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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.How many two-digit numbers have digits that add up to 8?[5]
A4
B7
C8
D9
E16
2.On the street grid shown you may only walk right or down. One junction is closed. How many routes go from the top-left corner to the bottom-right corner?[5]
Text description of the figure
A grid of streets with junctions arranged 4 across and 4 down. The junction one step right and one step down from the top-left corner is marked closed with a cross.
A8
B10
C12
D18
E20
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.How many three-digit whole numbers contain at least one digit 7?[5]
A243
B252
C271
D280
E648
5.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
6.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
7.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
8.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
9.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
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 8 · 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 — 8KA-0012Casework
AI counted each pair of digits once instead of counting both orders, such as 17 and 71.
BI listed the pairs starting from 1 and 7 and forgot the number 80, where the second digit is zero.
DI included 08 as a two-digit number, but a two-digit number cannot start with zero.
EI counted both orders of every pair and then counted the pairs that reverse to themselves twice as well.
2.A — 8KA-0031Counting paths on a grid
BI assumed closing one junction removes about half the routes and halved the total of 20.
CI counted the routes that pass through the closed junction and gave that instead of the ones that avoid it.
DI subtracted only the 2 routes that reach the closed junction, not all the routes that continue through it.
EI counted every route on the open grid and forgot to remove the ones through the closed junction.
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.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.
5.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.
6.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.
7.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.
8.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.
9.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.
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.