Mathematics (Core), 2009, WASSCE/WAEC MAY/JUNE, Exam, Paper 1 | Objectives

Paper Info

Year :

2009

Title :

Mathematics (Core)

Exam :

WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

31 - 40 of 46 Questions

#	Question	Ans
31.	Find the quadratic equation whose roots are -\(\frac{1}{2}\) and 3 A. 2x² - 2x + 3 = 0 B. 2x² - 2x - 3 = 0 C. 2x² - 5x - 3 = 0 D. 3x² - 5x - 3 = 0 Show Content Detailed Solution x = \(\frac{-1}{2}\) x = 3 2x = -1, x = 3; 2x + 1 = 0 x - 3 = 0 (2x + 1)(x - 3) = 0 2x2 - 6x + x - 3 = 0 2x2 - 5x - 3 = 0	C
32.	The following is the graph of a quadratic friction, find the co-ordinates of point P A. (0, 4) B. (4, 0) C. (0, -4) D. (-4, 0)	A
33.	The following is the graph of a quadratic friction, find the value of x when y = 0 A. 1, 2 B. 1, 4 C. 2, 3 D. 1, 6 Show Content Detailed Solution when y = 0, value of x = roots of the equation x² - 5x + 4 = 0 x = 1, 4 (see graph)	B
34.	PQRS is a trapezuim. QR//PS, /PQ/ = 5cm, /OR/ = 6cm, /PS/ = 10cm and angle QPS = 42^o. Calculate the perpendicular distance between the parallel sides A. 3.35cm B. 3.73cm C. 4.50cm D. 4.62cm Show Content Detailed Solution h = ? \(\frac{h}{5cm}\) = sin 45^o h = 5cm x 0.6691 = 3.35cm	A
35.	PQRS is a trapezium. QR//PS, /PQ/ = 5cm, /OR/ = 6cm, /PS/ = 10cm and angle QPS = 42^o. Calculate, correct to the nearest cm², the area of the trapezium (h = 3.35cm² ) A. 27cm² B. 30cm² C. 36cm² D. 37cm² Show Content Detailed Solution Area = \(\frac{1}{2}(a + b)\)h where a = QR = 6cm b = PA = 10cm Area = \(\frac{1}{2}(6 + 10)\) x 3.35cm² = \(\frac{1}{2}(16)\) x 3.35cm² = 26.8cm² = 27cm² (approx.)	A
36.	In the diagram, /TP/ = 12cm and it is 6cm from O, the centre of the circle, Calculate < TOP A. 120^o B. 90^o C. 60^o D. 45^o Show Content Detailed Solution Tan \(\theta = \frac{6}{6} = 1\) \(\theta\) = tab - 1(1) = 45^o < top = 20 = 2 x 45^o = 90^o	B
37.	In the diagram, IG is parallel to JE, JEF = 120^o and FHG = 130^o, fins the angle marked t A. 40^o B. 70^o C. 80^o D. 100^o Show Content Detailed Solution Now, t + 60^o + 50^o = 180^o t = 180^o - 110^o t = 70^o	B
38.	Find the value of x in the diagram A. 50^o B. 30^o C. 22^o D. 17^o Show Content Detailed Solution Sum of exterior angles of a polygon = 360^o (x + 64) + 2x + (3x - 54) + 5x + 4x + 3x + (2x + 10) = 360^o 20x + 20 = 360^o 20x + 360 - 20 = 340^o x = \(\frac{340}{20}\) x = 17^o	D
39.	In the diagram, \SQ\ = 4cm, \PT\ = 7cm. /TR/ = 5cm and ST//OR. If /SP/ = xcm, find the value of x A. 5.6 B. 6.5 C. 6.6 D. 6.8 Show Content Detailed Solution \(\frac{/PS/}{/PQ/} = \frac{/PT/}{/PR/}\) \(\frac{x}{4 + x} = \frac{7}{7 + 5}\) \(\frac{x}{4 + x} = \frac{7}{12}\) (2x = 7)(4 + x); 12x = 28 + 7x 12x - 7x = 28 5x = 28 x = \(\frac{28}{5}\) x = 5.6	A
40.	Using the diagram, find the bearing of X from Y A. 300^o B. 240^o C. 120^o D. 60^o Show Content Detailed Solution The bearing of X from Y = 360 - 120 = 240	B

31.	Find the quadratic equation whose roots are -\(\frac{1}{2}\) and 3 A. 2x² - 2x + 3 = 0 B. 2x² - 2x - 3 = 0 C. 2x² - 5x - 3 = 0 D. 3x² - 5x - 3 = 0 Show Content Detailed Solution x = \(\frac{-1}{2}\) x = 3 2x = -1, x = 3; 2x + 1 = 0 x - 3 = 0 (2x + 1)(x - 3) = 0 2x2 - 6x + x - 3 = 0 2x2 - 5x - 3 = 0	C
32.	The following is the graph of a quadratic friction, find the co-ordinates of point P A. (0, 4) B. (4, 0) C. (0, -4) D. (-4, 0)	A
33.	The following is the graph of a quadratic friction, find the value of x when y = 0 A. 1, 2 B. 1, 4 C. 2, 3 D. 1, 6 Show Content Detailed Solution when y = 0, value of x = roots of the equation x² - 5x + 4 = 0 x = 1, 4 (see graph)	B
34.	PQRS is a trapezuim. QR//PS, /PQ/ = 5cm, /OR/ = 6cm, /PS/ = 10cm and angle QPS = 42^o. Calculate the perpendicular distance between the parallel sides A. 3.35cm B. 3.73cm C. 4.50cm D. 4.62cm Show Content Detailed Solution h = ? \(\frac{h}{5cm}\) = sin 45^o h = 5cm x 0.6691 = 3.35cm	A
35.	PQRS is a trapezium. QR//PS, /PQ/ = 5cm, /OR/ = 6cm, /PS/ = 10cm and angle QPS = 42^o. Calculate, correct to the nearest cm², the area of the trapezium (h = 3.35cm² ) A. 27cm² B. 30cm² C. 36cm² D. 37cm² Show Content Detailed Solution Area = \(\frac{1}{2}(a + b)\)h where a = QR = 6cm b = PA = 10cm Area = \(\frac{1}{2}(6 + 10)\) x 3.35cm² = \(\frac{1}{2}(16)\) x 3.35cm² = 26.8cm² = 27cm² (approx.)	A

36.	In the diagram, /TP/ = 12cm and it is 6cm from O, the centre of the circle, Calculate < TOP A. 120^o B. 90^o C. 60^o D. 45^o Show Content Detailed Solution Tan \(\theta = \frac{6}{6} = 1\) \(\theta\) = tab - 1(1) = 45^o < top = 20 = 2 x 45^o = 90^o	B
37.	In the diagram, IG is parallel to JE, JEF = 120^o and FHG = 130^o, fins the angle marked t A. 40^o B. 70^o C. 80^o D. 100^o Show Content Detailed Solution Now, t + 60^o + 50^o = 180^o t = 180^o - 110^o t = 70^o	B
38.	Find the value of x in the diagram A. 50^o B. 30^o C. 22^o D. 17^o Show Content Detailed Solution Sum of exterior angles of a polygon = 360^o (x + 64) + 2x + (3x - 54) + 5x + 4x + 3x + (2x + 10) = 360^o 20x + 20 = 360^o 20x + 360 - 20 = 340^o x = \(\frac{340}{20}\) x = 17^o	D
39.	In the diagram, \SQ\ = 4cm, \PT\ = 7cm. /TR/ = 5cm and ST//OR. If /SP/ = xcm, find the value of x A. 5.6 B. 6.5 C. 6.6 D. 6.8 Show Content Detailed Solution \(\frac{/PS/}{/PQ/} = \frac{/PT/}{/PR/}\) \(\frac{x}{4 + x} = \frac{7}{7 + 5}\) \(\frac{x}{4 + x} = \frac{7}{12}\) (2x = 7)(4 + x); 12x = 28 + 7x 12x - 7x = 28 5x = 28 x = \(\frac{28}{5}\) x = 5.6	A
40.	Using the diagram, find the bearing of X from Y A. 300^o B. 240^o C. 120^o D. 60^o Show Content Detailed Solution The bearing of X from Y = 360 - 120 = 240	B