AMC WARM-UP PAPER

UPPER PRIMARY 8
SOLUTIONS

2009
c Australian Mathematics Trust

1. To walk around the edges will be 30 m + 40 m = 70 m.

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To walk diagonally is 50 m.
So walking diagonally is (70-50) m = 20 m shorter,
hence (B).

2. Since 3 2 5 + 6 = 21, we get that 3 2 5 = 15, and since 3 × 5 = 15, 2 must be ×,

hence (D).

3. Brett is 12 years old. Daina is half Brett’s age so is 6 years old. Omar is 13 years
older than Daina so is 6 + 13 = 19 years old,
hence (D).
 UP8 Solutions Page 2

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We can see that the Walk plus the Cycle parts are less than a semicircle, so (A) is
false.
The Car sector is less than a semicircle, so (B) is false.
The Walk sector is lass than a quarter-circle, so (C) is false.
The Bus sector plus the Cycle sector is greater than a semicircle, so (D) is true.
The Walk sector is less than the Car sector, so (E) is false.
The only correct alternative is (D),
hence (D).

5. Two numbers multiplying to get 60 could be 60 and 1; 30 and 2; 20 and 3; 15 and

4; 12 and 5 and 10 and 6. The differences between the numbers in these pairs is
59, 28, 17, 11, 7 and 4. The only one in the alternatives is 11,
hence (E).

6. Using the three cards, the numbers in order are:-

579; 597; 759; 795; 957; 975.
So the number 795 is in the fourth position,
hence (C).

7. The green block is above the red and so also above the purple. It is
also below the blue and hence also the orange. This means that the O
green block is in the middle and the order from the top is orange, blue,
green, red and purple, so the bottom block is purple, B
hence (E).
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 UP8 Solutions Page 3

2 1 3
8. Spinner (A) has = shaded, spinner (B) has shaded,
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2 1 1 1
spinner (C) has = shaded, spinner (D) has shaded and spinner (E) has
4 2 8 3
shaded, so the spinner with a 1 in 4 chance is (A),
hence (A).

9. Since the large square contains two small squares with areas 25 square centimetres
and 4 square centimetres, the sides of these squares are 5 cm and 2 cm as shown.
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7
This means that the side of the largest square is 7 cm and we can place the other
measurements, in centimetres, as shown.
The perimeter of the shaded region is then 2 + 5 + 7 + 2 + 5 + 3 = 24 cm,
hence 24.

10. There are 15 girls in the class and a smaller number of boys.
Since there are a smaller number of boys than girls in the class, the number of
students in the class must be somewhere from 16 to 29.
When they are split into 4 there are 2 students left, so the possibilities for the
number of students are:- 18, 22 and 26.
When they are split into groups of 5 there is 1 left, so the possibilities for the
number of students are :- 16, 21 and 26.
The only number satisfying both conditions is 26, so there are 26 students in the
class and so 26 − 15 = 11 boys.