kangaroo-atlas.vercel.app · seed 10 · 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.C — 16KA-0050Working backwards
AI added the 4 and the 6 and stopped, forgetting to undo the halving.
BI doubled the 6 first and then added the 4, reversing the steps in the wrong order.
DI doubled the total at the end rather than doubling only the amount left after the first spend.
EI doubled twice because spending half felt like it needed undoing more than once.
2.C — CalKA-0091Elimination grids
AI used Bo's clue and forgot the very first clue, which rules Ana out directly.
BI read Bo's clue as telling me what Bo has rather than what Bo does not have.
DI stopped after using each clue once, without going back to see what the ticks had ruled out.
EI applied Bo's clue but never came back to Ana's, so I left two people in the running.
3.A — Yes, alwaysKA-0102Informal proof by contradiction
BI found one pair that works and assumed the argument depended on that particular card.
CI tried a couple of selections, did not find a pair, and stopped looking.
DI looked for a pattern in the numbers rather than at how many pairs there are to avoid.
EI thought the answer depended on which six were taken, when the counting settles it for every choice.
4.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.
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.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.
7.B — 9KA-0095The extremal principle
AI divided 100 by 12 and rounded down instead of up.
CI rounded the division to a convenient number without checking that a packing with a smaller maximum exists.
DI gave the number of boxes as the answer rather than a number of apples.
EI described the case where one box holds everything, which is possible but not guaranteed.
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.D — DeeKA-0136Ordering and ranking from clues
AI ignored the very first clue, which rules Ana out of the front directly.
BI put Bo at the front, but Bo must have Ana immediately in front of him.
CI forgot that Cal is fixed at the back of the line.
EI stopped once two arrangements seemed possible, without testing whether the second one actually fits every clue.
10.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.