GRADE A Maths Assessment
A
1
Test 6 Answers
Number
(a) 90
(b) 82/3
(c) 100-1/2
1
4
Answer
2
√80 in form k√5
4√5
3
(a) Max of xy
(b) Min of x÷y
6.05x4.15=25.1075
5.95÷4.15=1.43373494
y = kx3
224 = k x 43 (when x = 4 & y = 224
k = 3.5
Formula: y = 3.5x3
4
(a) Find formula
Criteria
1
1
1
10
y is proportional to the cube of
x.
Mark
1
1
1
1
1
2
3
4
1
1
(b) Calculate x when y = 1792 y = 3.5x3
1792 = 3.5x3
X3 = 512
x=8
B
5
Algebra
(a) Factorise: x2 - 5x - 24
(b) Solve: x2 - 5x - 24= 0
6
Solve x2 + 6x - 4 = 0 by
fomula
1
Answer
(x - 8)(x + 3)
(x - 8)(x + 3)=0
x = 8 or x = -3
x=
Simplify:
7
x
8
-
5
x-4
Show that the equation:
6
 b  (b  4ac)
2a
5
=
x+2
4 – 3x
x–1
can be rearranged to make:
3x + 7x – 13 = 0
2
9
Solve: x2 + y2 = 18
y – 2x = 3
(by substitution)
2
2
7
7(x - 4) - 5x
x(x – 4)
= 7x - 28 - 5x
x(x – 4)
= 2x - 28
x(x – 4)
1
1
8
4 – 3x
x–1
5
=
x+2
Criteria
5
2
2
x = -6 ± √(62 – 4x1 x -4)
2
x = -6 + √52 or -6 - √52
2
2
x = 0.61
or -6.61 (2dp)
7
Mark
2
(4 – 2x)(x + 2)
5(x - 1)
=
5x – 5
= 8 – 2x – 3x
2
3x2 + 7x – 13 = 0
Substitute y = 2x + 3
x2 + (2x + 3)2 = 40
1
1
1
9
x2 + 4x2 + 12x + 9 = 18
1
5x2 + 12x - 9 = 0
1
(5x - 3)(x + 3) = 0
x = 0.6 or x = -3
When x = 0.6, y = 4.2
1
1
1
When x = -3, y = -3
10
y = pqx
(a)find value of p, q & k
10
y = pqx
6 = p x q1
(when x = 1 and y = 6)
3
24 = p x q (when x = 3 and y = 24)
24 = p x q3
6 p x q1
4 = q2  q = 2
and p = 3 (6 = p x q)
y = pqx
1
1
1
k = 3 x 24 (when y = k and x = 4)
k = 3 x 16 = 48
11
12
13
C
14
Solve: x2 + y2 = 25
y = 2x - 1
(by graphical method)
y = sinxº graph
Given cos 60º = √3
2
(i) Sin1200
(ii) Sin2400
Make a the subject
2(3a – c) = 5c + 1
Geometry & measures
(a) Scale factor
(b) Complete image
1
1
Draw line y = 2x - 1
Points of intersection:
x = 2.6 , y = 4.2
x = -1.8 , y = -4.6
11
1
1
12
√3
2
-√3
2
1
1
13
2(3a – c) = 5c + 1
6a – 2c = 5c + 1
6a = 7c + 1
a = 7c + 1
6
1
1
1
Answer
Mark
1
-2
Criteria
14
C
A
B
E
P
Q
1
15
16
Show that triangles ABD
and BCD are congruent.
Find height of larger tank
AB = CD (given)
Angle ABD = angle BDC (ALTERNATE angles)
BD is common
Condition of proof: SAS
Volume scale factor = 97.2 ÷ 3.6 = 27
Length scale factor = 3√27 = 3
Height of larger tank = 12cm x 3 = 36cm

36cm2 : 100cm2
15
2
16
1
1
17
Calculate length SR
QR = √(52 + 72) = √74
Q
5
R
7
SR = √(132 – √742) = 9.7cm
S
19
20
Find volume of wood
(a) Area of sector
(b) Length of arc
(a) Size of angle SOB
(b) Reason
D
21
Handling data
Find number of girls from y9
in sample
22
(a)Complete table
Area of cross-section = ½ πr2
= ½ x π x 0.62
= 0.5654866776
Volume = area of cross-section x length
= 0.5654866776 x 350
= 198 cm3
Length =120
x πx2
0
360
= ⅓ πx2
Length =120
x 2πx
3600
= 2πx
3
460
Angle between tangent and radus = 900
 Angle SOP = 680
Angle BOP = 220 (alternate angles)
Angle SOB = 68 – 22 = 460
Answer
167 x 50
1385
≈6
390
400
17
T
Q
13
18
1
√74
2
R
18
1
1
1
19
1
1
1
20
1
Mark
1
1
Criteria
21
1
1
22
(b)complete histogram
1
23
P(3 beads are all same
colour)
23
p(BBB or RRR or GGG)
2
2
2
3 3 3
5
5
5
=
x x +
x
x
+
x
x
10 10 10 10 10 10 10 10 10
27
125
8
=
+
+
1000 1000 1000
160
=
1000
1
1
1