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Mathematics
WAEC / WASSCE 2002
Indices, Logarithms & Surds
Solution
If \(M5_{ten} = 1001011_{two}\) find the value of M
A
5
B
6
C
7
D
8
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(M5_{ten} = 1001011_{two}\\
=1 \times 2^6 + 0\times 2^5 + 0\times 2^4 + 1\times 2^3 + 0\times 2^2 + 1\times 2^1 \\
=64+8+2+1=75_{ten}\\
∴ m = 7 \)
Source: Myschool Past Questions Archive (Record #26527)
Mathematics
WAEC / WASSCE 2002
Indices, Logarithms & Surds
Solution
If log 2 = 0.3010 and log 2\(^y\) = 1.8062, find; correct to the nearest whole number, the value of y.
A
6
B
5
C
4
D
-5
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(log2 = 0.3010\hspace{1mm}given\\
log2^y = 1.8062\\
∴ ylog2 = 1.8062\\
y=\frac{1.8062}{logy}=\frac{1.8062}{0.3010}=6\)
Source: Myschool Past Questions Archive (Record #26532)
Mathematics
WAEC / WASSCE 2002
Indices, Logarithms & Surds
Solution
Points X, Y and Z are located in the same horizontal plane such that Y is 12 km north of X and Z is on a bearing of 270^\circ from X. If |XZ| = 6km, calculate, correct to one decimal place, lYZl
A
18km
B
13.4km
C
13km
D
10.4km
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
ZY\(^2\) = 12\(^2\) + 6\(^2\)
ZY\(^2\) = 144 + 36 = 180
ZY = \(\sqrt{180}\)
= \(13.416\)
= 13.4 km
Source: Myschool Past Questions Archive (Record #26533)
Mathematics
WAEC / WASSCE 2002
Indices, Logarithms & Surds
Solution
Simplify \(\frac{2^{\frac{1}{2}}\times 8^{\frac{1}{2}}}{4}\)
A
1
B
2
C
4
D
16
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\frac{2^{\frac{1}{2}}\times 8^{\frac{1}{2}}}{4}\\=\frac{2^{\frac{1}{2}}\times 2^{\frac{3}{2}}}{2^2}\\=2^{\frac{1}{2}}+2^{\frac{3}{2}}-2=2^0 = 1\)
Source: Myschool Past Questions Archive (Record #26551)
Mathematics
WAEC / WASSCE 2002
Indices, Logarithms & Surds
Solution
Simplify \(3\sqrt{12} + 10\sqrt{3} - \frac{6}{\sqrt{3}}\)
A
\(7\sqrt{3}\)
B
\(10\sqrt{3}\)
C
\(14\sqrt{3}\)
D
\(18\sqrt{3}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(3\sqrt{12} + 10\sqrt{3} - \frac{6}{\sqrt{3}}\)
= \(3(\sqrt{4 \times 3}) + 10\sqrt{3} - (\frac{6}{\sqrt{3}})(\frac{\sqrt{3}}{\sqrt{3}})\)
= \(6\sqrt{3} + 10\sqrt{3} - 2\sqrt{3}\)
= \(14\sqrt{3}\)
Source: Myschool Past Questions Archive (Record #26555)
Mathematics
WAEC / WASSCE 2002
Indices, Logarithms & Surds
Diagram
Solution
In the diagram, QRT is a straight line. If angle PTR = 90^\circ, angle PRT = 60^\circ, angle PQR = 30^\circ and |PQ| = \(6\sqrt{3}cm\), calculate |RT|
Attached Figure / Diagram
A
0.3cm
B
\(\frac{\sqrt{3}}{2}cm\)
C
3cm
D
\(3\sqrt{3}cm\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
In \(\Delta\) QPT,
\(\frac{PT}{6\sqrt{3}} = \sin 30^\circ\)
PT = \(6\sqrt{3} \times \frac{1}{2} = 3\sqrt{3} cm\)
In \(\Delta\) RPT,
\(\frac{PT}{RT} = \tan 60^\circ\)
\(\frac{3\sqrt{3}}{RT} = \tan 60^\circ\)
\(RT = \frac{3\sqrt{3}}{\sqrt{3}} = 3 cm\)
Source: Myschool Past Questions Archive (Record #26558)
Mathematics
WAEC / WASSCE 2002
Indices, Logarithms & Surds
Diagram
Solution
In the diagram, |QR| = 10cm, PR⊥QS, angle PSR = 30^\circ and angle PQR = 45^\circ. Calculate in meters |QS|
Attached Figure / Diagram
A
\(10(1+\sqrt{3})\)
B
\(20\sqrt{3}\)
C
\(10\sqrt{3}\)
D
\((10+\sqrt{3})\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
In \(\Delta\) PQR,
\(\tan 45 = \frac{PR}{10} \implies PR = 10 \tan 45\)
= 10m
In \(\Delta\) PRS,
\(\tan 30 = \frac{10}{RS} \implies RS = \frac{10}{\tan 30}\)
= \(\frac{10}{\frac{1}{\sqrt{3}}\)
= \(10\sqrt{3}\)
PS = \(10 + 10\sqrt{3}\)
= \(10(1 + \sqrt{3}) cm\)
Source: Myschool Past Questions Archive (Record #26561)
Mathematics
WAEC / WASSCE 2002
Indices, Logarithms & Surds
Which of the following is not a rational number?
A
-5
B
\(\sqrt{4}\)
C
\(3\frac{3}{4}\)
D
\(\sqrt{90}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #26568)
Mathematics
NECO SSCE 2002
Indices, Logarithms & Surds
Solution
Evaluate 1.25 \times 10\(^2\) \times 0.125 \times 10\(^3\) \times 12.5 \times 10\(^{-4}\) leaving your answer in standard form
A
1.995 × 103
B
1.95 × 102
C
1.95 × 101
D
1.95 × 10-2
E
1.95 × 10-3
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Where 1.25 \times 10\(^2\) = 125
0.125 \times 10\(^3\) = 125
12.5 \times 10\(^{-4}\) = 0.00125
: 125 x 125 x 0.00125
= 19.53125 or 1.95 x 10\(^1\)
Source: Myschool Past Questions Archive (Record #42011)
Mathematics
NECO SSCE 2002
Indices, Logarithms & Surds
The angle of elevation of the top (T) of a palm tree from a point (R) on the ground is 30°. The distance from R to the base (B) of the palm tree is 120m. What is the distance between T and R?
A
60m
B
60\(\sqrt{3}\)m
C
80\(\sqrt{3}\)m
D
120\(\sqrt{3}\)m
E
240m
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option C.
Source: Myschool Past Questions Archive (Record #42070)
Mathematics
NECO SSCE 2002
Indices, Logarithms & Surds
Solution
If \(\frac{t}{L}\) = \(\sqrt{(\frac{4\pi^2}{g}})\), what is the value of g when L = 4.5; t = 5 secs
A
3.20 ms2
B
11.31 ms2
C
32.00 ms2
D
56.57 ms2
E
160.00 ms2
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\frac{t}{L}\) = \(\sqrt{(\frac{4\pi^2}{g}})\)
Take square of both sides
\(\frac{t^2}{L^2}\) = \(\frac{4\pi^2}{g}\)
Cross multiply
t\(^2\)g = L\(^2\)4\(\pi\)\(^2\)
g = \(\frac{4\pi^2L^2}{t^2}\)
g = \(\frac{4 \times 4.5 \times 4.5 \pi^2}{5^2}\)
g = \(\frac{799.44}{25}\) = 31.9776 ≈ 32m/s\(^2\)
Source: Myschool Past Questions Archive (Record #42078)
Mathematics
NECO SSCE 2002
Indices, Logarithms & Surds
If sin θ = \(\frac{\sqrt{3}}{2}\) and 0 < θ < 180°, what is cos θ ?
A
\(\frac{1}{\sqrt{3}}\)
B
\(\frac{2}{\sqrt{3}}\)
C
\(\frac{1}{2}\)
D
\(\frac{\sqrt{3}}{2}\)
E
\(\sqrt{3}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option C.
Source: Myschool Past Questions Archive (Record #42179)
Mathematics
WAEC / WASSCE 2002
Indices, Logarithms & Surds
Solution
(a) Simplify : \((2a + b)^{2} - (b - 2a)^{2}\)
(b) Given that \(S = K\sqrt{m^{2} + n^{2}}\); (i) make m the subject of the relations ; (ii) if S = 12.2, K = 0.02 and n = 1.1, find, correct to the nearest whole number, the positive value of m.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \((2a + b)^{2} - (b - 2a)^{2}\)
Using the method of difference of two squares,
\((2a + b)^{2} - (b - 2a)^{2} = ((2a + b) + (b - 2a))((2a + b) - (b - 2a))\)
= \((2b)(4a)\)
= \(8ab\).
(b)(i) \(S = K\sqrt{m^{2} + n^{2}}\)
\(S^{2} = K^{2}[m^{2} + n^{2}]\)
\(\frac{S^{2}}{K^{2}} = m^{2} + n^{2}\)
\(m^{2} = \frac{S^{2}}{K^{2}} - n^{2}\)
\(m = \sqrt{\frac{S^{2}}{K^{2}} - n^{2}}\)
(ii) When S = 12.2, K = 0.02, n = 1.1
\(m = \sqrt{\frac{12.2^{2}}{0.02^{2}} - (1.1^{2})}\)
\(m = \sqrt{\frac{148.84}{0.0004} - 1.21}\)
\(m = \sqrt{372,098.79}\)
= \(609.999 \approxeq 610\)
Source: Myschool Past Questions Archive (Record #53314)
Mathematics
WAEC / WASSCE 2002
Indices, Logarithms & Surds
Solution
(a) Without using calculator or mathematical tables, evaluate \(\frac{3}{\sqrt{3}}(\frac{2}{\sqrt{3}} - \frac{\sqrt{12}}{6})\)
(b) In the diagram, O is the centre of the circle. The side AB is produced to E, < ACB = 49^\circ and < CBE = 68^\circ. Calculate,
(i) the interior angle AOC ; (ii) < BOC.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \(\frac{3}{\sqrt{3}}(\frac{2}{\sqrt{3}} - \frac{\sqrt{12}}{6})\)
= \(\frac{6}{\sqrt{9}} - \frac{3\sqrt{12}}{6\sqrt{3}}\)
= \(2 - \frac{2\sqrt{3}}{2\sqrt{3}}\)
= \(2 - 1 = 1\)
(b)(i) \(< ABC = 180^\circ - 68^\circ = 112^\circ\)
\(< AOC = 2 \times 112^\circ = 224^\circ\)
\(\hat{O} = < AOC (obtuse)\)
= \(360^\circ - 224^\circ = 136^\circ\)
(ii) In \(\Delta AOC\), OA = OC (radii)
\(\therefore \Delta AOC = isosceles\)
\(\hat{A} = < OAC ; \hat{C} = < ACO ; \hat{A} = \hat{C} = x\)
\(\hat{O} + \hat{A} + \hat{C} = 180^\circ\)
\(136^\circ + x + x = 180^\circ \implies 2x = 44^\circ\)
\(x = 22^\circ\)
In \(\Delta BOC\), OB = OC (radii)
\(\therefore \Delta BOC = isosceles\)
\(< OCB = 49^\circ + 22^\circ= 71^\circ\)
\(< OBC = < OCB = 71^\circ\)
\(< OBC + < OCB + < BOC = 180^\circ\)
\(71^\circ + 71^\circ + < BOC = 180^\circ\)
\(< BOC = 180^\circ - 142^\circ = 38^\circ\)
Source: Myschool Past Questions Archive (Record #53319)
Mathematics
WAEC / WASSCE 2002
Indices, Logarithms & Surds
Solution
(a) Simplify : \(\frac{1}{2}\log_{10} 25 - 2\log_{10} 3 + \log_{10} 18\)
(b) If \(123_{y} = 83_{10}\), obtain an equation in y, hence find the value of y.
(c) Solve the equation \(\frac{9^{2x - 3}}{3^{x + 3}} = 1\)
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \(\frac{1}{2}\log_{10} 25 - 2\log_{10} 3 + \log_{10} 18\)
= \(\log_{10} (25^{\frac{1}{2}}) - \log_{10} (3^{2}) + \log_{10} 18\)
= \(\log_{10} (\frac{5 \times 18}{9})\)
= \(\log_{10} 10\)
= 1
(b) \(123_{y} = (1 \times y^{2}) + (2 \times y^{1}) + (3 \times y^{0})\)
\(y^{2} + 2y + 3 = 83\)
\(y^{2} + 2y + 3 - 83 = 0 \implies y^{2} + 2y - 80 = 0\)
Equation : \(y^{2} + 2y - 80 = 0\)
\(y^{2} - 8y + 10y - 80 = 0\)
\(y(y - 8) + 10(y - 8) = 0\)
\((y + 10)(y - 8) = 0 \implies y = -10 ; 8\)
y = 8 since y cannot be negative.
(c) \(\frac{9^{2x - 3}}{3^{x + 3}} = 1\)
\(\implies 9^{2x - 3} = 3^{x + 3}\)
\((3^{2})^{2x - 3} = 3^{x + 3}\)
\(3^{4x - 6} = 3^{x + 3}\)
\(\implies 4x - 6 = x + 3\)
\(4x - x = 3 + 6 \implies 3x = 9\)
\(x = 3\)
Source: Myschool Past Questions Archive (Record #53327)