Mathematics (Core), 2000, WASSCE/WAEC MAY/JUNE, Exam, Paper 1 | Objectives

Paper Info

Year :

2000

Title :

Mathematics (Core)

Exam :

WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

21 - 30 of 48 Questions

#	Question	Ans
21.	The probabilities that Kodjo and Adoga pass an examination are \(\frac{3}{4}\) and \(\frac{3}{5}\) respectively. Find the probability of both boys failing the examination A. \(\frac{1}{10}\) B. \(\frac{3}{10}\) C. \(\frac{9}{20}\) D. \(\frac{2}{3}\) Show Content Detailed Solution P(Kodjo passing) = \(\frac{3}{4}\); P(Adoga passing) = \(\frac{3}{5}\) P(Kodjo failing) = \(\frac{1}{4}\); P(Adoga failing) = \(\frac{2}{5}\) P(both fail) = \(\frac{1}{4} \times \frac{2}{5}\) = \(\frac{1}{10}\)	A
22.	Which of the following statement is not true about a rectangle? I.Each diagonal cuts the rectangle into two congruent triangles. II. A rectangle has four lines of symmetry III. The diagonals intersect at right angles A. I and II only B. III only C. II only D. II and III only	B
23.	In the diagram, PQRS is a circle center O. PQR is a diameter and ∠PRQ = 40°. Calculate ∠QSR. A. 30^o B. 40^o C. 45^o D. 50^o Show Content Detailed Solution < Q = < R (OQ = OR = radii) < QOR = 180° - 2(40°) = 100° < QSR = < RPQ = \(\frac{1}{2}\) < QOR = \(\frac{100}{2} = 50°\)	D
24.	Each side of a regular convex polygon subtends an angle of 30° at its center. Calculate each interior angle A. 75^o B. 150^o C. 160^o D. 68^o Show Content Detailed Solution Interior angle = 2(75°) = 150°	B
25.	If the interior angles of hexagon are 107°, 2x°, 150°, 95°, (2x-15)° and 123°, find x. A. \(57\frac{1}{2}^{\circ}\) B. \(65^{\circ}\) C. \(106^{\circ}\) D. \(120^{\circ}\) Show Content Detailed Solution Sum of interior angle in a hexagon = (6 - 2) x 180° = 720° \(\therefore\) 107° + 2x° + 150° + 95° + (2x - 15)° + 123° = 720° 460 + 4x = 720 \(\implies\) 4x = 720 - 460 4x = 260° \(\implies\) x = 65°	B
26.	In the diagram, POS and ROT are straight lines, OPQR is a parallelogram. \|OS\| = \|OT\| and ∠OST = 50°. Calculate ∠OPQ. A. 160^o B. 140^o C. 120^o D. 100^o Show Content Detailed Solution < T = < S = 50° (OS = OT) < SOT = 180° - 2(50°) = 80° < ROP = 80° (vertically opposite angle) \(\therefore\) < OPQ = 180° - 80° = 100° (adjacent angles)	D
27.	Given that \(x = -\frac{1}{2}and \hspace{1mm} y = 4 \hspace{1mm} evaluate \hspace{1mm} 3x^2y+xy^2\) A. -5 B. -1 C. 4 D. 11 Show Content Detailed Solution \(x = -\frac{1}{2}, y = 4\\ 3x^2y + xy^2\\ 3\left[-\frac{1}{2}\right]^2 \times 4 \times + \left(\frac{-1}{2}\right)(4)^2\\ 3\times \frac{1}{4} \times 4 -\frac{1}{2} \times 16\\ 3-8 = -5\)	A
28.	Given that \(27^{(1+x)}=9,)\ find x A. -3 B. \(\frac{-1}{3}\) C. \(\frac{5}{3}\) D. 2 Show Content Detailed Solution \(27^{(1+x)}=9\\ 3^{3(1+x)}=3^2\\ 3(1+x)=2\\ 3+3x = 2\\ 3x = -1 x = \frac{-1}{3}\)	B
29.	If x varies inversely as y and \(x = \frac{2}{3}\) when y = 9, find the value of y when \(x=\frac{3}{4}\) A. \(\frac{1}{18}\) B. \(\frac{8}{81}\) C. \(\frac{9}{2}\) D. 8 Show Content Detailed Solution \(x \propto \frac{1}{y}\) \(x = \frac{k}{y}\) \(\frac{2}{3} = \frac{k}{9}\) \(3k = 18 \implies k = 6\) \(x = \frac{6}{y}\) When y = \(\frac{3}{4}\), x = \(\frac{6}{\frac{3}{4}}\) = \(\frac{6 \times 4}{3}\) = 8	D
30.	Given that (2x + 7) is a factor of \(2x^2 + 3x - 14\), find the other factor A. x + 2 B. 2 - x C. x - 2 D. x + 1 Show Content Detailed Solution \(2x^2 + 3x - 14\) \(2x^2 + 7x - 4x - 14\) \(x(2x + 7) - 2(2x + 7)\) = \((x - 2)(2x + 7)\) The other factor = (x - 2).	C

21.	The probabilities that Kodjo and Adoga pass an examination are \(\frac{3}{4}\) and \(\frac{3}{5}\) respectively. Find the probability of both boys failing the examination A. \(\frac{1}{10}\) B. \(\frac{3}{10}\) C. \(\frac{9}{20}\) D. \(\frac{2}{3}\) Show Content Detailed Solution P(Kodjo passing) = \(\frac{3}{4}\); P(Adoga passing) = \(\frac{3}{5}\) P(Kodjo failing) = \(\frac{1}{4}\); P(Adoga failing) = \(\frac{2}{5}\) P(both fail) = \(\frac{1}{4} \times \frac{2}{5}\) = \(\frac{1}{10}\)	A
22.	Which of the following statement is not true about a rectangle? I.Each diagonal cuts the rectangle into two congruent triangles. II. A rectangle has four lines of symmetry III. The diagonals intersect at right angles A. I and II only B. III only C. II only D. II and III only	B
23.	In the diagram, PQRS is a circle center O. PQR is a diameter and ∠PRQ = 40°. Calculate ∠QSR. A. 30^o B. 40^o C. 45^o D. 50^o Show Content Detailed Solution < Q = < R (OQ = OR = radii) < QOR = 180° - 2(40°) = 100° < QSR = < RPQ = \(\frac{1}{2}\) < QOR = \(\frac{100}{2} = 50°\)	D
24.	Each side of a regular convex polygon subtends an angle of 30° at its center. Calculate each interior angle A. 75^o B. 150^o C. 160^o D. 68^o Show Content Detailed Solution Interior angle = 2(75°) = 150°	B
25.	If the interior angles of hexagon are 107°, 2x°, 150°, 95°, (2x-15)° and 123°, find x. A. \(57\frac{1}{2}^{\circ}\) B. \(65^{\circ}\) C. \(106^{\circ}\) D. \(120^{\circ}\) Show Content Detailed Solution Sum of interior angle in a hexagon = (6 - 2) x 180° = 720° \(\therefore\) 107° + 2x° + 150° + 95° + (2x - 15)° + 123° = 720° 460 + 4x = 720 \(\implies\) 4x = 720 - 460 4x = 260° \(\implies\) x = 65°	B

26.	In the diagram, POS and ROT are straight lines, OPQR is a parallelogram. \|OS\| = \|OT\| and ∠OST = 50°. Calculate ∠OPQ. A. 160^o B. 140^o C. 120^o D. 100^o Show Content Detailed Solution < T = < S = 50° (OS = OT) < SOT = 180° - 2(50°) = 80° < ROP = 80° (vertically opposite angle) \(\therefore\) < OPQ = 180° - 80° = 100° (adjacent angles)	D
27.	Given that \(x = -\frac{1}{2}and \hspace{1mm} y = 4 \hspace{1mm} evaluate \hspace{1mm} 3x^2y+xy^2\) A. -5 B. -1 C. 4 D. 11 Show Content Detailed Solution \(x = -\frac{1}{2}, y = 4\\ 3x^2y + xy^2\\ 3\left[-\frac{1}{2}\right]^2 \times 4 \times + \left(\frac{-1}{2}\right)(4)^2\\ 3\times \frac{1}{4} \times 4 -\frac{1}{2} \times 16\\ 3-8 = -5\)	A
28.	Given that \(27^{(1+x)}=9,)\ find x A. -3 B. \(\frac{-1}{3}\) C. \(\frac{5}{3}\) D. 2 Show Content Detailed Solution \(27^{(1+x)}=9\\ 3^{3(1+x)}=3^2\\ 3(1+x)=2\\ 3+3x = 2\\ 3x = -1 x = \frac{-1}{3}\)	B
29.	If x varies inversely as y and \(x = \frac{2}{3}\) when y = 9, find the value of y when \(x=\frac{3}{4}\) A. \(\frac{1}{18}\) B. \(\frac{8}{81}\) C. \(\frac{9}{2}\) D. 8 Show Content Detailed Solution \(x \propto \frac{1}{y}\) \(x = \frac{k}{y}\) \(\frac{2}{3} = \frac{k}{9}\) \(3k = 18 \implies k = 6\) \(x = \frac{6}{y}\) When y = \(\frac{3}{4}\), x = \(\frac{6}{\frac{3}{4}}\) = \(\frac{6 \times 4}{3}\) = 8	D
30.	Given that (2x + 7) is a factor of \(2x^2 + 3x - 14\), find the other factor A. x + 2 B. 2 - x C. x - 2 D. x + 1 Show Content Detailed Solution \(2x^2 + 3x - 14\) \(2x^2 + 7x - 4x - 14\) \(x(2x + 7) - 2(2x + 7)\) = \((x - 2)(2x + 7)\) The other factor = (x - 2).	C