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Mathematics
NECO SSCE 2004
Fractions, Decimals, Percentages & Approximations
Express the product of 0.007 and 0.057 to two significant figures
A
0.0004
B
0.00039
C
0.004
D
0.0039
E
0.04
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Myschool Past Questions Archive (Record #42334)
Mathematics
NECO SSCE 2004
Plane Geometry, Polygons & Circle Theorems
Find x if (x - 2)2 = 1\(\frac{7}{9}\)
A
\(\frac{2}{3}\) or \(\frac{2}{7}\)
B
3\(\frac{1}{3}\) or \(\frac{2}{3}\)
C
3\(\frac{7}{9}\) or \(\frac{2}{7}\)
D
3\(\frac{1}{3}\) or \(\frac{7}{9}\)
E
\(\frac{3}{7}\) or \(\frac{2}{9}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option B.
Source: Myschool Past Questions Archive (Record #42344)
Mathematics
WAEC / WASSCE 2004
Indices, Logarithms & Surds
Solution
(a) Solve \(\frac{1}{81^{(x - 2)}} = 27^{(1 - x)}\)
(b) Simplify \(\frac{5}{\sqrt{7} - \sqrt{3}} + \frac{1}{\sqrt{7} + \sqrt{3}}\), leaving your answer in surd form.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \(\frac{1}{81^{(x - 2)}} = 27^{(1 - x)}\)
\(81^{-(x - 2)} = 27^{(1 - x)}\)
\(3^{4[-(x - 2)]} = 3^{3(1 - x)}\)
\(3^{-4(x - 2)} = 3^{3(1 - x)}\)
\(-4(x - 2) = 3(1 - x)\)
\(-4x + 8 = 3 - 3x\)
\(8 - 3 = -3x + 4x \implies x = 5\)
(b) \(\frac{5}{\sqrt{7} - \sqrt{3}} + \frac{1}{\sqrt{7} + \sqrt{3}}\)
= \(\frac{5(\sqrt{7} + \sqrt{3}) + (\sqrt{7} - \sqrt{3})}{(\sqrt{7} - \sqrt{3})(\sqrt{7} + \sqrt{3})}\)
= \(\frac{5\sqrt{7} + 5\sqrt{3} + \sqrt{7} - \sqrt{3}}{7 - 3}\)
= \(\frac{6\sqrt{7} + 4\sqrt{3}}{4}\)
= \(\frac{3\sqrt{7} + 2\sqrt{3}}{2}\).
Source: Myschool Past Questions Archive (Record #53121)
Mathematics
WAEC / WASSCE 2004
Linear, Simultaneous & Quadratic Equations
Solution
(a) In the simultaneous equations : \(px + qy = 5 ; qx + py = -10\); p and q are constants. If x = 1 and y = -2 is a solution of the equations, find p and q.
(b) Solve : \(\frac{4r - 3}{6r + 1} = \frac{2r - 1}{3r + 4}\).
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \(px + qy = 5 \)
When x = 1 and y = -2, we have
\(p - 2q = 5 .... (1)\)
\(qx + py = -10 \)
When x = 1 and y = -2, we have
\(q - 2p = -10 .... (2)\)
Multiplying (1) by -2, we have
\(-2p + 4q = -10 ... (1a)\)
\(\equiv 4q - 2p = -10\)
(2) - (1a) : \(q - 4q = -10 + 10\)
\(-3q = 0\)
\(q = 0\)
Putting q = 0 in (1a), we have
\(-2p = -10 \implies p = 5\)
Hence, p = 5 and q = 0.
(b) \(\frac{4r - 3}{6r + 1} = \frac{2r - 1}{3r + 4}\)
\((4r - 3)(3r + 4) = (2r - 1)(6r + 1)\)
\(12r^{2} + 16r - 9r - 12 = 12r^{2} + 2r - 6r - 1\)
\(12r^{2} - 12r^{2} + 7r + 4r = - 1 + 12\)
\(11r = 11\)
\(r = 1\).
Source: Myschool Past Questions Archive (Record #53122)
Mathematics
WAEC / WASSCE 2004
Trigonometry, Elevation & Bearings
Solution
(a) AB is a chord of a circle centre O. If |AB| = 24.2 cm and the perimeter of \(\Delta\) AOB is 52.2 cm, calculate < AOB, correct to the nearest degree.
(b) A rectangular tank 60cm by 80cm by 100cm is half filled with water. How many litres of water is it holding?
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a)
\(2r + 24.2 = 52.2\)
\(2r = 52.2 - 24.2 = 28\)
\(r = 14 cm\)
\(|AB|^{2} = OA^{2} + OB^{2} - 2(OA)(OB) \cos
Source: Myschool Past Questions Archive (Record #53127)
Mathematics
WAEC / WASSCE 2004
Plane Geometry, Polygons & Circle Theorems
Solution
(a) The sides of an isosceles triangle triangle are in the ratio \(7 : 5 : 7\). Calculate, correct to the nearest degree, the angle included between the equal sides.
(b) The sum of the interior angles of a regular polygon is 1440^\circ. Calculate : (i) the number of sides ; (ii) the size of one exterior angle of the polygon.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a)
\(5^{2} = 7^{2} + 7^{2} - 2(7)(7) \cos \theta\)
\(25 = 98 - 98 \cos \theta\)
\(\cos \theta = \frac{98 - 25}{98}\)
\(\cos \theta = 0.7449\)
\(\theta = \cos^{-1} (0.7449)\)
\(\approxeq 42^\circ\).
(b) (i) \((2n - 4) \times 90^\circ = 1440^\circ\)
\(2n - 4 = \frac{1440}{90} = 16\)
\(2n = 16 + 4 = 20\)
\(n = 10\)
(ii) Size of exterior angle = \(\frac{360}{\text{Number of sides}}\)
= \(\frac{360}{10}\)
= \(36^\circ\)
Source: Myschool Past Questions Archive (Record #53133)
Mathematics
WAEC / WASSCE 2004
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
The table shows the marks scored by a group of students in a class test.
Marks
0
1
2
3
4
5
Frequency
1
4
9
8
5
3
(a)(i) Calculate the mean mark ; (ii) Find the median.
(b) If the information were to be represented in a pie chart, what would be the sectorial angle for the mark 2?
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a)
x
0
1
2
3
4
5
Total
f
1
4
9
8
5
3
30
fx
0
4
18
24
20
15
81
(i) Mean(\(\bar{x}\) = \(\frac{\sum fx}{\sum f}\)
= \(\frac{81}{30}\)
= 2.7
(ii) \(N = \sum f = 30\)
\(median = \frac{1}{2}(N + 1)\)
= \(\frac{1}{2}(30 + 1)\)
=15.5
= The 15th and 16th position.
\(\therefore median = \frac{3 + 3}{2} = 3\)
(b) Sectorial angle for the mark 2 = \(\frac{9}{30} \times 360^\circ\)
= 108^\circ.
Source: Myschool Past Questions Archive (Record #53139)
Mathematics
WAEC / WASSCE 2004
Trigonometry, Elevation & Bearings
Solution
(a) Simplify : \((\frac{x^{2}}{2} - x + \frac{1}{2})(\frac{1}{x - 1})\)
(b) A point P is 40km from Q on a bearing 061^\circ. Calculate, correct to one decimal place, the distance of P to (i) north of Q ; (ii) east of Q.
(c) A man left N5,720 to be shared among his son and three daughters. Each daughter's share was \(\frac{3}{4}\) of the son's share. How much did the son receive?
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \((\frac{x^{2}}{2} - x + \frac{1}{2})(\frac{1}{x - 1})\)
\(\frac{x^{2}}{2} - x + \frac{1}{2} = \frac{x^{2} - 2x + 1}{2}\)
= \(\frac{(x - 1)^{2}}{2}\)
\(\therefore (\frac{x^{2}}{2} - x + \frac{1}{2})(\frac{1}{x - 1}) = (\frac{(x - 1)^{2}}{2})(\frac{1}{x - 1})\)
= \(\frac{x - 1}{2}\)
(b)
(i) TQ = PR (North of Q)
\(\implies \sin 29 = \frac{PR}{40}\)
\(PR = 40 \sin 29 = 19.39 km\)
(ii) QR = East of Q
\(\frac{QR}{40} = \cos 29\)
\(QR = 40 \cos 29\)
= 34.98km
\(\approxeq\) 35km.
(c) Let the son's share = x.
Each daughter's share = \(\frac{3}{4}x\)
For the three daughters = \(3 \times \frac{3}{4} = \frac{9}{4}\)
\(x + \frac{9}{4}x = 5720 \implies \frac{13}{4}x = 5720\)
\(x = \frac{5720 \times 4}{13} = N1760\)
Source: Myschool Past Questions Archive (Record #53153)
Mathematics
WAEC / WASSCE 2004
Trigonometry, Elevation & Bearings
Solution
(a) The angles of depression of the top and bottom of a building are 51^\circ and 62^\circ respectively from the top of a tower 72m high. The base of the building is on the same horizontal level as the foot of the tower. Calculate the height of the building correct to 2 significant figures.
(b) In the diagram, PR is a chord of the circle centre O and radius 30cm, < POR = 120^\circ. Calculate correct to three significant figures : (i) the length of chord PR ; (ii) the length of arc PQR ; (iii) the perimeter of the shaded portion. (Take \(\pi = 3.142\)).
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a)
ST = QR ; PS = height of tower.
In \(\Delta\) PST,
\(\frac{72}{ST} = \tan 62^\circ\)
\(ST = \frac{72}{\tan 62}\)
= 38.283m \(\approxeq\) 38.3m
In \(\Delta\) PQR, \(\frac{h}{QR} = \tan 51^\circ\)
\(h = QR \times \tan 51^\circ\)
= \(38.3 \times 1.235\)
= \(47.3m\)
QS = RT (height of the building)
QS = PS - PQ
= 72 - 47.3 = 24.7m
(b)
(i) \(\Delta POR = \Delta QOR\) (isosceles triangle)
\(\frac{PT}{30} = \sin 60\)
\(PT = 30 \sin 60 = 30 \times \frac{\sqrt{3}}{2}\)
= \(15\sqrt{3} cm\)
PT = TR
PR = 2PT
= \(2 \times 15\sqrt{3} = 30\sqrt{3} cm\)
= \(30 \times 1.732 = 51.96 cm\)
\(\approxeq 52 cm\)
(ii) Length of minor arc, PQR = \(\frac{\theta}{360} \times 2\pi r\)
= \(\frac{120}{360} \times 2 \times 3.142 \times 30\)
= \(62.84 cm\)
(iii) Perimeter of shaded portion = chord PR + Arc PQR
= 51.96 + 62.84
= 114.8 cm \(\approxeq\) 115cm.
Source: Myschool Past Questions Archive (Record #53184)
Mathematics
WAEC / WASSCE 2004
Plane Geometry, Polygons & Circle Theorems
Solution
(a)
In the diagram, XY is a chord of a circle of radius 5cm. The chord subtends an angle 96^\circ at the centre. Calculate, correct to three significant figures, the area of the minor segment cut-off. (Take \(\pi = \frac{22}{7}\)).
(b) The figure shows a circle inscribed in a square. If a portion of the circle is shaded with some portions of the square, calculate the total area of the shaded portions. [Take \(\pi = \frac{22}{7}\)].
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) Area of minor sector = \(\frac{\theta}{360} \times \pi r^{2}\)
= \(\frac{96}{360} \times \frac{22}{7} \times 5^{2}\)
= \(20.95 cm^{2}\)
Area of triangle formed from the sector = \(\frac{1}{2} r^{2} \sin \theta\)
= \(\frac{1}{2} \times 5^{2} \times \sin 96\)
= \(12.43 cm^{2}\)
\(\therefore \text{The area of minor segment} = 20.95 - 12.43\)
= \(8.52 cm^{2}\)
(b) Area of minor sector = \(\frac{\theta}{360} \times \pi r^{2}\)
= \(\frac{80}{360} \times \frac{22}{7} \times 7^{2}\)
= \(34.22 cm^{2}\)
Area of the square = \((14)^{2} = 196 cm^{2}\)
Area of the circle = \(\pi r^{2} = \frac{22}{7} \times 7^{2} = 154 cm^{2}\)
Area of the shaded portion in the square = \(196 - 154 = 42 cm^{2}\)
Total area of the shaded portions = \(42 + 34.22\)
= \(76.22 cm^{2} \approxeq 76.2 cm^{2}\)
Source: Myschool Past Questions Archive (Record #53185)
Mathematics
WAEC / WASSCE 2004
Plane Geometry, Polygons & Circle Theorems
Using a ruler and a pair of compasses only,
(a) Construct : (i) \(\Delta PQR\) such that /PQ/ = 8cm, /PR/ = 7cm and < QPR = 105^\circ. (ii) locus \(L_{1}\) of points equidistant from P and Q. (iii) locus \(l_{2}\) of points equidistant Q and R.
(b)(i) Label the point T where \(l_{1}\) and \(l_{2}\) intersect ; (ii) With centre T and radius /TQ/, construct a circle \(l_{3}\). (iii) Complete quadrilateral PQSR such that /RS/ = /QS/ and /RQ/ = /TS/.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option .
Source: Myschool Past Questions Archive (Record #53186)
Mathematics
WAEC / WASSCE 2004
Mensuration (Areas, Perimeters & Volumes)
Solution
K(lat. 60^\circN, long. 50^\circW) is a point on the eart's surface. L is another point due East of K and the third point N is due South of K. The distance KL is 3520km and KN is 10951km.
(a) Calculate: (i) The longitude of L ; (ii) The latitude of N. (Take \(\pi = \frac{22}{7}\) and the radius of the earth = 6400km).
(b) A man was allowed 20% of his income as tax free. He then paid 25 kobo in the naira on the remainder. If he paid N1,200.00 as tax, calculate his total income.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a)
(i) Let the length of the longitude be \(\theta\).
Longitude difference = \(50 + \theta\)
Distance along the parallel of latitude (KL) = \(\frac{50 + \theta}{360} \times 2\pi R \cos 60\)
\(3520 = \frac{50 + \theta}{360} \times 2 \times \frac{22}{7} \times 6400 \cos 60\)
\(3520 = \frac{50 + \theta}{360} \times 2 \times \frac{22}{7} \times 3200\)
\(50 + \theta = \frac{3520 \times 360 \times 7}{22 times 2 \times 3200}\)
\(50 + \theta = 63^\circ\)
\(\theta = 63^\circ - 50^\circ = 13^\circ\)
(ii) Let the latitude of N be \(\alpha\).
The distance along the parallel of longitude (KN) = \(\frac{60 + \alpha}{360} \times 2 \times \frac{22}{7} \times 6400\)
\(10951 = \frac{60 + \alpha}{360} \times 2 \times \frac{22}{7} \times 6400\)
\(60 + \alpha = \frac{10951 \times 360 \times 7}{2 \times 22 \times 6400}\)
\(60 + \alpha = 98^\circ\)
\(\alpha = 98^\circ - 60^\circ = 38^\circ\)
The latitude of N = 38^\circS.
(b) Let the man's income tax be x.
80% of x = \(\frac{1200}{0.25} = N4800\)
\(0.8x = N4,800\)
\(x = \frac{4800}{0.8} = N6000\)
Source: Myschool Past Questions Archive (Record #53191)
Mathematics
WAEC / WASSCE 2004
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
The frequency distribution shows tha marks of 100 students in a Mathematics test.
Marks
1-10
11-20
21-30
31-40
41-50
51-60
61-70
71-80
81-90
91-100
No. of
Students
2
4
9
13
18
32
13
5
3
1
(a) Draw cumulative frequency curve for the distribution .
(b) Use your curve to estimate : (i) the median ; (ii) the lower quartile ; (iii) the 60th percentile.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a)
Marks
Frequency
Cum. freq
1 - 10
2
2
11 - 20
4
6
21 - 30
9
15
31 - 40
13
28
41 - 50
18
46
51 - 60
32
78
61 - 70
13
91
71 - 80
5
96
81 - 90
3
99
91 - 100
1
100
(b)(i) \(N = \sum f = 100\)
\(Median = \frac{1}{2}N = \frac{1}{2}(100)\)
= 50th percentile.
\(\therefore Median = 52\).
(ii) \(Q_{1} = \frac{1}{4} N = \frac{1}{4}(100)\)
= 25th percentile.
\(\therefore Q_{1} = 39\)
(iii) 60th percentile = 55.
Source: Myschool Past Questions Archive (Record #53193)
Mathematics
WAEC / WASSCE 2004
Indices, Logarithms & Surds
Solution
(a) Simplify : \(\sqrt{1001_{two}}\), leaving your answer in base two.
(b)
In the diagram, O is the centre of the circle radius x. /PQ/ = z, /OK/ = y and < OKP = 90^\circ. Find the value of z in terms of x and y.
(c)
In the diagram, P, Q, R and S are points of the circle centre O. \(\stackrel\frown{POQ} = 160^\circ\), \(\stackrel\frown{QSR} = 45^\circ\) and \(\stackrel\frown{PQS} = 40^\circ\). Calculate, (i) < QPS ; (ii) < RQS.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \(1001_{2} = 1 \times 2^{3} + 0 \times 2^{2} + 0 \times 2^{1} + 1 \times 2^{0}\)
= \(8 + 0 + 0 + 1\)
= 9
\(\sqrt{9} = 3_{10}\)
2
3
2
1 r 1
0 r 1
\(\therefore \sqrt{1001_{2}} = 11_{2}\)
(b) In \(\Delta QOP, OQ = OP\)
\(\therefore QK = KP\)
\(\Delta QOK = \Delta POK\) (right- angled triangle)
In \(\Delta POK\),
\(OP^{2} = PK^{2} + OK^{2}\)
\(x^{2} = y^{2} + (\frac{z}{2})^{2}\)
\(\frac{z^{2}}{4} = x^{2} - y^{2}\)
\(z^{2} = 4x^{2} - 4y^{2}\)
\(z = \sqrt{4(x^{2} - y^{2})}\)
\(z = 2\sqrt{x^{2} - y^{2}}\)
(c)(i) \(< PSQ = \frac{1}{2}(< POQ) = \frac{1}{2}(160^\circ)\)
= 80^\circ
In \(\Delta PSQ\),
\(< PQS + < QPS + < PSQ = 180^\circ\)
\(40^\circ + < QPS + 80^\circ = 180^\circ\)
\(< QPS = 180^\circ - 120^\circ = 60^\circ\)
(ii) \(< PSR = < PSQ + < QSR\)
= \(80^\circ + 45^\circ\)
= \(125^\circ\)
\(< PQS = < PQO + < RQS \)
= \(40^\circ + < RQS\)
\(< PSR + < PQS = 180^\circ\)
\(\therefore 125^\circ + 40^\circ + < RQS = 180^\circ\)
\(< RQS = 180^\circ - 165^\circ\)
= \(15^\circ\)
Source: Myschool Past Questions Archive (Record #53195)
Mathematics
WAEC / WASSCE 2004
Differentiation & Applications
Solution
(a) Copy and complete the following table of values for the relation \(y = 2x^{2} - 7x - 3\).
x
-2
-1
0
1
2
3
4
5
y
19
-3
-9
(b) Using 2 cm to 1 unit on the x- axis and 2 cm to 5 units on the y- axis, draw the graph of \(y = 2x^{2} - 7x - 3\) for \(-2 \leq x \leq 5\).
(c) From your graph, find the : (i) minimum value of y ;
(ii) gradient of the curve at x = 1.
(d) By drawing a suitable straight line, find the values of x for which \(2x^{2} - 7x - 5 = x + 4\).
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \(y = 2x^{2} - 7x - 3\)
x
-2
-1
0
1
2
3
4
5
\(2x^{2}\)
8
2
0
2
8
18
32
50
\(-7x\)
14
7
0
-7
-14
-21
-28
-35
\(-3\)
-3
-3
-3
-3
-3
-3
-3
-3
\(y\)
19
6
-3
-8
-9
-6
1
12
(b)
(c)(i) Minimum value of y = -9
(ii) \(y = 2x^{2} - 7x - 3\)
Gradient = \(\frac{\mathrm d y}{\mathrm d x} = 4x - 7\)
Gradient at x = 1 : \(4(1) - 7 = 4 - 7 = -3\).
(d) \(2x^{2} - 7x - 5 = x + 4\)
\(2x^{2} - 7x - 5 + 2 = x + 4 + 2\)
\(2x^{2} - 7x - 3 = x + 6\)
\(\implies y = x + 6\)
x
-2
-1
0
1
2
3
4
5
6
6
6
6
6
6
6
6
6
y
4
5
6
7
8
9
10
11
From the graph, x = -0.9 or 5.
Source: Myschool Past Questions Archive (Record #53197)