THE HARMONY SOUTH AFRICAN MATHEMATICS OLYMPIAD
SECOND ROUND 2004: JUNIOR SECTION: GRADES 8 AND 9 SOLUTIONS AND MODEL ANSWERS
PART A 1. If
6 1 2
5 =
,
, then the value of
0 06 0 5
, ,
is
A 1, 2
B 0,12 C 0,012 D 0,0012 E 0,00012
ANSWER: B EXPLANATION:
Given
6 1 2
5 =
,
0 06 0 6
0 5 5
6 10
5 1 2 10
0 12 =
= ÷
= ÷
= ,
, ,
, ,
2. If
x y
is defined to be the remainder when
x
is divided by
y
for example
8 5
3 =
, then the value of
13 11
3
is A
0 B 1 C
2 D 3 E
4 ANSWER: B
EXPLANATION:
x y
is defined to be the remainder when x is divided by y e.g.
8 5
3 =
.
13 11
3 1
=
since
11 3
3 =
remainder 2 and
13 6
2 =
remainder 1
3. If
6
10 10
10 10
⋅ ⋅
=
x y
z
, then the average of
, x y
and
z
is A 1
B
5 3
C 2 D
7 3
E 3
ANSWER: C EXPLANATION:
6 6
10 10
10 10
10 10
6 6
2 2
+ +
⋅ ⋅
= =
+ + = =
Average :
x y
z x y z
x y
z
4.
The length of the broken line, in metres, down the middle of a road is A 67 B 67,5
C 68 D 69 E 70 ANSWER: D
EXPLANATION:
1 1
34 3
7 1
3 20
2 2
69 + +
+ + +
= metres
OR
1 1
5 8 2
1 9
2 2
40 20
9 69
meters C
A B
C = + −
− =
+ +
=
5. In an isosceles triangle
ABC
,
2 =
. AB
BC
If the perimeter of triangle
ABC
is
300 mm
, then the length of
AC
in millimetres is. A 40 B 60 C 80 D 100
E 120 ANSWER: E
EXPLANATION:
Given isosceles
2 ∆
= with
ABC AB
BC
2 =
∴ =
Let BC x
AC x or x
but
2 =
AC x
to give a ∆ that is possible the sum of any 2 sides must be greater than the 3
rd
side.
2 2
300 5
300 60
2 60 120
+ + =
∴ =
∴ = =
∴ = ×
= mm
mm x
x x
x x
BC AC
PART B 6.
Half of
2004
2
is A 2
1002
B 2
2002
C 2
2003
D 1
2004
E 1
1002
ANSWER: C EXPLANATION:
2004 2004
1 2004 1
2003
1 2
2 2
2 2
2
−
× =
= =
7. You are given four fractions
5 12
; ;
; a
b c
Two fractions
a
and
b
are equally spaced between
5 12
and
c
. If
4 3
+ = a b
, then find the value of
. c
A
7 12
B
2 3
C
3 4
D
5 6
E
11 12
ANSWER: E EXPLANATION:
5 12
4 3
4 3
5 4
12 3
5 4
12 3
4 5
3 3
12 7
12 5
7 9
11 12
12 12
12 11
12
+ =
= −
−
− = − −
= + =
= ;
; ;
; ;
;
; ;
; a b c
a b
b a
a a
c a
a a a
a
c
OR
5 3
12 5
16 12
12 3
7 12
9 12
11 12
= + +
+ =
= =
= ÷
a a b
a
b c
8. What is the sum of the digits of the following product?
999 999
×
666 666 A
54 B 63 C
72 D 81 E
90 ANSWER: A
EXPLANATION:
999 999
×
666 666 9
×
6 = 54 Sum
1
= 5 + 4 = 9 = 1
×
9 99
×
66 = 66100-1
= 6600 - 66 = 6534
Sum
2
= 6 + 5 + 3 + 4 = 18 = 2
×
9 999
×
666 = 665334 Sum
3
= 6 + 6 + 5 + 3 + 3 + 4 = 27 = 3
×
9
∴
999 999
×
666 666 : Sum
4
= 6
×
9
= 54
9 A lady travels by car at a uniform speed, from
A
to
B
and then from
C
to
. D
Determine the average travelling speed of the vehicle from
A
to
D
in kmh.
A 92 B
96 C 100
D 104
E 120
ANSWER: B EXPLANATION:
Average speed from A to D
200 200
5 2
1 2
3 400
15 10 6
400 6
1 25
96 =
+ =
+ =
+ =
× =
Distance from A to D Total time
10. The maximum number of integer values that could be obtained
