Mathematics (Core), 2002, WASSCE/WAEC MAY/JUNE, Exam, Paper 1 | Objectives

Paper Info

Year :

2002

Title :

Mathematics (Core)

Exam :

WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

41 - 49 of 49 Questions

#	Question	Ans
41.	In the diagram, \|QR\| = 10cm, PR⊥QS, angle PSR = 30° and angle PQR = 45°. Calculate in meters \|QS\| A. \(10(1+\sqrt{3})\) B. \(20\sqrt{3}\) C. \(10\sqrt{3}\) D. \((10+\sqrt{3})\) Show Content Detailed Solution In \(\Delta\) PQR, \(\tan 45 = \frac{PR}{10} \implies PR = 10 \tan 45\) = 10m In \(\Delta\) PRS, \(\tan 30 = \frac{10}{RS} \implies RS = \frac{10}{\tan 30}\) = \(\frac{10}{\frac{1}{\sqrt{3}}\) = \(10\sqrt{3}\) PS = \(10 + 10\sqrt{3}\) = \(10(1 + \sqrt{3}) cm\)	A
42.	Calculate and correct to two significant figures, the percentage error in approximating 0.375 to 0.4 A. 2.0 B. 2.5 C. 6.6 D. 6.7 Show Content Detailed Solution Measured value = 0.375 Approximation = 0.4 Error = 0.4 - 0.375 = 0.025 Error% = \(\frac{0.025}{0.375}\) x 100% = 6.67% = 6.7%	D
43.	To arrive on schedule, a train is to cover a distance of 60km at 72km/hr. If it starts 10 minutes late, at what speed must it move to arrive on schedule? A. 60km/hr B. 80km/hr C. 90km/hr D. 108km/hr Show Content Detailed Solution \(speed = \frac{distance}{time}\\ 72 = \frac{60}{time}\\ t = \frac{60}{72} = \frac{5}{6}hr\) time lost = 10mis \(= \frac{10}{60}hr = \frac{1}{6}\) Time required for the journey \(=\frac{5}{6}-\frac{1}{6} = \frac{2}{3}\\ speed \hspace{1mm}=60 \div \frac{2}{3} = 90km/hr\)	C
44.	In the diagram, \(QR\|\|TP and W\hat{P}T = 88^{\circ} \). Find the value of x A. 92^o B. 68^o C. 67^o D. 23^o Show Content Detailed Solution Sum of the angles in a triangle = 180° 3x - 180° + 92° + x = 180° 4x - 88° = 180° 4x = 268° x = 67°	C
45.	In the diagram O is the center of the circle, ∠SOR = 64° and ∠PSO = 36°. Calculate ∠PQR A. 100^o B. 96^o C. 94^o D. 86^o Show Content Detailed Solution < OSR = < ORS = \(\frac{180° - 64°}{2}\) = 58° < PSR = 36° + 58° = 94° < PSR + < PQR = 180° 94° + < PQR = 180° \(\implies\) < PQR = 180° - 94° = 86°	D
46.	Given that \(p = x-\frac{1}{x} and\hspace{1mm}q = x^2 + \frac{1}{x^2}\) express q in terms of p. A. (p² + 2) B. (p - 2) ² C. (p + 2) ² D. (p² - 2) Show Content Detailed Solution Given \(p = x - \frac{1}{x}\); \(q = x^2 + \frac{1}{x^2}\). \(p^2 = (x - \frac{1}{x})(x - \frac{1}{x})\) \(p^2 = x^2 + \frac{1}{x^2} - 2\) \(p^2 = q - 2 \implies q = p^2 + 2\)	A
47.	The number of goals scored by a school team in 10 netball matches are as follows: 3, 5, 7, 7, 8, 8, 8, 11, 11, 12. Find the probability that in a match, the school team will score at most 8 goals. A. \(\frac{7}{10}\) B. \(\frac{2}{5}\) C. \(\frac{3}{5}\) D. \(\frac{1}{5}\) Show Content Detailed Solution Number of at most 8 goals = 7 P(at most 8 goals) = \(\frac{7}{10}\)	A
48.	Which of the following is not a rational number? A. -5 B. \(\sqrt{4}\) C. \(3\frac{3}{4}\) D. \(\sqrt{90}\)	D
49.	In the diagram, LMT is a straight line. lf O is the centre of circle LMN, OMN = 20°, LTN = 32° and \|NM\| = \|MT\|, find LNM. A. 44^o B. 46^o C. 52^o D. 70^o Show Content Detailed Solution < MNT = < MTN = 32° < NMT = 180° - 2(32°) = 116° < OMN + < NMT + < LMO = 180° 20° + 116° + < LMO = 180° \(\implies\) < LMO = 180° - 136° = 44° < LMN = < LMO + < OMN = 44° + 20° = 64° < NOM = 180° - 2(20°) = 140° < NLM = \(\frac{1}{2} \times < NOM = 70°\) < LNM + < LMN + < NLM = 180° < LNM + 64° + 70° = 180° < LNM = 180° - 134° = 46&a	B

41.	In the diagram, \|QR\| = 10cm, PR⊥QS, angle PSR = 30° and angle PQR = 45°. Calculate in meters \|QS\| A. \(10(1+\sqrt{3})\) B. \(20\sqrt{3}\) C. \(10\sqrt{3}\) D. \((10+\sqrt{3})\) Show Content Detailed Solution In \(\Delta\) PQR, \(\tan 45 = \frac{PR}{10} \implies PR = 10 \tan 45\) = 10m In \(\Delta\) PRS, \(\tan 30 = \frac{10}{RS} \implies RS = \frac{10}{\tan 30}\) = \(\frac{10}{\frac{1}{\sqrt{3}}\) = \(10\sqrt{3}\) PS = \(10 + 10\sqrt{3}\) = \(10(1 + \sqrt{3}) cm\)	A
42.	Calculate and correct to two significant figures, the percentage error in approximating 0.375 to 0.4 A. 2.0 B. 2.5 C. 6.6 D. 6.7 Show Content Detailed Solution Measured value = 0.375 Approximation = 0.4 Error = 0.4 - 0.375 = 0.025 Error% = \(\frac{0.025}{0.375}\) x 100% = 6.67% = 6.7%	D
43.	To arrive on schedule, a train is to cover a distance of 60km at 72km/hr. If it starts 10 minutes late, at what speed must it move to arrive on schedule? A. 60km/hr B. 80km/hr C. 90km/hr D. 108km/hr Show Content Detailed Solution \(speed = \frac{distance}{time}\\ 72 = \frac{60}{time}\\ t = \frac{60}{72} = \frac{5}{6}hr\) time lost = 10mis \(= \frac{10}{60}hr = \frac{1}{6}\) Time required for the journey \(=\frac{5}{6}-\frac{1}{6} = \frac{2}{3}\\ speed \hspace{1mm}=60 \div \frac{2}{3} = 90km/hr\)	C
44.	In the diagram, \(QR\|\|TP and W\hat{P}T = 88^{\circ} \). Find the value of x A. 92^o B. 68^o C. 67^o D. 23^o Show Content Detailed Solution Sum of the angles in a triangle = 180° 3x - 180° + 92° + x = 180° 4x - 88° = 180° 4x = 268° x = 67°	C
45.	In the diagram O is the center of the circle, ∠SOR = 64° and ∠PSO = 36°. Calculate ∠PQR A. 100^o B. 96^o C. 94^o D. 86^o Show Content Detailed Solution < OSR = < ORS = \(\frac{180° - 64°}{2}\) = 58° < PSR = 36° + 58° = 94° < PSR + < PQR = 180° 94° + < PQR = 180° \(\implies\) < PQR = 180° - 94° = 86°	D

46.	Given that \(p = x-\frac{1}{x} and\hspace{1mm}q = x^2 + \frac{1}{x^2}\) express q in terms of p. A. (p² + 2) B. (p - 2) ² C. (p + 2) ² D. (p² - 2) Show Content Detailed Solution Given \(p = x - \frac{1}{x}\); \(q = x^2 + \frac{1}{x^2}\). \(p^2 = (x - \frac{1}{x})(x - \frac{1}{x})\) \(p^2 = x^2 + \frac{1}{x^2} - 2\) \(p^2 = q - 2 \implies q = p^2 + 2\)	A
47.	The number of goals scored by a school team in 10 netball matches are as follows: 3, 5, 7, 7, 8, 8, 8, 11, 11, 12. Find the probability that in a match, the school team will score at most 8 goals. A. \(\frac{7}{10}\) B. \(\frac{2}{5}\) C. \(\frac{3}{5}\) D. \(\frac{1}{5}\) Show Content Detailed Solution Number of at most 8 goals = 7 P(at most 8 goals) = \(\frac{7}{10}\)	A
48.	Which of the following is not a rational number? A. -5 B. \(\sqrt{4}\) C. \(3\frac{3}{4}\) D. \(\sqrt{90}\)	D
49.	In the diagram, LMT is a straight line. lf O is the centre of circle LMN, OMN = 20°, LTN = 32° and \|NM\| = \|MT\|, find LNM. A. 44^o B. 46^o C. 52^o D. 70^o Show Content Detailed Solution < MNT = < MTN = 32° < NMT = 180° - 2(32°) = 116° < OMN + < NMT + < LMO = 180° 20° + 116° + < LMO = 180° \(\implies\) < LMO = 180° - 136° = 44° < LMN = < LMO + < OMN = 44° + 20° = 64° < NOM = 180° - 2(20°) = 140° < NLM = \(\frac{1}{2} \times < NOM = 70°\) < LNM + < LMN + < NLM = 180° < LNM + 64° + 70° = 180° < LNM = 180° - 134° = 46&a	B