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Unformatted text preview: SOLUTIONS — SENIOR DIVISION . (4x5]+(2><10}=20+20=1,
hence . x + 70 = 130 (exterior an—
gle of the triangle), so 3 = 60,
hence 13DD
. (A1501?)
1 2 9
1+3ié=1+f=1+?=?’
2 2 hence . The straight line 9": a: + 9 passes throught the point (2,3).
Then3:2+gandg=1,
hence . (Also 18)
The number £811. 2 lUUt + 10 X 8 + u = 100i + 80 + u,
hence . $13K Q is a right angled triangle with sides of length 3 and 7 as
Shown. ' 3032+72=PQ2and PQ=¢E R :9 use 7
o .. P 3 X 58 AreaofﬁhPQR:éxPQxQR=éxﬂxﬂ=éx58=20 hence s. The total number of marks gained was 70 x 20 + 80 x 30 = 1400 + 2400 = 3800.
3800 _ y _ k ‘ th
The average mar overall 15 en 50 5 76,
hence (C). gAlso 110)
11116 wheel with the larger diameter will rotate less than the wheel
'h the smaller diameter, so the difference in the number of rev elations of each wheel in a journey of 1800 km is the number of
etations of the smaller wheel (200cm diameter) minus the num—
r of rotations of the larger wheel (diameter 225 cm} This difference
' 1800 x 1000 x 100 1800 X 1000 x 100 200 225
= 900 000 — 800 000 = 100000, hence angle sum of a pentagon is 5 X 180 — 360 = 540 degrees.
e remaining angle, in degrees, is then 540 — 400 = 140,
hence hence (A). ' 2 2 . .
1 1 81 _ _1681
'G+(El)"e_(ill*JL‘w0+4‘4m
1 #1681 41 ‘+E=VZﬁ=%}
hence [3). 59 12. 13. 14. Att:0, 100
Q: (1+0), _100.
So, half the gas is 50 cubic units.
Then
100
50 = (1 + 2t)2
(1 + 2:]2 = 2
1 + 4t + 4:2 = 2
t _ —4 :l: m
s
—4 i 4J5
= 8
z — ﬂ _ 1
2 I
hence (Also [14) The outcomes for the digits to be a perfect square are:— (1,6), (2,5), (3,6), (4,6), (5,2), (6.1), (6,3) and (6,4)é 8 i112 all. There are 36 possible outcomes, so the probability is 56 = g, hence (B). (Also 11?) I
Let the area. of the shaded portion be m, so the area of the white
portion is :1: and the part of the square not visible also has area 9:. Then 33': =12 and a: = 4.
P Q 60 t the side of the shaded rectangle be 3;. Then % x y2 = 4 and 8 length of the fold line U V is the hypotenuse l of the isosceles
 gle with two equal sides length 3;.
y2+y2 = 23;2 =16andl= 4, hence(A).
P
%asUTSP.
' T ssSTHRT.
Q u 3 R
+———— 4 ———————o 4—24—————+
4 _40 _ 5
2.4 “24" 3'
 5 5
=EsndsoQUzéx4=2.5, hence (B). in from Canberra be at a point P at 12:40 pm, when the
In leaves Sydney. " from Sydney will reach P at the same time the train
.erra arrives at Sydney, that is at 3:30pm, which is 170
as: than the ﬁrst train arrived at P.  rains travelled at the same speed, they passed at the
between P and Sydney, 85 minutes after 12:40 pm, at hence (C). angle that each triangle has at the origin is 45°, it will 61‘ be the hypotenuse of the eighth triangle which will overlap OX 1. X4 X3 X5 0 X1 The length of the hypotenuse increases by a factor of V5 each
tima so the eighth, OX3, will have length = 16. Thus Xle (where k = 8) = 16 — = 15,
hence (D). 18. {Also 122) _
Consider 5—digit numbers starting with 2. The following tree dia— gram sh0ws there are 4 such numbers starting with 2. /8 /2—5\2
2—5 \8 5'/8 _ \2 In a similar manner, we get 4 starting with I, 8 with 3, 4 with 4, 4
with 5, 8 with 6, 4 with 7, 4 with 8 and 5 with 9, giving a total of 4+4+8+4+4+8+4+4+5=45 such numbers,
hence[D). 19. Let PQ = I. 62 cos 60 = [I cos 45 ating l we obtain v? 1 =2(v’§+l). 1'1"?“ .2—v’§ ﬁ—i _ x leR from the triangle PQT 1
2
1
E from the triangle PBS '=\/§(x+1]=\,/§z+x/§, —x/§+1. hence (B). _the radius of the quarter circle. of the are rolled to form the circumference of the base of the base of the cone is (e) K?) = +51"! 63 21. 22. so that hence (A). ' 552+ 3:3——1 : 1—:r
x2+ $3~—1 — 1—23;+::32
m3——1 — 1—23: 323—4 2 1—43+4x2
x3—4x2+4x : 0
3(x2—4a:+4) : 0
a:(:r—2)(:c2) = 0 :1: : Oor 2. As we have squared both sides, there are possibly extraneous roots,
so checking for each value: x : 0 gives ﬂ = 1 which is true.
x:2gives2+ \/4+\/§=2+v/79£1whichisfalse. So there is one solution, :0 = 0,
hence (Also I23) Using the fact that the four
outer triangles are similar, we
obtain the additional dimensions
as shown on the ﬁgure. The area of the shaded rectangle
is 321;. From the triangle on the top we
get mac Horn the triangle on the right we get 2_3+3_E§
I "16 64—64' 64 v/EX%_15_7.5 my 2 T 8 _ 3—2 _ E
{' i'reater than i and less than 3
g 16 16‘ hence (D). rig, (1—293)3(1+lc33)2 =(1—6$+12w2—8:c3)(1+2k:r+k:r2)
eﬂicient of 9:2 is then 392 — 12k + 12.
'12k + 12: 40 and k2 — 12k — 28 = 0. values of k which satisfy are the sum of the roots of this
' (12} _
1 — l2, tion, that is £71 + k2 = hence (D). of + = 4 is that part of the lines a: +3; = 4,
: —$+y=4 and —:c—y:4 lihgciand—zlgygél. 1
;ea is 2 x E x 8 X 4 = 32 square units, hence x 32 = (1024)200 x 32 a: 1060“ x 102 = 105“2 210 = 1024 > 103. hence 22000 > 1050“. 65 Also 2005 < 13 x 155 and 213 = 8192 < 10“. 80
22005 < (104)55 and 213 = 8192 < 10" Hence 22005 < (104)155 : 10620‘ So the number of digits in 22005 is closest to 600,
hence (C). (Alternative 2)
The number of digits in N is the integer [logm N]. Note that
213 = 8192 < 10‘1 and 210 =11124 > 103 whence 4/13 > loglﬂ 2 > 3/10. Hence 200 .
601 < —150—X—§ < toglo 220‘J5 < m So the number of digits in logm 22005 is closest to 600,
hence ' . (Also I26) Label the three pieces as shown. Depending on the angle of the
out, A may be bigger than C or vice versa, but whatever the angle,
it 15 clear that one of them is going to be smaller than B. So, if the cut does not go through the inside corner: it will be
posslble to make my piece bigger by moving the cut a small distance
to the left, but parallel to the old one. So, for my best solution. I know the cut must go through the
corner.  66 Which edge the cut may pass through, there are now _ lities:— _ _ either of the ﬁrst two types, it is easy to see that my
9 and that this can be improved by making a small
tment to the direction of the cut. So my optimum .elockwise and see what happens. At the starting
smaller than A and so is my piece. As the cut
_ bigger and A gets smaller {smoothly}, approaching
_ ere A is very 5111111111 and clearly smaller than C, in
'my piece. The best solutiOn is when A and C are
If we write the dimension ;L' as shown:— 67 27. 23. We see that the area of A is 20m and the area of C is 100 — 5x.
Equating these gives :1: = 4 and my piece is either A or C, of equal
minimum area 80 cm2. So my largest possible piece is 80 c1112. We have f (f (ti) =
f (t) = Thus 5: ﬂ 2005 = f{f(t)) = f(6t — 2005). Hence f[f(t)) = f(6t — 2005) 2 6t — 2005. But, from (2}, Les 2‘ 0(05 — 2005) — 2005 = 6t 4 2005 = HHS Thus 30: = 5 x 2005 and 2:: 401‘ St — 2005 (1) and 6t — 2005. (2} [Also I29) Using the symmetry of the ﬁgure, we can see that the octahedron
is made up of 6 shapes which are identical to the one shown. 68 3 12
__e of that portion is then % = 20 cm3. ss+y+z = 5 (1]
3:2413”? = 15 (2)
are = z”, (3)
cc+y = 5—2
$2+2$y+y2 = 25—103+22.
mg (2) and {3) in this gives
" 152"‘+2z2 = 25—10z+32
10z z 10
z = 1
l 1 1 33+ 1
if; = wk;
5—; 1
Z 2:2 +3
— 4+1 =5. jannot be odd. Otherwise all its divisors would be
11111 of the four squares of its divisors even, yielding
Hence the smallest two divisors must be 1 and 2.
est divisor must be 4 or a prime p. It cannot. be
e sum would include exactly two odd squares (to
sand would be a number divisible by 2 but not by 4)
tsediction. liest three divisors are 1, 2 and an odd prime p. Since
, the remaining divisor is 239. Thus the number is l+4+p2+4p2=5(1+p2). 69 Since p does not divide 1 + 332, it must divide (and so be equal to)
5. The number is 5 x 26 = 130 2 l x 2 x 5 x 13, and so the largest
prime divisor is 13. ...
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 Spring '10
 ChaoHue
 triangle, shaded rectangle, best solution, right angled triangle, following tree dia—

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