south african mathematics olympiad

336 Pages·2013·10.19 MB·English

SOUTH AFRICAN MATHEMATICS OLYMPIAD
2015 FIRST ROUND
GRADE 8 SOLUTIONS

1. E 2 – (0 – (1 – 5)) = 2 – (0 – (– 4)) = 2 – (0 + 4) = 2 – 4 = – 2

2. C 1 cm per month = 10 mm per month = 120 mm per year = 1 200 mm in ten years

3. C x must be a divisor of 12, and the possibilities are 1, 2, 3, 4, 6, 12, which is 6 in all.

4. D If the numerator has been multiplied by 6 and the fraction stays the same, the
denominator is also multiplied by 6 and becomes 3 6 = 18.

5. E The last digit of 2011 × 2013 × 2015 is a five, and the last digit of
2010 × 2012 × 2014 is zero, so the last digit of the difference will be
5 – 0 = 5.

6. A Every multiple of 7 represents a full week. Since today is Thursday, in one
day’s time it will be Friday. Thus each full week after today starts on a Friday and
ends on a Thursday. 150 days divided by 7 equals 21 full weeks with a remainder
of 3 days. The 147th day from now will thus be a Thursday (end of 21st full week),
and consequently the 150th day from now will be a Sunday.

7. A Each fold doubles the number of layers that will be pierced. There will be 25 layers
and therefore 25 = 32 holes.

8. D The person on the extreme left can be any one of the four people that is not Alfred;
the second left can be any one of the remaining three; the first person on the right
of centre… and so on. The number of possibilities is 4 × 3 × 2 × 1 = 24.

9. A By definition of , we know 5  x = 5x + 5 + x = 6x + 5. If this is 35,
then 6x = 30, so x = 5

10. C Since BO = AO (radii) and AB = BO (given), we must have
ˆ
AB = AO = BO, i.e. Δ ABO equilateral, so that AOB = 60°.
ˆ ˆ ˆ ˆ
Now OBC and BCOare equal and total the exteriorAOB; soBCO= 30°

11. A Every cube is joined to an adjacent cube on two faces, leaving the other four
exposed to paint.
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12. C There were 6 + 5 + 1 + 3 = 15 children altogether, of whom 3 were in the third-
3
largest group. That group therefore requires 360  324  72
15
300
13. D abcd 230360  abcd 300 so the average is 75
4
5
14. D The shaded area must be 8050 cm2 , so the area of the square is
8
2 × 50 = 100 cm2, and then the side length of the square is 100 = 10 cm.

15. E Let the tick be placed in any one of the 16 blocks. Then the cross can go
in any of three other rows or three other columns, which gives 9 possible positions.
That makes 16 × 9 = 144 ways.

16. D Joining P to C we see that
Δ PCM is identical to Δ ABM. That
means that P, C, B are vertices of a
square, and the required angle is the
one between a diagonal of a square
and its side, i.e. 45°

17. C
area ΔABP 1 1BP.AB 1 1BP 1 BP 2
= so 2 = so 2 = and therefore =
area ΔABCD 3 BC.AB 3 BC 3 BC 3
2 1
and then BP : PC is :  2:1
3 3

18. B Draw the diagonal of the squares that
bisects area a with each half being c. Then
clearly b = c as each is half the parallelogram
MQPN. It follows that a = 2b.

19. E Between (and including) 98 and 200 there are 51 multiples of 2; between 98 and
199 there are 34 multiples of 3. Between 102 and 198 there are 17 multiples of 6.
The number we seek is 51 + 34 – 17 = 68

c
20. B Since t toffees cost c cents, each toffee costs cents.
t
r rands equals 100 r cents.
c 100rt
The number of toffees that can be bought for 100 r cents is thus 100r  =
t c

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SOUTH AFRICAN MATHEMATICS OLYMPIAD
2015 FIRST ROUND
GRADE 9 SOLUTIONS

1. E 2 – (0 – (1 – 5)) = 2 – (0 – (– 4)) = 2 – (0 + 4) = 2 – 4 = – 2

2. C 1 cm per month = 10 mm per month = 120 mm per year = 1 200 mm in ten years

3. A There are ten factors: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. Four of them are not
multiples of 4, viz. 1, 2, 3, 6. The probability is therefore 4/10 or 40 %

4. A Every cube is joined to an adjacent cube on two faces, leaving the other four
exposed to paint.

5. C There were 6 + 5 + 1 + 3 = 15 children altogether, of whom 3 were in the third-
3
largest group. That group therefore requires 360  324  72
15

6. A Each fold doubles the number of layers that will be pierced. There will be 25 layers
and therefore 25 = 32 holes.

7. D The person on the extreme left can be any one of the four people that is neither
Alfred nor Mollie; the second left can be any one of the remaining three; the first
person on the right of centre… and so on. For every arrangement of the people
around them, Alfred and Mollie can swap places to make a new arrangement. So
the number of possibilities is (4 × 3 × 2 × 1) × 2 = 48.

8. B The digits 1 to 9 total 45, and 460 = 10×45 + 10. So the complete set 1…9 will
appear ten times, and then as any more digits as total 10, viz. 1+2+3+4. So the last
digit is 4.

9. C wz45180 and x y45180.
We thus have w x y z90360 and hence w x y z  270

10. B Horizontal distance travelled:
1353 (i.e. 3 km East)
Vertical distance travelled:
2464 (i.e. 4 km North)
By Pythagoras, the straight-line distance
from the starting point is thus 5 km.

6 4
11. C 120% of a is a while 80% of b is b.
5 5
6 4 a 4 5 2
We thus have a  b from which it follows that   
5 5 b 5 6 3

12. A Let the tick be placed in any one of the 16 blocks. Then the cross can go
in any of three other rows or three other columns, which gives 9 possible positions.
That makes 16 × 9 = 144 ways.

13. E Given a + 2b = 13 and 5a – 2b = 5, we can add both left-hand sides and both
right-hand sides to find 6a = 18. Thus a = 3, and then since a + 2b = 13, we must
have 2b = 10, i.e. b = 5.

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14. D BAP = 60° while BAC = 90°, so PAC = 30°. With BPA = 60° that means C = 30°,
and so Δ PAC is isosceles. Then BC has length 4, and so by Pythagoras the length
of AC is 42 22

area ΔABP 1 1BP.AB 1 1BP 1 BP 2
15. C = so 2 = so 2 = and therefore =
area ΔABCD 3 BC.AB 3 BC 3 BC 3
2 1
and then BP : PC is :  2:1
3 3

16. D Multiplying gives us 9n < 4n + 16, i.e. 5n < 16, so n = 1 or 2 or 3.

17. D Joining P to C we see that
Δ PCM is identical to Δ ABM. That
means that P, C, B are vertices of a
square, and the required angle is the
one between a diagonal of a square
and its side, i.e. 45°

18. E Between (and including) 98 and 200 there are 51 multiples of 2; between 98 and
199 there are 34 multiples of 3. Between 102 and 198 there are 17 multiples of 6.
The number we seek is 51 + 34 – 17 = 68

19. A ABC90 since AB and BC are both

diagonals of a square,

AB 52 52  50 5 2

BC 72 72  98 7 2

1  
area of ΔABC is 5 2 7 2 35 cm2
2

20. B Suppose angle AOB is θ, and let PA = x.
.(1 x)2 .(1)2

Then the shaded area is
360 360
.(1 x)2
and the whole sector AOB has area .
360
.(1 x)2 . 1 .(1 x)2
So we must have   .
360 360 4 360
3 .(1 x)2 . 4
and therefore .  and so (1 x)2  .
4 360 360 3
2 2 3
Then x = 1or
3 3

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2015 SOUTH AFRICAN MATHEMATICS OLYMPIAD
JUNIOR ROUND TWO – SOLUTIONS

1. 2 4 + (–2) = 4 – 2 = 2

2. 4 The factors of 128 are all powers of 2, i.e. 20, 21, …, 27. The largest power of 2 that
divides 120 is 8 = 23. The powers of 2 dividing 128 and exceeding 8 are 24, 25, 26
and 27.

3. 6 If the length of the strip is 12 units, marks will be made at 3 units, 6 units and 9
units for four equal lengths, and at 4 and 8 units for the three equal lengths. This
gives 5 marks in all, meaning there will be 6 pieces when the cuts are made.

4. 15 The smallest value of n is 20 (for which 1 n is 10) and the largest value of n is 49
2
(for which 2n is 98). From 20 to 49 inclusive represents 30 integers, half of which
are even (so that 1 n is an integer).
2

2 2 2 2 15
5. 30     2  30
1 1 1 1 1
0 0 0
2 2 510 15
5 5
1 1
0
5 5

6. 36 27 people must have a pen and 36 must have a pencil. If all the people with pens
also have pencils, this could be as few as 36 people in all.

7. 2040 In order to be divisible by 2, 3, 4, 5 and 6, the number only needs to be
divisible by 2235, i.e. it must be a multiple of 60. The smallest multiple of 60
bigger than 2015 is 2040.

n3
8. 10 Suppose  k ; then n = 13k – 3 or n = 13(k –1) + 10. This shows that n
13
leaves remainder 10 on division by 13.

9. 32 The area of the whole circle is π.42 = 16π, but the area
1π 1
of the sector is 1 π, so the probability is 2 
2 16π 32

10. 9 Area of shaded trapezium
1
= (48)1,5 = 9
2

11. 75 If W is the number of litres the tank can hold, then when it is 30% empty it is
holding 0,7W litres, and so we have 0,7W = 0,3W + 30, i.e. 0,4W = 30 and
then W = 30 ÷ 0,4 = 300 ÷ 4 = 75.

12. 808 Each row, apart from the bottom row and top row, contributes 4 cm to the
perimeter. The top and bottom rows each contribute 8 cm. Since there are
1000 ÷ 5 = 200 rows we have Perimeter = (200 – 2) × 4 + 2 × 8
= 99 × 8 + 2 × 8 = 101 × 8 = 808.
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13. 6 By Pythagoras, OA = 17 . Then OA = 18, OA = 19and so on, with
2 3 4
OA = n15, and thus OA = 36 = 6.
n 21

14. 2 The prime numbers smaller than 20 are 2, 3, 5, 7, 11, 13, 17 and 19. In order
for the fraction to be as big as possible we need to maximise the numerator and
195 14 192 17
minimise the denominator. We thus try  2, and  and
2(2)3 7 2(3)5 11
193 16
 which are smaller.
2(2)5 9

15. 4 109876543215.2.32.23.7.2.3.5.22  7.52.34.28.
Since 28 44 we have x  4

16. 32 9 – 4 = 5, and the length marked bold in
the diagram must be 12 by Pythagoras.
That means the length of the rectangle is
4 + 12 + 12 + 4 = 32

x y
17. 6 We can re-write  1 in the form 3x + 4y = 60. Then we can see that
20 15
3x must be divisible by 4, so x must be, and trying successive possible values
we see that only the following combinations of x- and y-values are acceptable:
(0; 15), (4; 12), (8; 9), (12; 6), (16; 3), (20; 0).

18. 120 For a square with diagonal d and side length s we have the relationship
d
s2 s2 d2 and thus s  . The respective side lengths of the three squares are
2
2
3 4 5  3  9
thus , and respectively. So grey area =    while
2 2 2  2 2
2 2
 5   4  25 16 9
black area =        = grey area.
 2  2 2 2 2
Thus the black area must also require 120 bricks.

19. 16 To have a sum of 75, the five consecutive numbers must be 13, 14, 15, 16 and 17.
Since the numbers in the triangle total 29, the two possibilities are 13 & 16 or 14 &
15. Since the numbers in the square total 30, the two possibilities are 14 & 16 or 13
& 17. Since the numbers in the circle total 47, the only possibility is 14, 16 & 17.
The only possible letter that could represent 15 is thus A, from which it follows
that B = 14, E = 13, D = 17 and finally C = 16.

20. 31 Join AC ; then ΔABC has base 2 cm and
height 10 cm, while Δ ADC has base 6 cm
and height 7 cm. So the shaded area is
1 2.10 + 1 6.7 = 10 + 21 = 31 cm2.
2 2

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South African Mathematics Olympiad
Junior Third Round 2015
Solutions
1. Label the 5×5 grid as shown:
Consider the diagonal A5 - E1: the letter T must appear in the diagonal, but it cannot appear in D2 or
C3, since columns C and D already contain a T. Hence A5 = T. The letter A must also appear in the
diagonal, but cannot be in C3, since column C already contains an A, so D2 = A and hence C3 = H.
Considering the other diagonal, B2 cannot be an A since B1 = A, hence E5 = A which forces B2 = S.
Looking at row 2 and letter T, columns A and C already contain Ts, which implies that E2 = T. This
then forces C2 = M and A2 = H.
The only place in row 3 where A can occur is in A3 and T can only occur in B3. The rest of the grid can
now be easily completed in a similar fashion.
1
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2. If the average of four numbers is 8, their sum must be 8×4 = 32. To maximize the largest one of these
numbers,wechoosethesmallestpossiblevaluesfortheotherthree,whichare1,2and3(sincetheintegers
must be positive and different). The remaining number is then 32−(1+2+3)=26.
3. 1110222222003333333330004444444444440000 = 10002002000030030030000040040040040000, which contains 28
111
zeroes.
4. There are 7×24 = 168 hours in a week and there are 4×2.5 = 10 hours during which load shedding
may occur. Since there is a 60% chance of load shedding, the probability that there is load shedding at a
particular moment during the week is
4×2.5 10 60 1
×60%= × = .
7×24 168 100 28
5. Placing a 3 at both ends of the number increases it by 3372, a four-digit number. Hence the original
number is a two-digit number. Suppose the original number is x. Then the number formed by placing a
3 at both ends is equal to 3003+10x. This is 3372 more than x, that is,
3003+10x−x=3372 =⇒ 9x=369 =⇒ x=41.
6. (a) Suppose the side length of the cube is n; we wish to find the largest n such that n3 ≤ 500. Now,
73 =343 and 83 =512, so the largest value of n is 7.
(b) Supposethesidelengthofthehollowcubeisn. Thehollowcubeisobtainedbyremovingthesmaller
interior cube from the solid cube, so the shell contains n3−(n−2)3 cubes, so we want to find the
largest n such that n3−(n−2)3 ≤500. Now
n3−(n−2)3 ≤500
=⇒ 2(n2+n(n−2)+(n−2)2)≤500
=⇒ n2+n2−2n+n2−4n+4≤250
=⇒ 3n2−6n≤246
=⇒ n(n−2)≤82.
Now, 10×8=80<82 while 11×9=99>82, so the largest value for n is 10.
7. We will compute the angle sizes of quadrilateral DEGB and use the fact that the sum of the interior
angles of a quadrilateral is equal to 360◦.
Since AFC is an equilateral triangle, (cid:54) FAC = 60◦ and so (cid:54) EAF = 90◦ − 60◦ = 30◦, which yields
(cid:54) BAE =90◦+30◦ =120◦. Since AB =AC =AF, triangle BAF is isosceles, and so
1 1
(cid:54) GBC =(cid:54) FBA= (180◦−(cid:54) BAF)= (180◦−120◦)=30◦.
2 2
Next, AE = AB = AC, and so triangle EAC is a right-angled isosceles triangle, which means that
(cid:54) BCG=(cid:54) ACE =45◦. Hence,
(cid:54) BGE =(cid:54) GBC+(cid:54) GCB =30◦+45◦ =75◦.
8. Suppose that the rectangular sheet of paper has dimensions a and b, with b being the longer side. We
calculate the volume of the two cylinders formed by joining the long sides and short sides, respectively.
• Supposetheshortsidesaregluedtogether. Thentheheightofthecylinderisaandthecircumference
of the cylinder is b. If r is the radius of the cylinder, it means that 2πr = b, or r = b . Hence the
2π
cylinder’s volume is πr2h=π(cid:0) b (cid:1)2·a= ab2.
2π 4π
2
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• Supposethelongsidesaregluedtogether. Thentheheightofthecylinderisbandthecircumference
of the cylinder is a. If r is the radius of the cylinder, it means that 2πr = a, or r = a . Hence the
2π
cylinder’s volume is πr2h=π(cid:0) a (cid:1)2·b= a2b.
2π 4π
Comparing these two numbers, we see that the first volume can be written as b(cid:0)ab(cid:1), while the second
4π
volume is a(cid:0)ab(cid:1). Since the number in brackets is the same for both and b > a, it means that the first
4π
volume b(cid:0)ab(cid:1) is the largest. Hence the largest volume is obtained when the short sides of the sheet of
4π
paper are glued together.
9. In any arrangement of cubes, each edge is aligned in one of three directions, and note that for each
direction, the number of edges is the same. Hence to count the total number of visible edges, we only
need to count the number of edges in a particular direction, and multiply it by 3. For argument’s sake,
let’s consider the vertical edges only.
Notethatinthetoplayer, thereare3verticaledges, inthesecondlayerthereare5andinthethirdthere
are 7. Each next layer can be formed by copying the last layer, and adding one extra cube, which adds
two extra vertical edges. Hence the sequence of vertical edges in the layers is 3,5,7,9,..., the sequence
of odd integers and so the nth layer has 2n+1 vertical edges.
Now, we know that 1+3+5+···+(2n−1)=n2, so
3+5+7+···+(2n+1)=[1+3+5+···+(2(n+1)−1)]−1=(n+1)2−1.
Hence in total there are 3((n+1)2−1) visible edges.
(a) If n=3, this is equal to 3(42−1)=45.
(b) If n=20, this is equal to 3(212−1)=1320.
10. (a) The chessboard contains an odd number of squares and every time a 2×1 tile is placed, the number
ofopensquaresdecreaseby2, anevennumber. Hencethenumberofremainingsquaresisalwaysan
odd number (odd minus even is odd).
(b) Beforeanytileisplaced,thereare13blacksquaresand12whitesquares. Everytimeatileisplaced,
thetilecoversoneblacksquareandonewhitesquare, sothenumberofopenblacksquaresandopen
white squares both decrease by 1. Since there is one more black square than white squares at the
start, there will be one more black square than white squares at the end.
(c) The following arrangement shows 7 empty 1×1 squares.
Assume for a contradiction that it is possible to have more than 7 empty squares. Since there must
be an odd number, there are at least 9 empty squares.
3
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