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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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4 on the sheet, 4 match these filtersReshuffle Start overseed 9

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Math Kangaroo practice

Name: ______________________Date: ____________4 questions · 18 points

1.A committee of 4 must be formed from 5 boys and 4 girls, and it must contain exactly 2 boys and 2 girls. How many different committees are possible?[4]
- A60
- B126
- C20
- D240
- E40
2.Each of the four edges of a square is painted either black or white. Two paintings count as the same if one can be rotated onto the other. How many genuinely different paintings are there?[4]
- A6
- B16
- C4
- D8
- E5
3.What is the smallest number n such that ANY collection of n whole numbers must contain two of them whose difference is divisible by 7?[5]
- A8
- B7
- C14
- D15
- E4
4.How many diagonals does a convex polygon with 12 sides have? A diagonal joins two vertices that are not already joined by a side.[5]
- A54
- B66
- C108
- D120
- E42

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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.A — 60KA-0172Selections of a few objects
- BI chose any 4 from all 9 people and ignored the two-and-two requirement.
- CI added the two counts instead of multiplying them.
- DI treated the choices as ordered, counting the same committee several times.
- EI used 5 x 4 x 2 or a similar shortcut rather than counting each selection properly.
2.A — 6KA-0173Counting up to symmetry
- BI counted every colouring of the four edges and forgot that rotations make some of them identical.
- CI divided 16 by 4, but that only works when no painting is left unchanged by a rotation, and some are.
- DI halved 16, treating only the 180 degree turn as a symmetry.
- EI listed by how many edges are black but forgot that two black edges can be adjacent or opposite.
3.A — 8KA-0175The pigeonhole principle
- BI used the number of possible remainders without adding one for the pair that must collide.
- CI doubled 7, thinking I needed two full sets of remainders.
- DI doubled 7 and added one, applying the pigeonhole idea to the wrong number of boxes.
- EI guessed from small cases without identifying what the boxes actually are.
4.A — 54KA-0177Overcount, then correct
- BI counted every line joining two vertices and forgot to remove the 12 sides.
- CI counted each diagonal from both of its endpoints and forgot to halve.
- DI used 12 x 10 without halving, double counting every diagonal.
- EI subtracted 24 rather than 12, removing each side twice.

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