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Question 1 Report
Question 2 Report
The pie chart above shows the statistical distribution of 80 students in five subjects in an examination. Calculate how many student offer Mathematics.
360x∘ - 24 + 12 + 12 = 360∘
36x∘ = 360∘
x∘ = 360036
= 10∘
Thus, the angle of sector representing Mathematics is 5 x 10∘ = 50∘ . Hence the number of students who offer mathematics is
55o36×80≈11
Question 3 Report
Question 4 Report
Find the inverse \( \begin{vmatrix} 5 & 3 \\ 6 & 4 \end{vmatrix} \)
Question 5 Report
Evaluate \( \frac{1.25 \times 0.025}{0.05} \), correct to 1 decimal place
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
Length of chord = 2rsin(θ2)
= 2×8×sin(902)
= 16×2√2
= 82–√cm
Question 8 Report
Question 9 Report
Question 10 Report
Integrate \( \frac{1+x}{x^3}\,\mathrm{d}x \)
∫1+xx3dx
= ∫(1x3+xx3)dx
= ∫(x−3+x−2)dx
= −12x2−1x+k
Question 11 Report
| Age | 20 | 25 | 30 | 35 | 40 | 45 |
| Number of people | 3 | 5 | 1 | 1 | 2 | 3 |
Find the median age of the frequency distribution in the table above.
Question 12 Report
Simplify \( \frac{3^{-5n}}{9^{1-n}} \times 27^{n+1} \)
3−5n−2(1−n)+3(n+1)
3−5n−2+2n+3n+3
3−5n+5n+3−2
31
= 3
Question 13 Report
What is the probability that an integer \( (1 \le x \le 25) \) chosen at random is divisible by both 2 and 3?
Question 14 Report
P, Q and R are subsets of the universal set U. The Venn diagram showing the relationship \( (P \cap Q) \cup R \) is
Answer Details
Question 15 Report
If \(y = (2x + 2)^3\), find \(\frac{\delta x}{\delta y}\).
Question 16 Report
The bar chart above shows the allotment of time(in minutes) per week for selected subjects in a certain school. What is the total time allocated to the six subjects per week?
Question 17 Report
Question 18 Report
In the diagram above, < CDE = < CED = 15o
(base < s of isos. △)
< ECD = 180o - (15 + 15)o
= 180o - 30o = 150o
But x + 110o = 150o
(Sum of opp. interior < s of a△ = opp. exterior < )
x = 150o - 110o = 40o
Question 19 Report
Question 20 Report
Question 21 Report
If \(P = \begin{vmatrix} 5 & 3 \\ 2 & 1 \end{vmatrix}\) and \(Q = \begin{vmatrix} 4 & 2 \\ 3 & 5 \end{vmatrix}\), find \(2P + Q\)
Question 22 Report
Question 23 Report
If \( \tan \theta = \frac{3}{4} \), find the value of \( \sin \theta + \cos \theta \).
tanθ=oppadj=34
hyp2=opp2+adj2
hyp=32+42−−−−−−√
= 5
sinθ=35;cosθ=45
sinθ+cosθ=35+45
= 75=125
Question 25 Report
Question 26 Report
Question 27 Report
Convert 2710 to another number in base three
Question 28 Report
Evaluate \( \int_{0}^{\frac{\pi}{2}} \sin x \, dx \)
Question 29 Report
Question 30 Report
If y = x sin x, find \( \frac{\delta y}{\delta x} \)
Question 31 Report
If \(S = \sqrt{t^2 - 4t + 4}\), find t in terms of S
Question 32 Report
Question 33 Report
Question 34 Report
P varies jointly as m and u, and varies inversely as q. Given that p = 4, m = 3 and u = 2 and q = 1, find the value of p when m = 6, u = 4 and q = \( \frac{8}{5} \)
Question 35 Report
The nth term of the progression \( \frac{4}{2}, \frac{7}{3}, \frac{10}{4}, \frac{13}{5} \) is ..
Question 36 Report
In triangle PQR, \(q = 8\text{ cm}\), \(r = 6\text{ cm}\) and \(\cos P = \frac{1}{12}\). Calculate the value of \(p\).
Using the cosine rule, we have
p2=q2+r2−2qrcosP
p2=82+62−2(8)(6)(112)
= 64+36−8
p2=92∴p=92−−√cm
Question 37 Report
A chord of a circle subtends an angle of 120° at the centre of a circle of diameter \(4\sqrt{3}\,cm\). Calculate the area of the major sector.
Angle of major sector = 360° - 120° = 240°
Area of major sector : θ360×πr2
r = 43√2=23–√cm
Area : 240360×π×(23–√)2
= 8πcm2
Question 38 Report
The graph is shown is correctly represented by
Question 39 Report
Question 40 Report
Question 41 Report
Question 42 Report
| Score | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| Frequency | 1 | 0 | 7 | 5 | 2 | 3 | 1 | 1 |
The table above shows the scores of 20 students in further mathematics test. What is the range of the distribution?
Question 43 Report
Question 44 Report
Question 45 Report
Simplify \( \frac{\sqrt{5}(\sqrt{147}-\sqrt{12})}{\sqrt{15}} \)
Question 46 Report
Question 47 Report
Question 48 Report
Given P(2, -3) and Q(-5, 1)
Midpoint = (2+(−5)2,−3+12)
= (−32,−1)
Slope of the line PQ = 1−(−3)−5−2
= −47
The slope of the perpendicular line to PQ = −1−47
= 74
The equation of the perpendicular line: y=74x+b
Using a point on the line (in this case, the midpoint) to find the value of b (the intercept).
−1=(74)(−32)+b
−1+218=138=b
∴ The equation of the perpendicular bisector of the line PQ is y=74x+138
≡8y=14x+13⟹8y−14x−13=0
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