Paper 1  Objectives  49 Questions
WASSCE/WAEC MAY/JUNE
Year: 2002
Level: SHS
Time:
Type: Question Paper
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#  Question  Ans 

1. 
In the diagram O is the center of the circle. Reflex angle XOY = 210° and the length of the minor arc is 5.5m. Find, correct to the nearest meter, the length of the major arc. A. 8m B. 9m C. 10m D. 13m
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Detailed SolutionGiven, Length of minor arc = 5.5mAngle subtended by minor arc = 360°  210° = 150° \(\therefore 5.5 = \frac{150}{360} \times 2 \times \frac{22}{7} \times r \) \(\frac{55r}{21} = 5.5\) \(r = \frac{5.5 \times 21}{55}\) r = 2.1m Length of major arc = \(\frac{210}{360} \times 2 \times \frac{22}{7} \times 2.1\) = \(7.7m \approxeq 8m\) (to the nearest metre) 

2. 
A right pyramid is on a square base of side 4cm. The slanting side of the pyramid is \(2\sqrt{3}\) cm. Calculate the volume of the pyramid A. \(5\frac{1}{3}cm^3\) B. \(10\frac{2}{3}cm^3\) C. \(16cm^3\) D. \(32cm^3\)
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Detailed Solution\(BD^2 = 4^2 + 4^2\) \(BD = \sqrt{16 + 16} = \sqrt{32}\) \(BD = 4\sqrt{2} cm\) \((2\sqrt{3})^2 = (2\sqrt{2})^2 + h^2\) \(h^2 = 12  8 = 4\) \(h = \sqrt{4} = 2 cm\) Volume of pyramid = \(\frac{a^2 h}{3}\) = \(\frac{4^2 \times 2}{3}\) = \(\frac{32}{3} = 10\frac{2}{3} cm^3\) 

3. 
The height of a right circular cone is 4cm. The radius of its base is 3cm. Find the curved surface area A. \(9\pi cm^2\) B. \(15\pi cm^2\) C. \(16\pi cm^2\) D. \(20\pi cm^2\)
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Detailed SolutionCurved surface area or a cone \(=\pi rl\)from the information \(l^2 = 4^2 + 3^2 = 16+9\\ l = \sqrt{25} = 5; ∴ C.S.A\hspace{1mm} = \frac{22}{7}\times 3 \times 5\\ Since \frac{22}{7}=\pi ∴ C.S.A\hspace{1mm} =\hspace{1mm}15\pi\) 

4. 
In the diagram above, ∠PQU=36°, ∠QRT = 29°, PQRT. Find ∠PQR A. 94^{o} B. 65^{o} C. 61^{o} D. 54^{o}
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Detailed Solution< UQR = 29° (alternate angles)< PQR = < PQU + < UQR = 36° + 29° = 65° 

5. 
Simplify \(5\frac{1}{4}\div \left(1\frac{2}{3} \frac{1}{2}\right)\) A. \(1\frac{3}{4}\) B. \(3\frac{1}{2}\) C. \(4\frac{1}{2}\) D. \(8\frac{1}{2}\)
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Detailed Solution\(5\frac{1}{4}\div \left(1\frac{2}{3} \frac{1}{2}\right)\\\frac{21}{4}\div \left(1\frac{43}{6}\right)\\ \frac{21}{4}\div \left(1\frac{1}{6}\right)\\ \frac{21}{4} \times \frac{6}{7}= 4\frac{1}{2}\) 

6. 
Find the value of x in 0.5x + 2.6 = 5x + 0.35 A. 0.5 B. 2 C. 2.6 D. 5
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Detailed Solution\(0.5x + 2.6 = 5x + 0.35\\0.5x  5x = 0.352.6\\ 4.5x = 2.25\\ x = \frac{2.25}{4.5}\\ 0.5\) 

7. 
Find the value of x in the diagram A. 31^{o} B. 35^{o} C. 37^{o} D. 41^{o}
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Detailed SolutionSum of exterior angle of any polygon is 360^{o}(2x+5)^{o} + 2x^{o} + (x20)^{o} + x^{o} + (3x+10)^{o} + (x + 15)^{o} = 360^{o}; 10x = 350 x = 35 

8. 
If \(M5_{ten} = 1001011_{two}\) find the value of M A. 5 B. 6 C. 7 D. 8
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Detailed Solution\(M5_{ten} = 1001011_{two}\\=1 \times 2^6 + 0\times 2^5 + 0\times 2^4 + 1\times 2^3 + 0\times 2^2 + 1\times 2^1 \\ =64+8+2+1=75_{ten}\\ ∴ m = 7 \) 

9. 
The diagram is the graph of \(y = 6 + x  x^2\). The graph intercepts the x axis at P and R and the y axis at Q.What is the value of y at Q? A. \(6\frac{1}{3}\) B. 6 C. 3 D. zero 
A 
10. 
The diagram is the graph of \(y = 6 + x  x^2\). The graph intercepts the x axis at P and R and the y axis at Q.When \(y = 3\frac{1}{3}\), what is the positive value of x? A. \( 2\frac{1}{2}\) B. \( 2\frac{1}{5}\) C. \( 1\frac{1}{5}\) D. zero 
B 
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