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Question 1 Report
Question 2 Report
\( \dfrac{0.0001432}{1940000} = k \times 10^n \) where \( 1 \le k < 10 \) and n is a whole number. The values K and n are
0.00014321940000
= k x 10n
where 1 ≤
k ≤
10 and n is a whole number. Using four figure tables, the eqn. gives 7.38 x 10-11
k = 7.381, n = -11
Question 3 Report
If \(pq + 1 = q^2\) and \(t = \frac{1}{p} - \frac{1}{pq}\) express \(t\) in terms of \(q\)
Question 4 Report
If \( \sin \theta = \frac{x}{y} \) and \(0^o < 90^o\), then find \( \frac{1}{\tan \theta} \).
Question 5 Report
Question 6 Report
In the diagram, find the size of the angle marked ao
S = 280o2
= 140
< O = 360 - 280 = 80o
60 + 80 + 140 + a = 360o
(< in a quad); 280 = a = 360
a = 360 - 280
a = 80o
Question 7 Report
What is the volume of this regular three dimensional figure?
A = 12 x 4 x 3
= 6cm2
V = 6 x 8
= 48cm2
Question 8 Report
Question 9 Report
Question 10 Report
Question 11 Report
Question 12 Report
If the price of oranges was raised by \( \frac{1}{2} \)k per orange. The number of oranges a customer can buy for ?2.40 will be less by 16. What is the present price of an orange?
Question 15 Report
The table below is drawn for a graph \(y = x^3 - 3x + 1\).
| \(x\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) |
| \(y = x^3 - 3x + 1\) | \(1\) | \(-1\) | \(3\) | \(1\) | \(-1\) | \(3\) | \(1\) |
From x = -2 to x = 1, the graph crosses the x-axis in the range(s)
Question 16 Report
Question 17 Report
Rationalize \( \frac{5\sqrt{7}-7\sqrt{5}}{\sqrt{7}-\sqrt{5}} \)
Question 18 Report
Question 19 Report
Using \( \triangle \) XYZ in the figure, find XYZ.
sin 120∘ = sin 60∘
5 sin y = 3 sin 60∘
sin y = 3sin60o5
3×0.8665
= 2.5985
y = sin-1 0.5196 = 30∘ 18'
Question 20 Report
The sides of a triangle are \( (x + 4)\text{cm} \), \( x\text{cm} \) and \( (x - 4)\text{cm} \), respectively If the cosine of the largest angle is \( \frac{1}{5} \), find the value of \( x \)
Cosine B = 15
= 0.2 given
b2 - a2 + c2 - 2a Cos B
Cos B = a2+c2−b22ac
15
= x2+?(x−4)2−(x+4)22x(x−4)
15
= x(x−16)2x(x−4)
15
= x−162x−8
= 5(x - 16)
= 2x - 8
3x = 72
x = 723
= 24
Question 21 Report
Question 22 Report
Find the area of the shaded portion of the semicircular figure.
= 16πr2
A△ = 12r2sin60o
12r2×√32=r2√34
Ashaded portion
= Asector -
A△
= (16πr2−r2√34)3
= πr22−3r2√34
= r24(2π−3√3)
Question 23 Report
The quadratic equation whose roots are \(1 - \sqrt{13}\) and \(1 + \sqrt{13}\) is
Question 24 Report
In the figure, PQRSTW is a regular hexagon. QS intersects RT at V. Calculate TVS.
Hexagon is a six sided polygon.
Sum of interior angles of polygon = (2n - 4)90∘ = [2 x 6 - 4] x 90 = 8 x 90 = 720∘
each angle = 720o6=120o and TVS = 1202=60o
Question 25 Report
Answer Details
Question 26 Report
If \(a = \frac{2x}{1-x}\) and \(b = \frac{1+x}{1-x}\), then \(a^2 - b^2\) in the simplest form is \(\frac{3x+1}{1-x}\)
Question 27 Report
In the figure, TSP = PRQ, QR = 8cm, PR = 6cm and ST = 12cm. Find the length SP.
612=8PS
PS = 12×86
= 16cm
Question 28 Report
Find, without using logarithm tables, the value of \( \frac{\log_{3} 27-\log_{\frac{1}{4}} 64}{\log_{3} \frac{1}{81}} \)
Question 29 Report
Simplify \( \frac{3^n - 3^{n-1}}{3^3 \times 3^n - 27 \times 3^{n-1}} \)
Question 30 Report
Question 31 Report
Find the x co-ordinates of the points of intersection of the two equations in the graph.
then x2 - 2x + 1 = 2x + 1
x2 - 4x = 0
x(x - 4) = 0
x = 0 or 4
Question 32 Report
6x < 108
= x < 18
0 < x < 18 = 0 ≤ x < 18
Question 33 Report
Question 34 Report
Simplify \( (1 + \frac{\frac{x - 1}{1}}{\frac{1}{x + 1}})(x + 2) \)
Answer Details
Question 35 Report
Answer Details
Question 36 Report
Measurements of the diameters, in centimeters, in centimeters, of 20 copper spheres are distributed as shown below:
| Class boundary in cm | frequency |
| 3.35 − 3.45 | 3 |
| 3.45 − 3.55 | 6 |
| 3.55 − 3.65 | 7 |
| 3.65 − 3.75 | 4 |
What is the mean diameter of the copper spheres?
Question 37 Report
Find a factor which is common to all 3 binomial expressions \(4a^2 - 9b^2,\; 8a^3 + 27b^3,\; (4a + 6b)^2\).
Answer Details
Question 38 Report
Question 39 Report
Question 40 Report
XYZ is a triangle and XW is perpendicular to YZ at = W. If XZ = 5cm and WZ = 4cm, Calculate XY.
Also by Pythagoras theorem, XY2 = 62 + 32
XY2 = 36 + 9 = 45
XY = √45=3√3
Question 41 Report
Question 42 Report
In the figure, 0 is the centre of circle PQRS and PS//RT. If PRT = 135, then PSO is
< R = < P = 45∘ (corresponding angles)
< PSO = < P = 45∘ (△ PSO is a right angle)
Question 43 Report
Question 44 Report
Simplify \( \frac{\left(\frac{2}{3}-\frac{1}{5}\right)-\frac{1}{3}\text{ of }\frac{2}{5}}{3-\frac{1}{1\frac{1}{2}}} \)
23−15
= 10−315
= 715
13
Of 25
= 13
x 25
= 25
(23−15
) - 13
of 25
= 715−215
= 13
3 - 1112
= 3 - 23
= 73
23−15of2153−1112
= 1373
= 13
x 37
= 17
Question 45 Report
The cumulative frequency function of the data below is given below by the equation \(y = cf(x)\). What is \(cf(5)\)?
| Score(\(n\)) | Frequency(\(f\)) |
| 3 | 30 |
| 4 | 32 |
| 5 | 30 |
| 6 | 35 |
| 8 | 20 |
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