Mathematics (Core), 2019, JAMB Exam, Exam, Paper 1 | Objectives

Paper Info

Year :

2019

Title :

Mathematics (Core)

Exam :

JAMB Exam

Paper 1 | Objectives

41 - 50 of 80 Questions

#	Question	Ans
41.	Simplify \(\frac{0.0839 \times 6.381}{5.44}\) to 2 significant figures. A. 0.2809 B. 2.51 C. 3.5 D. 0.098 Show Content Detailed Solution There is an explanation video available below.	D
42.	Find the value of x and y in the simultaneous equation: 3x + y = 21; xy = 30 A. x = 3 or 7, y = 12 or 8 B. x = 6 or 1, y = 11 or 5 C. x = 2 or 5, y = 15 or 6 D. x = 1 or 5, y = 10 or 7 Show Content Detailed Solution 3x + y = 21 ... (i); xy = 30 ... (ii) From (ii), \(y = \frac{30}{x}\). Putting the value of y in (i), we have 3x + \(\frac{30}{x}\) = 21 \(\implies\) 3x\(^2\) + 30 = 21x 3x\(^2\) - 21x + 30 = 0 3x\(^2\) - 15x - 6x + 30 = 0 3x(x - 5) - 6(x - 5) = 0 (3x - 6)(x - 5) = 0 3x - 6 = 0 \(\implies\) x = 2. x - 5 = 0 \(\implies\) x = 5. If x = 2, y = \(\frac{30}{2}\) = 15; If x = 5, y = \(\frac{30}{5}\) = 6. There is an explanation video available below.	C
43.	Points X and Y are 20km North and 9km East of point O, respectively. What is the bearing of Y from X? Correct to the nearest degree. A. 24° B. 56° C. 127° D. 156° Show Content Detailed Solution \(\tan \theta = \frac{9}{20} = 0.45\) \(\theta = \tan^{-1} (0.45) \) = 24.23° \(\therefore\) The bearing of Y from X = 180° - 24.23° = 155.77° = 156° (to the nearest degree) There is an explanation video available below.	D
44.	If \(P = (\frac{Q(R - T)}{15})^{\frac{1}{3}}\), make T the subject of the formula. A. \(T = \frac{15R - Q}{P^3}\) B. \(T = R - \frac{15P^3}{Q}\) C. \(T = \frac{R - 15P^3}{Q}\) D. \(T = \frac{R + P^3}{15Q}\) Show Content Detailed Solution \(P = (\frac{Q(R - T)}{15})^{\frac{1}{3}}\) \(P^3 = \frac{Q(R - T)}{15}\) \(Q(R - T) = 15P^3\) \(R - T = \frac{15P^3}{Q}\) \(T = R - \frac{15P^3}{Q}\) There is an explanation video available below.	B
45.	In the diagram above, O is the centre of the circle ABC, < ABO = 26° and < BOC = 130°. Calculate < AOC. A. 26° B. 13° C. 80° D. 102° Show Content Detailed Solution < BAC = \(\frac{130}{2}\) (angle subtended at the centre) < BAC = 65° Also, x = 26° (theorem) y = 65° - 26° = 39° < AOC = 180° - (39° + 39°) = 102° There is an explanation video available below.	D
46.	Each of the interior angles of a regular polygon is 140°. Calculate the sum of all the interior angles of the polygon. A. 1080° B. 1260° C. 2160° D. 1800° Show Content Detailed Solution Since each interior angle = 140°; Each exterior angle = 180° - 140° = 40° Number of sides of the polygon = \(\frac{360°}{40°}\) = 9 Sum of angles in the polygon = 140° x 9 = 1260° There is an explanation video available below.	B
47.	A man bought a car newly for ₦1,250,000. He had a crash with the car and later sold it at the rate of ₦1,085,000. What is the percentage gain or loss of the man? A. 43.7% loss B. 13.2% gain C. 13.2% loss D. 43.7% gain Show Content Detailed Solution Cost price of the car = N 1,250.00 Selling price = N 1,085.00 Loss = N (1250 - 1085) = N 165.00 % loss = \(\frac{165}{1250} \times 100%\) = 13.2% loss There is an explanation video available below.	C
48.	If the volume of a frustrum is given as \(V = \frac{\pi h}{3} (R^2 + Rr + r^2)\), find \(\frac{\mathrm d V}{\mathrm d R}\). A. \(\frac{\pi h}{3} (2R + r)\) B. \(2R + r + \frac{\pi h}{3}\) C. \(\frac{\pi h}{3} (2R^2 + r + 2r)\) D. \(\frac{2R^2}{3} \pi h\) Show Content Detailed Solution \(V = \frac{\pi h}{3} (R^2 + Rr + r^2)\) \(V = \frac{\pi R^2 h}{3} + \frac{\pi Rr h}{3} + \frac{\pi r^2 h}{3}\) \(\frac{\mathrm d V}{\mathrm d R} = \frac{2 \pi R h}{3} + \frac{\pi r h}{3}\) = \(\frac{\pi}{3} (2R + r)\) There is an explanation video available below.	A
49.	Express \((0.0439 \div 3.62)\) as a fraction. A. \(\frac{21}{100}\) B. \(\frac{21}{1000}\) C. \(\frac{12}{1000}\) D. \(\frac{12}{100}\) Show Content Detailed Solution \((0.0439 \div 3.62)\) = 0.01213 \(\approxeq\) 0.012 = \(\frac{12}{1000}\) There is an explanation video available below.	C
50.	If \(25^{1 - x} \times 5^{x + 2} \div (\frac{1}{125})^{x} = 625^{-1}\), find the value of x. A. x = -4 B. x = 2 C. x = -2 D. x = 4 Show Content Detailed Solution \(25^{1 - x} \times 5^{x + 2} \div (\frac{1}{125})^{x} = 625^{-1}\) \((5^2)^{(1 - x)} \times 5^{(x + 2)} \div (5^{-3})^x = (5^4)^{-1}\) \(5^{2 - 2x} \times 5^{x + 2} \div 5^{-3x} = 5^{-4}\) \(5^{(2 - 2x) + (x + 2) - (-3x)} = 5^{-4}\) Equating bases, we have \(2 - 2x + x + 2 + 3x = -4\) \(4 + 2x = -4 \implies 2x = -4 - 4\) \(2x = -8\) \(x = -4\) There is an explanation video available below.	A

41.	Simplify \(\frac{0.0839 \times 6.381}{5.44}\) to 2 significant figures. A. 0.2809 B. 2.51 C. 3.5 D. 0.098 Show Content Detailed Solution There is an explanation video available below.	D
42.	Find the value of x and y in the simultaneous equation: 3x + y = 21; xy = 30 A. x = 3 or 7, y = 12 or 8 B. x = 6 or 1, y = 11 or 5 C. x = 2 or 5, y = 15 or 6 D. x = 1 or 5, y = 10 or 7 Show Content Detailed Solution 3x + y = 21 ... (i); xy = 30 ... (ii) From (ii), \(y = \frac{30}{x}\). Putting the value of y in (i), we have 3x + \(\frac{30}{x}\) = 21 \(\implies\) 3x\(^2\) + 30 = 21x 3x\(^2\) - 21x + 30 = 0 3x\(^2\) - 15x - 6x + 30 = 0 3x(x - 5) - 6(x - 5) = 0 (3x - 6)(x - 5) = 0 3x - 6 = 0 \(\implies\) x = 2. x - 5 = 0 \(\implies\) x = 5. If x = 2, y = \(\frac{30}{2}\) = 15; If x = 5, y = \(\frac{30}{5}\) = 6. There is an explanation video available below.	C
43.	Points X and Y are 20km North and 9km East of point O, respectively. What is the bearing of Y from X? Correct to the nearest degree. A. 24° B. 56° C. 127° D. 156° Show Content Detailed Solution \(\tan \theta = \frac{9}{20} = 0.45\) \(\theta = \tan^{-1} (0.45) \) = 24.23° \(\therefore\) The bearing of Y from X = 180° - 24.23° = 155.77° = 156° (to the nearest degree) There is an explanation video available below.	D
44.	If \(P = (\frac{Q(R - T)}{15})^{\frac{1}{3}}\), make T the subject of the formula. A. \(T = \frac{15R - Q}{P^3}\) B. \(T = R - \frac{15P^3}{Q}\) C. \(T = \frac{R - 15P^3}{Q}\) D. \(T = \frac{R + P^3}{15Q}\) Show Content Detailed Solution \(P = (\frac{Q(R - T)}{15})^{\frac{1}{3}}\) \(P^3 = \frac{Q(R - T)}{15}\) \(Q(R - T) = 15P^3\) \(R - T = \frac{15P^3}{Q}\) \(T = R - \frac{15P^3}{Q}\) There is an explanation video available below.	B
45.	In the diagram above, O is the centre of the circle ABC, < ABO = 26° and < BOC = 130°. Calculate < AOC. A. 26° B. 13° C. 80° D. 102° Show Content Detailed Solution < BAC = \(\frac{130}{2}\) (angle subtended at the centre) < BAC = 65° Also, x = 26° (theorem) y = 65° - 26° = 39° < AOC = 180° - (39° + 39°) = 102° There is an explanation video available below.	D

46.	Each of the interior angles of a regular polygon is 140°. Calculate the sum of all the interior angles of the polygon. A. 1080° B. 1260° C. 2160° D. 1800° Show Content Detailed Solution Since each interior angle = 140°; Each exterior angle = 180° - 140° = 40° Number of sides of the polygon = \(\frac{360°}{40°}\) = 9 Sum of angles in the polygon = 140° x 9 = 1260° There is an explanation video available below.	B
47.	A man bought a car newly for ₦1,250,000. He had a crash with the car and later sold it at the rate of ₦1,085,000. What is the percentage gain or loss of the man? A. 43.7% loss B. 13.2% gain C. 13.2% loss D. 43.7% gain Show Content Detailed Solution Cost price of the car = N 1,250.00 Selling price = N 1,085.00 Loss = N (1250 - 1085) = N 165.00 % loss = \(\frac{165}{1250} \times 100%\) = 13.2% loss There is an explanation video available below.	C
48.	If the volume of a frustrum is given as \(V = \frac{\pi h}{3} (R^2 + Rr + r^2)\), find \(\frac{\mathrm d V}{\mathrm d R}\). A. \(\frac{\pi h}{3} (2R + r)\) B. \(2R + r + \frac{\pi h}{3}\) C. \(\frac{\pi h}{3} (2R^2 + r + 2r)\) D. \(\frac{2R^2}{3} \pi h\) Show Content Detailed Solution \(V = \frac{\pi h}{3} (R^2 + Rr + r^2)\) \(V = \frac{\pi R^2 h}{3} + \frac{\pi Rr h}{3} + \frac{\pi r^2 h}{3}\) \(\frac{\mathrm d V}{\mathrm d R} = \frac{2 \pi R h}{3} + \frac{\pi r h}{3}\) = \(\frac{\pi}{3} (2R + r)\) There is an explanation video available below.	A
49.	Express \((0.0439 \div 3.62)\) as a fraction. A. \(\frac{21}{100}\) B. \(\frac{21}{1000}\) C. \(\frac{12}{1000}\) D. \(\frac{12}{100}\) Show Content Detailed Solution \((0.0439 \div 3.62)\) = 0.01213 \(\approxeq\) 0.012 = \(\frac{12}{1000}\) There is an explanation video available below.	C
50.	If \(25^{1 - x} \times 5^{x + 2} \div (\frac{1}{125})^{x} = 625^{-1}\), find the value of x. A. x = -4 B. x = 2 C. x = -2 D. x = 4 Show Content Detailed Solution \(25^{1 - x} \times 5^{x + 2} \div (\frac{1}{125})^{x} = 625^{-1}\) \((5^2)^{(1 - x)} \times 5^{(x + 2)} \div (5^{-3})^x = (5^4)^{-1}\) \(5^{2 - 2x} \times 5^{x + 2} \div 5^{-3x} = 5^{-4}\) \(5^{(2 - 2x) + (x + 2) - (-3x)} = 5^{-4}\) Equating bases, we have \(2 - 2x + x + 2 + 3x = -4\) \(4 + 2x = -4 \implies 2x = -4 - 4\) \(2x = -8\) \(x = -4\) There is an explanation video available below.	A

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