Mathematics (Core), 1993, WASSCE/WAEC MAY/JUNE, Exam, Paper 1 | Objectives

Paper Info

Year :

1993

Title :

Mathematics (Core)

Exam :

WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

11 - 20 of 48 Questions

#	Question	Ans
11.	Write as a single fraction: \(\frac{5}{6r} - \frac{3}{4r}\) A. r/12 B. 12/r C. 1/6r D. r/6 E. 1/12r Show Content Detailed Solution \(\frac{5}{6r} - \frac{3}{4r}\) = \(\frac{10 - 9}{12r}\) = \(\frac{1}{12r}\)	E
12.	Factorize 2x\(^2\) - 21x + 45 A. (2x - 9)(x - 5) B. (2x-15)(x + 3) C. (2x +15)(x - 3) D. (2x -15(x - 3) E. (2x - 9)(x + 5) Show Content Detailed Solution 2x\(^2\) - 21x + 45 = 2x\(^2\) - 15x - 6x + 45 = x(2x - 15) - 3(2x - 15) = (x - 3)(2x - 15)	D
13.	Solve the simultaneous equations y = 3x; 4y - 5x =14 A. .-2,-6 B. 2,6 C. 2,-6 D. -2,6 E. -1,-3 Show Content Detailed Solution y = 3x ... (i); 4y - 5x =14 ... (ii) Put 3x for y in (ii). 4(3x) - 5x = 14 12x - 5x = 14 7x = 14 \(\implies\) x = 2. y = 3x = 3(2) = 6 (x, y) = (2, 6)	B
14.	A sector of a circle of radius 9cm subtends angle 120° at the centre of the circle. Find the area of the sector to the nearest cm\(^2\) [Take π = 22/7] A. 75cm² B. 84cm² C. 85cm² D. 86cm² E. 95cm² Show Content Detailed Solution Area of a sector = θ/360 x πr\(^2\) = 120/360 x 22/7 x 81/1 = 84.86 cm\(^2\) \(\approxeq\) 85cm\(^2\) (to the nearest cm)	C
15.	Calculate the total surface area of a cone of height 12cm and base radius 5cm. [Take π = 22/7] A. 180 5/7cm² B. 240 2/7cm² C. 235 5/7cm² D. 282 6/7cm² E. 361 3/7cm² Show Content Detailed Solution πrl + πr² = πr(l + r) ²²/₇ x 5(13 + 5) 282 6/7cm²	D
16.	A cone is 14cm deep and the base radius is 41/2cm. Calculate the volume of water that is exactly half the volume of the cone.[Take π = 22/7] A. 49.5cm³ B. 99cm³ C. 148.5cm³ D. 297cm³ E. 445.5cm³ Show Content Detailed Solution Volume of a cone = \(\frac{1}{3} \pi r^2 h\) r = 4\(\frac{1}{2}\) cm; h = 14 cm Volume of cone = \(\frac{1}{3} \times \frac{22}{7} \times \frac{9}{2} \times \frac{9}{2} \times 14\) = 297 cm\(^3\) When half- filled, the volume of the water = \(\frac{297}{2} = 148.5 cm^3\)	C
17.	The area and a diagonal of a rhombus are 60 cm\(^2\) and 12 cm respectively. Calculate the length of the other diagonal. A. 5cm B. 6cm C. 10cm D. 12cm E. 15cm Show Content Detailed Solution Area of rhombus = \(\frac{pq}{2}\) where p and q are the two diagonals of the rhombus. \(\therefore 60 = \frac{12 \times q}{2}\) 6q = 60 \(\implies\) q = 10 cm	C
18.	The angle of a sector of a circle radius 10.5cm is 120°. Find the perimeter of the sector [Take π = 22/7] A. 22cm B. 33.5cm C. 43cm D. 66cm E. 115.5cm Show Content Detailed Solution Perimeter of a sector = \(2r + \frac{\theta}{360} \times 2\pi r\) = \(2(10.5) + \frac{120}{360} \times 2 \times \frac{22}{7} \times 10.5\) = \(21 + 22\) = 43 cm	C
19.	In the diagram, PQR is a tangent to the circle QST at Q. If \|QT\| = \|ST\| and ∠SQR = 68°, find ∠PQT. A. 34^o B. 48^o C. 56^o D. 68^o E. 73^o Show Content Detailed Solution < STQ = < SQR = 68° (alternate segment) \(\therefore\) < STQ = 68° < TQS = \(\frac{180° - 68°}{2}\) = \(\frac{112}{2} = 56°\) \(\therefore\) < PQT = 180° - (68° + 56°) = 180° - 124° = 56°	C
20.	The sum of an interior angles of a regular polygon is 30 right angles. How many sides has the polygon? A. 34 sides B. 30 sides C. 26 sides D. 17 sides E. 15 sides Show Content Detailed Solution Sum of interior angles in a polygon = \((2n - 4) \times 90°\) \(\therefore (2n - 4) \times 90° = 30 \times 90°\) \(\implies 2n - 4 = 30 \) \(2n = 34 \implies n = 17\) The polygon has 17 sides.	D

11.	Write as a single fraction: \(\frac{5}{6r} - \frac{3}{4r}\) A. r/12 B. 12/r C. 1/6r D. r/6 E. 1/12r Show Content Detailed Solution \(\frac{5}{6r} - \frac{3}{4r}\) = \(\frac{10 - 9}{12r}\) = \(\frac{1}{12r}\)	E
12.	Factorize 2x\(^2\) - 21x + 45 A. (2x - 9)(x - 5) B. (2x-15)(x + 3) C. (2x +15)(x - 3) D. (2x -15(x - 3) E. (2x - 9)(x + 5) Show Content Detailed Solution 2x\(^2\) - 21x + 45 = 2x\(^2\) - 15x - 6x + 45 = x(2x - 15) - 3(2x - 15) = (x - 3)(2x - 15)	D
13.	Solve the simultaneous equations y = 3x; 4y - 5x =14 A. .-2,-6 B. 2,6 C. 2,-6 D. -2,6 E. -1,-3 Show Content Detailed Solution y = 3x ... (i); 4y - 5x =14 ... (ii) Put 3x for y in (ii). 4(3x) - 5x = 14 12x - 5x = 14 7x = 14 \(\implies\) x = 2. y = 3x = 3(2) = 6 (x, y) = (2, 6)	B
14.	A sector of a circle of radius 9cm subtends angle 120° at the centre of the circle. Find the area of the sector to the nearest cm\(^2\) [Take π = 22/7] A. 75cm² B. 84cm² C. 85cm² D. 86cm² E. 95cm² Show Content Detailed Solution Area of a sector = θ/360 x πr\(^2\) = 120/360 x 22/7 x 81/1 = 84.86 cm\(^2\) \(\approxeq\) 85cm\(^2\) (to the nearest cm)	C
15.	Calculate the total surface area of a cone of height 12cm and base radius 5cm. [Take π = 22/7] A. 180 5/7cm² B. 240 2/7cm² C. 235 5/7cm² D. 282 6/7cm² E. 361 3/7cm² Show Content Detailed Solution πrl + πr² = πr(l + r) ²²/₇ x 5(13 + 5) 282 6/7cm²	D

16.	A cone is 14cm deep and the base radius is 41/2cm. Calculate the volume of water that is exactly half the volume of the cone.[Take π = 22/7] A. 49.5cm³ B. 99cm³ C. 148.5cm³ D. 297cm³ E. 445.5cm³ Show Content Detailed Solution Volume of a cone = \(\frac{1}{3} \pi r^2 h\) r = 4\(\frac{1}{2}\) cm; h = 14 cm Volume of cone = \(\frac{1}{3} \times \frac{22}{7} \times \frac{9}{2} \times \frac{9}{2} \times 14\) = 297 cm\(^3\) When half- filled, the volume of the water = \(\frac{297}{2} = 148.5 cm^3\)	C
17.	The area and a diagonal of a rhombus are 60 cm\(^2\) and 12 cm respectively. Calculate the length of the other diagonal. A. 5cm B. 6cm C. 10cm D. 12cm E. 15cm Show Content Detailed Solution Area of rhombus = \(\frac{pq}{2}\) where p and q are the two diagonals of the rhombus. \(\therefore 60 = \frac{12 \times q}{2}\) 6q = 60 \(\implies\) q = 10 cm	C
18.	The angle of a sector of a circle radius 10.5cm is 120°. Find the perimeter of the sector [Take π = 22/7] A. 22cm B. 33.5cm C. 43cm D. 66cm E. 115.5cm Show Content Detailed Solution Perimeter of a sector = \(2r + \frac{\theta}{360} \times 2\pi r\) = \(2(10.5) + \frac{120}{360} \times 2 \times \frac{22}{7} \times 10.5\) = \(21 + 22\) = 43 cm	C
19.	In the diagram, PQR is a tangent to the circle QST at Q. If \|QT\| = \|ST\| and ∠SQR = 68°, find ∠PQT. A. 34^o B. 48^o C. 56^o D. 68^o E. 73^o Show Content Detailed Solution < STQ = < SQR = 68° (alternate segment) \(\therefore\) < STQ = 68° < TQS = \(\frac{180° - 68°}{2}\) = \(\frac{112}{2} = 56°\) \(\therefore\) < PQT = 180° - (68° + 56°) = 180° - 124° = 56°	C
20.	The sum of an interior angles of a regular polygon is 30 right angles. How many sides has the polygon? A. 34 sides B. 30 sides C. 26 sides D. 17 sides E. 15 sides Show Content Detailed Solution Sum of interior angles in a polygon = \((2n - 4) \times 90°\) \(\therefore (2n - 4) \times 90° = 30 \times 90°\) \(\implies 2n - 4 = 30 \) \(2n = 34 \implies n = 17\) The polygon has 17 sides.	D