1993 - WAEC Mathematics Past Questions and Answers - page 2
Write as a single fraction: \(\frac{5}{6r} - \frac{3}{4r}\)
\(\frac{5}{6r} - \frac{3}{4r}\)
= \(\frac{10 - 9}{12r}\)
= \(\frac{1}{12r}\)
Factorize 2x\(^2\) - 21x + 45
2x\(^2\) - 21x + 45
= 2x\(^2\) - 15x - 6x + 45
= x(2x - 15) - 3(2x - 15)
= (x - 3)(2x - 15)
Solve the simultaneous equations y = 3x; 4y - 5x =14
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)
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]
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)
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]
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 = \(\frac{297}{2} = 148.5 cm^3\)
The area and a diagonal of a rhombus are 60 cm\(^2\) and 12 cm respectively. Calculate the length of the other diagonal.
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
The angle of a sector of a circle radius 10.5cm is 120°. Find the perimeter of the sector [Take π = 22/7]
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
In the diagram, PQR is a tangent to the circle QST at Q. If |QT| = |ST| and ∠SQR = 68°, find ∠PQT.
< 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°
The sum of an interior angles of a regular polygon is 30 right angles. How many sides has the polygon?
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.