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

31 - 40 of 49 Questions

#	Question	Ans
31.	Simplify \(\frac{2^{\frac{1}{2}}\times 8^{\frac{1}{2}}}{4}\) A. 1 B. 2 C. 4 D. 16 Show Content Detailed Solution \(\frac{2^{\frac{1}{2}}\times 8^{\frac{1}{2}}}{4}\\=\frac{2^{\frac{1}{2}}\times 2^{\frac{3}{2}}}{2^2}\\=2^{\frac{1}{2}}+2^{\frac{3}{2}}-2=2^0 = 1\)	A
32.	In the diagram, \|SR\| = \|RQ\| and ∠PRQ = 58^o ∠VQT = 19^o, PQT, SQV and PSR are straight lines. Find ∠QPS A. 42^o B. 39^o C. 38^o D. 30^o	A
33.	The angle of depression of a point Q from a vertical tower PR, 30m high, is 40°. If the foot P of the tower is on the same horizontal level as Q, find, correct to 2 decimal places, \|PQ\|. A. 35.75m B. 25.00m C. 22.98m D. 19.28 Show Content Detailed Solution \(\tan 40° = \frac{30}{\|PQ\|}\) \(\|PQ\| = \frac{30}{\tan 40°} = \frac{30}{0.839}\) = 35.75m	A
34.	The volume of a cylinder of radius 14cm is 210cm³. What is the curved surface area of the cylinder? A. 15cm² B. 30cm² C. 616cm² D. 1262cm² Show Content Detailed Solution \(V = \pi r^2 h\\ 210 = \frac{22}{7} \times 14^2 \times h \\ h = \frac{210}{22 \times 28}\) Curved surface area \(= 2r\pi h\\ = 2 \times \frac{22}{7} \times 14 \times \frac{210}{22 \times 26} = 30cm^2\)	B
35.	Simplify \(3\sqrt{12} + 10\sqrt{3} - \frac{6}{\sqrt{3}}\) A. \(7\sqrt{3}\) B. \(10\sqrt{3}\) C. \(14\sqrt{3}\) D. \(18\sqrt{3}\) Show Content Detailed Solution \(3\sqrt{12} + 10\sqrt{3} - \frac{6}{\sqrt{3}}\) = \(3(\sqrt{4 \times 3}) + 10\sqrt{3} - (\frac{6}{\sqrt{3}})(\frac{\sqrt{3}}{\sqrt{3}})\) = \(6\sqrt{3} + 10\sqrt{3} - 2\sqrt{3}\) = \(14\sqrt{3}\)	C
36.	Factorize m(2a-b)-2n(b-2a) A. (2a-b)(2n-m) B. (2a+b)(m-2n) C. (2a-b)(m+2n) D. (2a-b)(m-2n) Show Content Detailed Solution m(2a - b) - 2n(b - 2a) = m(2a - b) - (-2n)(2a - b) = m(2a - b) + 2n(2a - b) = (m + 2n)(2a - b)	C
37.	If q oranges are sold for t Naira, how many oranges can be bought for p naira? A. \(\frac{p}{2}t\) B. \(\frac{qt}{p}\) C. \(\frac{q}{pt}\) D. \(\frac{pq}{t}\) Show Content Detailed Solution q oranges = t naira 1 naira = \(\frac{q}{t}\) p naira = \(p(\frac{q}{t})\) = \(\frac{pq}{t}\) oranges	D
38.	In the diagram, QRT is a straight line. If angle PTR = 90°, angle PRT = 60°, angle PQR = 30° and \|PQ\| = \(6\sqrt{3}cm\), calculate \|RT\| A. 0.3cm B. \(\frac{\sqrt{3}}{2}cm\) C. 3cm D. \(3\sqrt{3}cm\) Show Content Detailed Solution In \(\Delta\) QPT, \(\frac{PT}{6\sqrt{3}} = \sin 30°\) PT = \(6\sqrt{3} \times \frac{1}{2} = 3\sqrt{3} cm\) In \(\Delta\) RPT, \(\frac{PT}{RT} = \tan 60°\) \(\frac{3\sqrt{3}}{RT} = \tan 60°\) \(RT = \frac{3\sqrt{3}}{\sqrt{3}} = 3 cm\)	C
39.	In the diagram, calculate the value of x A. 35^o B. 80^o C. 100^o D. 115^o Show Content Detailed Solution x - 35° = 65° (corresponding angles) x = 65° + 35° = 100°	C
40.	In the diagram, \(\bar{PS}\hspace{1mm} and \hspace{1mm}\bar{QT}\) are two altitudes of ∆PQR. Which of the following is equal to ∠RQT? A. ∠PQT B. ∠SRP C. ∠PQR D. ∠SPR	D

31.	Simplify \(\frac{2^{\frac{1}{2}}\times 8^{\frac{1}{2}}}{4}\) A. 1 B. 2 C. 4 D. 16 Show Content Detailed Solution \(\frac{2^{\frac{1}{2}}\times 8^{\frac{1}{2}}}{4}\\=\frac{2^{\frac{1}{2}}\times 2^{\frac{3}{2}}}{2^2}\\=2^{\frac{1}{2}}+2^{\frac{3}{2}}-2=2^0 = 1\)	A
32.	In the diagram, \|SR\| = \|RQ\| and ∠PRQ = 58^o ∠VQT = 19^o, PQT, SQV and PSR are straight lines. Find ∠QPS A. 42^o B. 39^o C. 38^o D. 30^o	A
33.	The angle of depression of a point Q from a vertical tower PR, 30m high, is 40°. If the foot P of the tower is on the same horizontal level as Q, find, correct to 2 decimal places, \|PQ\|. A. 35.75m B. 25.00m C. 22.98m D. 19.28 Show Content Detailed Solution \(\tan 40° = \frac{30}{\|PQ\|}\) \(\|PQ\| = \frac{30}{\tan 40°} = \frac{30}{0.839}\) = 35.75m	A
34.	The volume of a cylinder of radius 14cm is 210cm³. What is the curved surface area of the cylinder? A. 15cm² B. 30cm² C. 616cm² D. 1262cm² Show Content Detailed Solution \(V = \pi r^2 h\\ 210 = \frac{22}{7} \times 14^2 \times h \\ h = \frac{210}{22 \times 28}\) Curved surface area \(= 2r\pi h\\ = 2 \times \frac{22}{7} \times 14 \times \frac{210}{22 \times 26} = 30cm^2\)	B
35.	Simplify \(3\sqrt{12} + 10\sqrt{3} - \frac{6}{\sqrt{3}}\) A. \(7\sqrt{3}\) B. \(10\sqrt{3}\) C. \(14\sqrt{3}\) D. \(18\sqrt{3}\) Show Content Detailed Solution \(3\sqrt{12} + 10\sqrt{3} - \frac{6}{\sqrt{3}}\) = \(3(\sqrt{4 \times 3}) + 10\sqrt{3} - (\frac{6}{\sqrt{3}})(\frac{\sqrt{3}}{\sqrt{3}})\) = \(6\sqrt{3} + 10\sqrt{3} - 2\sqrt{3}\) = \(14\sqrt{3}\)	C

36.	Factorize m(2a-b)-2n(b-2a) A. (2a-b)(2n-m) B. (2a+b)(m-2n) C. (2a-b)(m+2n) D. (2a-b)(m-2n) Show Content Detailed Solution m(2a - b) - 2n(b - 2a) = m(2a - b) - (-2n)(2a - b) = m(2a - b) + 2n(2a - b) = (m + 2n)(2a - b)	C
37.	If q oranges are sold for t Naira, how many oranges can be bought for p naira? A. \(\frac{p}{2}t\) B. \(\frac{qt}{p}\) C. \(\frac{q}{pt}\) D. \(\frac{pq}{t}\) Show Content Detailed Solution q oranges = t naira 1 naira = \(\frac{q}{t}\) p naira = \(p(\frac{q}{t})\) = \(\frac{pq}{t}\) oranges	D
38.	In the diagram, QRT is a straight line. If angle PTR = 90°, angle PRT = 60°, angle PQR = 30° and \|PQ\| = \(6\sqrt{3}cm\), calculate \|RT\| A. 0.3cm B. \(\frac{\sqrt{3}}{2}cm\) C. 3cm D. \(3\sqrt{3}cm\) Show Content Detailed Solution In \(\Delta\) QPT, \(\frac{PT}{6\sqrt{3}} = \sin 30°\) PT = \(6\sqrt{3} \times \frac{1}{2} = 3\sqrt{3} cm\) In \(\Delta\) RPT, \(\frac{PT}{RT} = \tan 60°\) \(\frac{3\sqrt{3}}{RT} = \tan 60°\) \(RT = \frac{3\sqrt{3}}{\sqrt{3}} = 3 cm\)	C
39.	In the diagram, calculate the value of x A. 35^o B. 80^o C. 100^o D. 115^o Show Content Detailed Solution x - 35° = 65° (corresponding angles) x = 65° + 35° = 100°	C
40.	In the diagram, \(\bar{PS}\hspace{1mm} and \hspace{1mm}\bar{QT}\) are two altitudes of ∆PQR. Which of the following is equal to ∠RQT? A. ∠PQT B. ∠SRP C. ∠PQR D. ∠SPR	D