Mathematics (Core), 1999, WASSCE/WAEC MAY/JUNE, Exam, Paper 1 | Objectives

Paper Info

Year :

1999

Title :

Mathematics (Core)

Exam :

WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

21 - 30 of 43 Questions

#	Question	Ans
21.	From the diagram above. ABC is a triangle inscribed in a circle center O. ∠ACB = 40^o and \|AB\| = x cm. calculate the radius of the circle. A. \(\frac{x}{sin 40^o}\) B. \(\frac{x}{cos 40^o}\) C. \(\frac{x}{2 sin 40^o}\) D. \(\frac{x}{2 cos 40^o}\)	C
22.	The diagram shows the position of three ships A, B and C at sea. B is due north of C such that \|AB\| = \|BC\| and the bearing of B from A = 040°. What is the bearing of A from C A. 040^o B. 070^o C. 110^o D. 290^o Show Content Detailed Solution < ABC = 40° (alternate angles) \(\therefore\) < ACB = \(\frac{180° - 40°}{2}\) = 70° \(\therefore\) Bearing of A from C = 360° - 70° = 290°	D
23.	A pole of length L leans against a vertical wall so that it makes an angle of 60^o with the horizontal ground. If the top of the pole is 8m above the ground, calculate L. A. \(16\sqrt{3}m\) B. \(4\sqrt{3}m\) C. \(\frac{\sqrt{3}}{16}\) D. \(\frac{16\sqrt{3}}{3}\) Show Content Detailed Solution \(sin 60 = \frac{8}{L} = 8 \div \frac{\sqrt{3}}{2}; L = \frac{8}{sin 60}=\frac{16}{\sqrt{3}}\) When we rationalize gives \(\frac{16\sqrt{3}}{3}\)	D
24.	In the diagram, PQST and QRST are parallelograms. Calculate the area of the trapezium PRST. A. 60cm² B. 50cm² C. 40cm² D. 30cm² Show Content Detailed Solution PR = 4 + 4 = 8 cm Area of trapezium = \(\frac{1}{2} (a + b) h\) = \(\frac{1}{2} (4 + 8) \times 5\) = 30 cm\(^2\)	D
25.	The height of a pyramid on square base is 15cm. If the volume is 80cm\(^3\), find the length of the side of the base A. 3.3cm B. 5.3cm C. 4.0cm D. 8.0cm Show Content Detailed Solution Volume = \(\frac{1}{3} b^2 h\) \(\therefore b = \sqrt{\frac{3v}{h}}\) \(b = \sqrt{\frac{3(80)}{15}} \) \(b = \sqrt{16} = 4.0 cm\)	C
26.	Simplify 3.72 x 0.025 and express your answer in the standard form A. \(9.3\times 10^3\) B. \(9.3\times 10^2\) C. \(9.3\times 10^{-2}\) D. \(9.3\times 10^{-3}\) Show Content Detailed Solution \(3.72 \times 0.025 = 0.093 = 9.3\times 10^{-2}\)	C
27.	Two numbers 24\(_{x}\) and 31\(_y\) are equal in value when converted to base ten. Find the equation connecting x and y A. 2x = 3(y - 1) B. 4x - y = 1 C. 3y + 2x = 3 D. 3y = 2 (x + 3) Show Content Detailed Solution 24\(_x\) = 31\(_y\) \(2 \times x^1 + 4 \times x^0 = 3 \times y^1 + 1 \times y^0\) \(2x + 4 = 3y + 1 \implies 2x = 3y + 1 - 4\) \(2x = 3y - 3 = 3(y - 1)\)	A
28.	A car travel at x km per hour for 1 hour and at y km per hour for 2 hours. Find its average speed A. \(\frac{2x + 2y}{3}kmh^{-1}\) B. \(\frac{x + y}{3}kmh^{-1}\) C. \(\frac{x + 2y}{3}kmh^{-1}\) D. \(\frac{2x + y}{3}kmh^{-1}\) Show Content Detailed Solution Travelled x km/h for 1 hour \(\therefore\) traveled x km in the first hour. Traveled y km/h for 2 hours \(\therefore\) traveled 2y km in the next 2 hours. Average speed = \(\frac{x + 2y}{1 + 2}\) = \(\frac{x + 2y}{3} kmh^{-1}\)	C
29.	The height and base of a triangle are in ratio 1:3 respectively. If the area of the triangle is 216 cm\(^2\), find the length of the base. A. 24cm B. 36cm C. 72cm D. 144cm Show Content Detailed Solution Area = \(\frac{1}{2} \times base \times height\) \(height : base = 1 : 3\) \(\implies base = 3 \times height\) Let height = h; Area = \(\frac{1}{2} \times 3h \times h = 216\) \(3h^2 = 216 \times 2 = 432\) \(h^2 = \frac{432}{3} = 144\) \(h = \sqrt{144} = 12.0 cm\) \(\therefore base = 3 \times 12 = 36 cm\)	B
30.	Water flows from a tap into cylindrical container at the rate 5πcm\(^3\) per second. If the radius of the container is 3cm, calculate the level of water in the container at the end of 9 seconds. A. 2cm B. 5cm C. 8cm D. 15cm Show Content Detailed Solution Volume of water after 9 seconds = \(5\pi \times 9 = 45\pi cm^3\) Volume of cylinder = \(\pi r^2 h\) \(\therefore \pi r^2 h = 45\pi\) \(\pi \times 3^2 \times h = 45\pi\) \(\implies 9h = 45 \) \(h = 5 cm\) (where h = height of the water after 9 secs)	B

21.	From the diagram above. ABC is a triangle inscribed in a circle center O. ∠ACB = 40^o and \|AB\| = x cm. calculate the radius of the circle. A. \(\frac{x}{sin 40^o}\) B. \(\frac{x}{cos 40^o}\) C. \(\frac{x}{2 sin 40^o}\) D. \(\frac{x}{2 cos 40^o}\)	C
22.	The diagram shows the position of three ships A, B and C at sea. B is due north of C such that \|AB\| = \|BC\| and the bearing of B from A = 040°. What is the bearing of A from C A. 040^o B. 070^o C. 110^o D. 290^o Show Content Detailed Solution < ABC = 40° (alternate angles) \(\therefore\) < ACB = \(\frac{180° - 40°}{2}\) = 70° \(\therefore\) Bearing of A from C = 360° - 70° = 290°	D
23.	A pole of length L leans against a vertical wall so that it makes an angle of 60^o with the horizontal ground. If the top of the pole is 8m above the ground, calculate L. A. \(16\sqrt{3}m\) B. \(4\sqrt{3}m\) C. \(\frac{\sqrt{3}}{16}\) D. \(\frac{16\sqrt{3}}{3}\) Show Content Detailed Solution \(sin 60 = \frac{8}{L} = 8 \div \frac{\sqrt{3}}{2}; L = \frac{8}{sin 60}=\frac{16}{\sqrt{3}}\) When we rationalize gives \(\frac{16\sqrt{3}}{3}\)	D
24.	In the diagram, PQST and QRST are parallelograms. Calculate the area of the trapezium PRST. A. 60cm² B. 50cm² C. 40cm² D. 30cm² Show Content Detailed Solution PR = 4 + 4 = 8 cm Area of trapezium = \(\frac{1}{2} (a + b) h\) = \(\frac{1}{2} (4 + 8) \times 5\) = 30 cm\(^2\)	D
25.	The height of a pyramid on square base is 15cm. If the volume is 80cm\(^3\), find the length of the side of the base A. 3.3cm B. 5.3cm C. 4.0cm D. 8.0cm Show Content Detailed Solution Volume = \(\frac{1}{3} b^2 h\) \(\therefore b = \sqrt{\frac{3v}{h}}\) \(b = \sqrt{\frac{3(80)}{15}} \) \(b = \sqrt{16} = 4.0 cm\)	C

26.	Simplify 3.72 x 0.025 and express your answer in the standard form A. \(9.3\times 10^3\) B. \(9.3\times 10^2\) C. \(9.3\times 10^{-2}\) D. \(9.3\times 10^{-3}\) Show Content Detailed Solution \(3.72 \times 0.025 = 0.093 = 9.3\times 10^{-2}\)	C
27.	Two numbers 24\(_{x}\) and 31\(_y\) are equal in value when converted to base ten. Find the equation connecting x and y A. 2x = 3(y - 1) B. 4x - y = 1 C. 3y + 2x = 3 D. 3y = 2 (x + 3) Show Content Detailed Solution 24\(_x\) = 31\(_y\) \(2 \times x^1 + 4 \times x^0 = 3 \times y^1 + 1 \times y^0\) \(2x + 4 = 3y + 1 \implies 2x = 3y + 1 - 4\) \(2x = 3y - 3 = 3(y - 1)\)	A
28.	A car travel at x km per hour for 1 hour and at y km per hour for 2 hours. Find its average speed A. \(\frac{2x + 2y}{3}kmh^{-1}\) B. \(\frac{x + y}{3}kmh^{-1}\) C. \(\frac{x + 2y}{3}kmh^{-1}\) D. \(\frac{2x + y}{3}kmh^{-1}\) Show Content Detailed Solution Travelled x km/h for 1 hour \(\therefore\) traveled x km in the first hour. Traveled y km/h for 2 hours \(\therefore\) traveled 2y km in the next 2 hours. Average speed = \(\frac{x + 2y}{1 + 2}\) = \(\frac{x + 2y}{3} kmh^{-1}\)	C
29.	The height and base of a triangle are in ratio 1:3 respectively. If the area of the triangle is 216 cm\(^2\), find the length of the base. A. 24cm B. 36cm C. 72cm D. 144cm Show Content Detailed Solution Area = \(\frac{1}{2} \times base \times height\) \(height : base = 1 : 3\) \(\implies base = 3 \times height\) Let height = h; Area = \(\frac{1}{2} \times 3h \times h = 216\) \(3h^2 = 216 \times 2 = 432\) \(h^2 = \frac{432}{3} = 144\) \(h = \sqrt{144} = 12.0 cm\) \(\therefore base = 3 \times 12 = 36 cm\)	B
30.	Water flows from a tap into cylindrical container at the rate 5πcm\(^3\) per second. If the radius of the container is 3cm, calculate the level of water in the container at the end of 9 seconds. A. 2cm B. 5cm C. 8cm D. 15cm Show Content Detailed Solution Volume of water after 9 seconds = \(5\pi \times 9 = 45\pi cm^3\) Volume of cylinder = \(\pi r^2 h\) \(\therefore \pi r^2 h = 45\pi\) \(\pi \times 3^2 \times h = 45\pi\) \(\implies 9h = 45 \) \(h = 5 cm\) (where h = height of the water after 9 secs)	B