Assessor’s
use only
QUESTION ONE
(a)
(3x + 2)(x – 4)2
Expand and simplify:
(3x + 2)(x2 – 8x + 16)
(x2 – 8x + 16)
(x2 – 8x + 16)
3x ×
2×
(b)
A
𝟏𝟎𝒙𝟐 − 𝒙 − 𝟐
Simplify:
𝟔𝒙 − 𝟑
10𝑥 2 − 𝑥 − 2
6𝑥 − 3
(c)
= 3x3 – 24x2 + 48x
=
2x2 – 16x + 32
2
= 3x – 22x2 + 32x + 32
(5𝑥 + 2)(2𝑥 − 1)
=
3(2𝑥 – 1)
=
𝟓𝒙 + 𝟐
𝟑
top line factorised
A
fully cancelled
M
Katya is going use a picture frame that is 15 cm by 20 cm to hold a picture.
The width of the coloured edging around the outside of the
picture (to the edge of the frame) is the same all the way round.
The area of the coloured edging inside the frame is 150 cm2.
Calculate the width of the coloured edging.
Area of strip = 20 × 15 – (20 – 2x)(15 – 2x) = 150
or 20x + 15x + 20x + 15x – 4x2 = 150
or 150 = 2 × 20x + 2 × (15 – 2x)x etc
Equation to solve is 150 = 70x – 4x2
A
4x2 – 70x + 150 = 0
Solves to x = 15 or 2.5
15 makes no sense, as the picture is only 15cm wide, so strip is 2.5 cm wide
(d)
𝟐
Simplify:
𝒙−𝟏
𝟐
𝒙−𝟏
=
–
–
𝟓
𝒙+𝟑
=
M
𝟓
𝒙+𝟑
𝟐(𝒙 + 𝟑)
(𝒙 – 𝟏)(𝒙 + 𝟑)
+
−𝟓(𝒙 − 𝟏)
(𝒙 + 𝟑)(𝒙 − 𝟏)
=
𝟐𝒙 + 𝟔 − 𝟓𝒙 + 𝟓
(𝒙 – 𝟏)(𝒙 + 𝟑)
−𝟑𝒙 + 𝟏𝟏
A
(𝒙 − 𝟏)(𝒙 + 𝟑)
or numerator may be 11 – 3x or denominator may be x2 + 4x + 3
1
Assessor’s
use only
(e)
A rectangle is twice as wide as it is high.
The height is then increased by the half the amount the width is decreased.
→
The result is that the area of the second rectangle is 4.5 smaller than that of the
first rectangle.
How much higher is the second rectangle than the first?
Original rectangle is x × 2x. New rectangle is (x + k)(2x – 2k) or equivalent
A
Equation is 2x2 – (x + k)(2x – 2k) = 4.5
2x2 – (2x2 – 2kx + 2kx – 2k2) = 4.5
2x2 = 4.5
k = ± 1.5
M
The question states that the second rectangle is taller, so only the positive
value applies in this case. (Or can regard negative answer as negative “increase”)
The second rectangle is 1.5 higher
E
(Cannot get E without at least some indication of the negative answer)
(f)
A isosceles triangular support is
14 metres wide, and 7 metres high.
A diagonal supporting beam is 10
metres long.
How far off the ground is the high
end of the beam?
The right sloping line has the equation y = – x + 14
By Pythagoras, 102 = x2 + y2 = x2 + (14 – x) 2
A
Expanding, 100 = x2 + 196 – 28x + x2, which gives 2x2 – 28x + 96 = 0
Solving gives x = 6 or 8
M
x = 6 is outside the triangle, so x = 8. Using y = –x + 14, the height is 6 m
E
2
Assessor’s
use only
QUESTION TWO
(a)
Solve:
4x2 – 9 = 35x
4x2 – 35x – 9 = 0
x = 9 or –0.25
(b)
Solve:
A
4(x + 4) > 9(x – 2)
4x + 16 > 9x – 18
16 + 18 > 9x – 4x
34 > 5x
x < 6.8
(c)
(or
34
5
)
A
Katya buys fish and chips for her family.
She buys four pieces of fish and three scoops of chips.
She spends $25.60 altogether.
Each scoop of chips costs 60 cents more than a piece of fish.
Calculate the price of piece of fish.
4f + 3s = 25.60
4f + 3(f + 0.6) = 25.60
A
7f + 1.8 = 25.6
f = 3.4
A piece of fish is $3.40
M
No marks if solved by guess and check without equations!
(d)
Where does the graph of y = 12x2 – 17x + 6 cross the x-axis?
When 12x2 – 17x + 6 = 0
A
x = 0.75 or 0.6666...
M
also award A if factorised y = (3x –2)(4x – 3) as a partial solution
3
Assessor’s
use only
(e)
Find k so that x2 + k x + 16 has only one root.
Only one root when discriminant Δ = 0
b2 – 4ac = 0
(f)
k2 – 4 × 1 × 16 = 0
A
k2 = 64
M
k = ±8
E
Find k so that 2x2 + k x + 50 has one root four times as large as the other.
2(x + r)(x + 4r) = 2x2 + k x + 50
A
2x2 + 10r x + 8r2 = 2x2 + k x + 50
So 8r2 = 50
which means r = ±2.5
so k = 10r
k = ± 25
M
E
or
2r2 + k r + 50 = 0
and
so 2r2 + k r + 50 = 0
2(4r)2 + k (4r) + 50 = 0
32r2 + 4k r + 50 = 0
and
so 8r2 + 4k r + 200 = 0
and
32r2 + 4k r + 50 = 0
r = ±2.5 which we put back into 2r2 + k r + 50 = 0 to get
M
k = ±25
E
or
roots are given by
−𝑏±√𝑏2 −4𝑎𝑐
2𝑎
Designate them as r =
So as 4r = 4(
−𝑘−√𝑘 2 −4×2×50
2×2
−𝑘−√𝑘 2 −4×2×50
2×2
)=
and 4r =
−𝑘+√𝑘 2 −4×2×50
2×2
−𝑘+√𝑘 2 −4×2×50
A
M
2×2
–4k – 4√𝑘 2 − 400 = –k + √𝑘 2 − 400
5√𝑘 2 − 400 = –3k
squaring both sides
25(k2 – 400) = 9k2
k = ± 25
16k2 – 10,000 = 0
E
4
Assessor’s
use only
QUESTION THREE
(a)
Write as the logb of a single number: 2 logb 6 – logb 4
36
2 log 6 – log 4 = log 62 – log 4 = log ( 4 )
= log 9
(b)
Solve:
must be fully simplified
A
logx125 = 3
log x 125 = 3 so x3 = 125
3
x = √125 = 5
(c)
Solve:
must have some working, not just the answer
A
34 = 4x
34 = 4x
log(34) = log(4x) = x log(4)
x = log(34) ÷ log(4)
x = 2.544
A
must have at least one stage (any of above) of working, not just the answer
(d) A fully charged car battery has 800 amps. However, the lights have been left
on and so the end of each hour, the amperage of the battery is only 95% of
what it was at the end of the previous hour.
The amperage of the battery is given by the formula;
Amps = 800 × 0.95t
where t = time in hours
The car will not start if the amperage remaining is less than 200 amps.
Calculate how long before the car will not start
200 = 800× 0.95t
0.25 = 0.95t
log(0.25) = log(0.95t) = t log(0.95)
A
t = log(0.25) ÷ log(0.95)
It takes 27.0268 hours (= 27 hours 2 minutes)
must have at least one stage (any of above) of working, not just the answer.
5
M
Assessor’s
use only
(e)
If 16x + 1 + 16x = 68, find 22x
16x × 16 + 16x × 1 = 68
16x × 17 = 68
16x = 4
M
42x = 4
2x = 1
22x = 21 = 2
E
(x = ½ but you don’t need to solve to answer the question)
(f)
Solve: 2k = 13 +
𝟕
𝒌
The easiest way is to multiply through by k.
so 2k2 – 13k – 7 = 0
2k2 = 13k + 7
k = 7 or –0.5
E
or
2k =
13𝑘
7
+
𝑘
𝑘
2k2 = 13k + 7
so 2k = +
13𝑘 + 7
𝑘
etc
For each question:
N0: No relevant evidence
N1: Attempt at one question
N2: 1 A
A3: 2 A
A4: 3A or 1M with 1 A
M
or 1 M
M5: 1 M
M6: 2 M
E7: 1E
E8: 2E
Overall grade
0–8 = Not Achieved, 9–14 = Achieved, 15–19 = Merit, 20+ = Excellence
6