Year :
2021
Title :
Mathematics (Core)
Exam :
WASSCE/WAEC MAY/JUNE
Paper 1 | Objectives
21 - 30 of 49 Questions
| # | Question | Ans |
|---|---|---|
| 21. |
Find The quadratic Equation Whose Roots Are -2q And 5q. A. 3x\(^2\) + 3qx - 10q\(^2\) B. x\(^2\) + 3qx + 10q\(^2\) C. x\(^2\) - 3qx + 10q\(^2\) D. x\(^2\) - 3qx - 10q\(^2\) Detailed Solutionx\(^2\) - (sum of roots)x + (products of roots) = 0x\(^2\) - (-2q + 5q) + (-2q * 5q) = 0 x\(^2\) -(3q) + (-10q\(^2\)) = 0 x\(^2\) -3q - 10q\(^2\) = 0 |
|
| 22. |
If tanθ = \(frac{3}{4}\), 180° < θ < 270°, find the value of cosθ. A. \(\frac{4}{5}\) B. \(\frac{3}{5}\) C. -\(\frac{4}{5}\) D. -\(\frac{3}{5}\) Detailed Solutiontanθ = \(frac{3}{4}\) → tanθ = 0.75θ = tan\(^{-1}\)[0.75] → 36.8698° cosθ = cos[36.8698°] → 0.800 or \(frac{4}{5}\) However; in the third quadrant Cosine is negative i.e -\(frac{4}{5}\) |
|
| 23. |
If \(\frac{2}{x-3}\) - \(\frac{3}{x-2}\) = \(\frac{p}{(x-3)(x -2)}\), find p. A. 5 - x B. - (x + 5) C. 13 - x D. - (5x - 13) Detailed Solution\(\frac{2}{x-3}\) - \(\frac{3}{x-2}\)= \(\frac{2(x -2) -3(x - 3)}{(x-3)(x -2)}\) = \(\frac{2x -4 -3x + 9}{(x-3)(x -2)}\) = \(\frac{5 - x}{(x-3)(x -2)}\) |
|
| 24. |
The diagonal of a rhombus are 12cm and 5cm. calculate its perimeter A. 26cm B. 24cm C. 17cm D. 34cm Detailed SolutionPerimeter of a Rhombus = 4L or 2√( d1\(^2\) + d2\(^2\))P = 2√(12\(^2\) + 5\(^2\)) P = 2√(144 + 25) P = 2√169 = 2 * 13 P = 26cm |
|
| 25. |
A. 110° B. 130° C. 140° D. 180° Detailed SolutionIsosceles Triangle. Two equal sides. Two equal angles ;∠ X = ∠ Z ∠Y + ∠X + Z = 180 --> 40° + ∠X + ∠X = 180° ∠2X = 180° - 40° = 140° ∠X = 70°, i.e ∠X = 70 and ∠Z = 70°. The exterior angle of triangle is the sum of two opposite internal angles : ∠X + ∠Y = ∠XZT =70° + 40° = 110° |
|
| 26. |
A solid brass cube is melted and recast as a solid cone of height h and base radius r. If the height of the cube is h, find r in terms of h. A. r = h B. r = √\(\frac{3h}{π}\) C. r = πh D. r = h √\(\frac{3}{h}\) Detailed SolutionVolume of cube = H. H. H. = H\(^3\)the volume of a cone is, V=\(\frac{1}{3}\)πr\(^2\)h. Volume of cube = the volume of a cone H\(^3\) = \(\frac{1}{3}\)πr\(^2\)h. \(\frac{3{H}^3}{πh}\) = r\(^2\) √\(\frac{3{H}^3}{πh}\) = r h √\(\frac{3}{π}\) = r |
|
| 27. |
Which of the following is not an exterior angle of a regular polygon? A. 66° B. 72° C. 24° D. 15° |
A |
| 28. |
From a point T, a man moves 12km due west and then moves 12km due south to another point Q. Calculate the bearing of T from Q. A. 225° B. 315° C. 045° D. 135° |
C |
| 29. |
In the diagram O is the centre of the circle PQRS, ∠PQR = 72° and OR is parallel to PS. Find . A. 18° B. 108° C. 54° D. 36° Detailed SolutionAngle at the centre O is twice the angle at the circumference.i.e O = 72° * 2 = 144° ∠OPS = 180 - 144 = 36° |
|
| 30. |
A trapezium of parellel sides 10cm and 21cm and height 8cm is inscribed in a circle of radius 7cm. calculate the area of the region not covered by the trapezium.π =\(\frac{22}{7}\) A. 84cm\(^2\) B. 80cm\(^2\) C. 30cm\(^2\) D. 94cm\(^2\) Detailed SolutionArea of circle = πr\(^2\) = \(\frac{22}{7}\) * 7\(^2\) = 154cm\(^2\)Area o Trapezium = \(\frac{(a + b) h}{2}\) = \(\frac{(10 + 21)8}{2}\) = 124cm\(^2\) Area of the region not covered = Area of (Circle - Trapezium) = 154 - 124 = 30cm\(^2\) |
| 21. |
Find The quadratic Equation Whose Roots Are -2q And 5q. A. 3x\(^2\) + 3qx - 10q\(^2\) B. x\(^2\) + 3qx + 10q\(^2\) C. x\(^2\) - 3qx + 10q\(^2\) D. x\(^2\) - 3qx - 10q\(^2\) Detailed Solutionx\(^2\) - (sum of roots)x + (products of roots) = 0x\(^2\) - (-2q + 5q) + (-2q * 5q) = 0 x\(^2\) -(3q) + (-10q\(^2\)) = 0 x\(^2\) -3q - 10q\(^2\) = 0 |
|
| 22. |
If tanθ = \(frac{3}{4}\), 180° < θ < 270°, find the value of cosθ. A. \(\frac{4}{5}\) B. \(\frac{3}{5}\) C. -\(\frac{4}{5}\) D. -\(\frac{3}{5}\) Detailed Solutiontanθ = \(frac{3}{4}\) → tanθ = 0.75θ = tan\(^{-1}\)[0.75] → 36.8698° cosθ = cos[36.8698°] → 0.800 or \(frac{4}{5}\) However; in the third quadrant Cosine is negative i.e -\(frac{4}{5}\) |
|
| 23. |
If \(\frac{2}{x-3}\) - \(\frac{3}{x-2}\) = \(\frac{p}{(x-3)(x -2)}\), find p. A. 5 - x B. - (x + 5) C. 13 - x D. - (5x - 13) Detailed Solution\(\frac{2}{x-3}\) - \(\frac{3}{x-2}\)= \(\frac{2(x -2) -3(x - 3)}{(x-3)(x -2)}\) = \(\frac{2x -4 -3x + 9}{(x-3)(x -2)}\) = \(\frac{5 - x}{(x-3)(x -2)}\) |
|
| 24. |
The diagonal of a rhombus are 12cm and 5cm. calculate its perimeter A. 26cm B. 24cm C. 17cm D. 34cm Detailed SolutionPerimeter of a Rhombus = 4L or 2√( d1\(^2\) + d2\(^2\))P = 2√(12\(^2\) + 5\(^2\)) P = 2√(144 + 25) P = 2√169 = 2 * 13 P = 26cm |
|
| 25. |
A. 110° B. 130° C. 140° D. 180° Detailed SolutionIsosceles Triangle. Two equal sides. Two equal angles ;∠ X = ∠ Z ∠Y + ∠X + Z = 180 --> 40° + ∠X + ∠X = 180° ∠2X = 180° - 40° = 140° ∠X = 70°, i.e ∠X = 70 and ∠Z = 70°. The exterior angle of triangle is the sum of two opposite internal angles : ∠X + ∠Y = ∠XZT =70° + 40° = 110° |
| 26. |
A solid brass cube is melted and recast as a solid cone of height h and base radius r. If the height of the cube is h, find r in terms of h. A. r = h B. r = √\(\frac{3h}{π}\) C. r = πh D. r = h √\(\frac{3}{h}\) Detailed SolutionVolume of cube = H. H. H. = H\(^3\)the volume of a cone is, V=\(\frac{1}{3}\)πr\(^2\)h. Volume of cube = the volume of a cone H\(^3\) = \(\frac{1}{3}\)πr\(^2\)h. \(\frac{3{H}^3}{πh}\) = r\(^2\) √\(\frac{3{H}^3}{πh}\) = r h √\(\frac{3}{π}\) = r |
|
| 27. |
Which of the following is not an exterior angle of a regular polygon? A. 66° B. 72° C. 24° D. 15° |
A |
| 28. |
From a point T, a man moves 12km due west and then moves 12km due south to another point Q. Calculate the bearing of T from Q. A. 225° B. 315° C. 045° D. 135° |
C |
| 29. |
In the diagram O is the centre of the circle PQRS, ∠PQR = 72° and OR is parallel to PS. Find . A. 18° B. 108° C. 54° D. 36° Detailed SolutionAngle at the centre O is twice the angle at the circumference.i.e O = 72° * 2 = 144° ∠OPS = 180 - 144 = 36° |
|
| 30. |
A trapezium of parellel sides 10cm and 21cm and height 8cm is inscribed in a circle of radius 7cm. calculate the area of the region not covered by the trapezium.π =\(\frac{22}{7}\) A. 84cm\(^2\) B. 80cm\(^2\) C. 30cm\(^2\) D. 94cm\(^2\) Detailed SolutionArea of circle = πr\(^2\) = \(\frac{22}{7}\) * 7\(^2\) = 154cm\(^2\)Area o Trapezium = \(\frac{(a + b) h}{2}\) = \(\frac{(10 + 21)8}{2}\) = 124cm\(^2\) Area of the region not covered = Area of (Circle - Trapezium) = 154 - 124 = 30cm\(^2\) |