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
Correct Option: Answer is 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
Correct Option: Answer is 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°
Correct Option: Answer is D
\(\tan \theta = \frac{9}{20} = 0.45\)
\(\theta = \tan^{-1} (0.45) \)
= 24.23°
\(\therefore\) The bearing of Y from X = 180° – 24.23°
= 155.77°
= 156° (to the nearest degree)
- A. \(T = \frac{15R – Q}{P^3}\)
- B. \(T = R – \frac{15P^3}{Q}\)
- C. \(T = \frac{R – 15P^3}{Q}\)
- D. \(T = \frac{R + P^3}{15Q}\)
Correct Option: Answer is B
\(P = (\frac{Q(R – T)}{15})^{\frac{1}{3}}\)
\(P^3 = \frac{Q(R – T)}{15}\)
\(Q(R – T) = 15P^3\)
\(R – T = \frac{15P^3}{Q}\)
\(T = R – \frac{15P^3}{Q}\)
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°
Correct Option: Answer is D
< BAC = 130/2 (angle subtended at the centre)
< BAC = 65°
Also, x = 26° (theorem)
y = 65° – 26° = 39°
< AOC = 180° – (39° + 39°)
= 102°
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°
Correct Option: Answer is B
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°
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
Correct Option: Answer is C
Cost price of the car = N 1,250.00
Selling price = N 1,085.00
Loss = N (1250 – 1085)
= N 165.00
% loss = 165/1250×100
= 13.2% loss
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. \(\frac{2R^2}{3} \pi h\)
Correct Option: Answer is A
\(V = \frac{\pi h}{3} (R^2 + Rr + r^2)\)
\(V = \frac{\pi R^2 h}{3} + \frac{\pi Rr h}{3} + \frac{\pi r^2 h}{3}\)
\(\frac{\mathrm d V}{\mathrm d R} = \frac{2 \pi R h}{3} + \frac{\pi r h}{3}\)
= \(\frac{\pi}{3} (2R + r)\)
49. Express \((0.0439 \div 3.62)\) as a fraction.
- A.\(\frac{21}{100}\)
- B.\(\frac{21}{1000}\)
- C.\(\frac{12}{1000}\)
- D.\(\frac{12}{100}\)
Correct Option: Answer is C
(0.0439÷3.62)
= 0.01213
≊$\approxeq $ 0.012
= 121000
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
Correct Option: Answer is A
\(25^{1 – x} \times 5^{x + 2} \div (\frac{1}{125})^{x} = 625^{-1}\)
\((5^2)^{(1 – x)} \times 5^{(x + 2)} \div (5^{-3})^x = (5^4)^{-1}\)
\(5^{2 – 2x} \times 5^{x + 2} \div 5^{-3x} = 5^{-4}\)
\(5^{(2 – 2x) + (x + 2) – (-3x)} = 5^{-4}\)
Equating bases, we have
\(2 – 2x + x + 2 + 3x = -4\)
\(4 + 2x = -4 \implies 2x = -4 – 4\)
\(2x = -8\)
\(x = -4\)