This document contains solutions to an intermediate math paper with 4 questions: 1) The solution is 100, hence choice C. 2) By calculating fractions, the solution is 3, hence choice B. 3) With 4 more girls, the new fraction is 18/34, hence choice E. 4) In an isosceles triangle with angles of 25° and 130°, the third angle must be 120°, hence the solution.

1. AMC WARM-UP PAPER
INTERMEDIATE PAPER 1
SOLUTIONS
c 2009 Australian Mathematics Trust
20 200
1. = = 100,
0.2 2
hence (C).
2. 1 1 1 1 1 1 5 3
+ + + − + + +
2 4 5 6 5 2 6 4
1 1 1 3 1 1 1 5
= + + + + − + +
2 2 4 4 5 5 6 6
= 1+1+0+1
= 3,
hence (B).
18 9
3. With 4 more girls, the fraction becomes = ,
34 17
hence (E).
P .
..
...
.. . ..
4. In the isosceles triangle P SR, RP S = 25◦ and ...
...
... . .
... . ...
... . ..
x◦
.
.
. ..
.... ..
.
...
... ....
. ..
P SR = 130◦ . ..
. .
. ◦..
..
..
...
... 25 .
.
.
.
. ...
.
...
... S 130 .
.
. ◦... ..
...
...
.
.
Then P SQ = 360◦ − 240◦ = 120◦ . .
...
.
..
..
. ...
...
...
... .....
.....
...
..
.....
110
....
..... .... .
..... .... .
◦ ....... ◦...
.
. . 25
... ..
...
.
.
... ◦ ....... ..... ..
... ..
25
... .........
So, in P SQ, x = 180 − 25 − 120, ..
.
........
.. ...
... .........
...... ...
............
... ..
.. ..
... ..
... .
.. .
...
......
..... ...
.
i.e. x = 35, Q R
hence (D).
5. Consider the subtraction:–
a 4 b 7 c
− 5 d 8 e 6
2 8 4 9 9
Working from right to left, c must be 5, e + 1 +9 results in a last digit of 7, so
e = 7, 9 + 4 gives a last digit b, so b = 3, d + 1 + 8 gives a last digit 4, so d = 5
and a − 6 = 2 so a = 8.
The largest is 8,
hence (A).

2. Intermediate 1 Solutions Page 2
1 5
6. The next palindrome is 36063, i.e. 110 km further on. The time taken is 1 =
4 4
hours.
The average speed is then
5 440
110 ÷ = = 88,
4 5
hence (A).
7. Let the pineapple, banana and mango be worth p, b and m units respectively.
Then
7p = 9b + 8m (1)
5p = 6b + 6m (2)
(1) × 2 14p = 18b + 16m (3)
(2) × 3 15p = 18b + 18m (4)
(4) − (3) p = 2m,
so a pineapple is worth two mangos.
As there are two equations with three variables, we must check each of the other
alternatives to exclude them as possibilities.
If b = 2x, then m = 3x and p = 6x. The other alternatives are then b + 2m =
8x, 3b + m = 9x, b + 2m = 5x and 4b = 8x, none equal to 6x,
hence (A).
8. Let the distance be x metres.
x 3x
Then there is m on open ground and m on rough ground.
4 4
Now paces on open ground + paces on rough ground = 600.
60 x 80 3x
So × + × = 600
100 4 100 4
6x + 24x = 40 × 600
x = 800,
hence (E).
9. Assume that U(na) is one place behind T(erry). The possibilities for finishing order
are then SVRTU or TUSVR. In both cases V(iv) is in an even position, so neither
satisfies.
Assume U(na) is two places behind T(erry). Then S(am) is four places ahead
of R(obin), and this only occurs when S(am) is ahead of T(erry) and R(obin) is
behind U(na), which places V(iv) third, i.e. the order is STVUR, and R(obin) is
fifth,
hence (E).

3. Intermediate 1 Solutions Page 3
10. After the first go, 100 − 20 = 80 people are left, after the second 80 − 16 = 64
remain and so on.
From this process, we get
100 − 20 = 80∗ , 80 − 16 = 64∗ , 64 − 12 = 52, 52 − 10 = 42, 42 − 8 = 34,
34 − 6 = 28, 28 − 5 = 23, 23 − 4 = 19, 19 − 3 = 16, 16 − 3 = 13,
13 − 2 = 11, 11 − 2 = 9, 9 − 1 = 8, 8 − 1 = 7, 7 − 1 = 6,
6 − 1 = 5, 5 − 1 = 4∗
So, there are 4 people remaining after this procedure is carried out 17 times.
We can see that in each step, for the last person in the line to go, the number in
the line is divisible by 5, and this occurs in the three steps marked ∗ .
So, person in position 100 goes in the first go, person number 99 in the second
and person number 98 in the final go. Thus the last person to go is then the one
originally in position 98,
hence (D).