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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.Suppose it is true that some birds cannot fly. Which of these must also be true?[3]
ANo birds can fly
BAll birds can fly
CMost birds cannot fly
DExactly one bird cannot fly
ENot all birds can fly
3.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
4.Ana says: "Bo and I are both liars." Each child is either always truthful or always a liar. Who is telling the truth?[4]
AAna only
BBo only
Cboth of them
Dneither of them
Eit cannot be decided
5.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
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 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.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
9.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
10.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
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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.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.E — Not all birds can flyKA-0146Reading a logical statement precisely
AI read some cannot as meaning none can, which is a far stronger claim.
BI chose a statement that directly contradicts the one I was given.
CI read some as meaning a majority, when it only guarantees at least one.
DI read some as meaning exactly one, when it means at least one.
3.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.
4.B — Bo onlyKA-0096Truth-tellers and liars
AI took Ana's statement at face value without checking whether a truthful person could say it.
CI decided nobody was lying without testing Ana's statement against that assumption.
DI believed Ana's statement even after concluding she was a liar, which is what a liar's statement cannot be.
EI gave up when Ana's statement looked circular, instead of testing one assumption all the way through.
5.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.
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 — 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 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.
9.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.
10.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.