21 - 30 of 47 Questions
# | Question | Ans |
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21. |
Solve the equation 7y\(^2\) = 3y A. y = 3 or 7 B. y = 0 or 7 C. y = 0 or 3/7 D. y = 0 or 9 E. y = 0 or 10 Detailed Solution7y\(^2\) = 3y7y\(^2\) - 3y = 0 y(7y - 3) = 0 y = 0 or y = \(\frac{3}{7}\) |
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22. |
Find the value of m which makes x\(^2\) + 8 + m a perfect square A. 2 B. 4 C. 8 D. 12 E. 16 Detailed Solutionx\(^2\) + 8 + mTo make it a perfect square, we add \((\frac{b}{2})^2\) = \((\frac{8}{2})^2\) = 16 |
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23. |
Solve the equation 2a\(^2\) - 3a - 27 = 0 A. 3/2, 9 B. -2/3, 9 C. 3, 9/2 D. -3, -9/2 E. -3, 9/2 Detailed Solution2a\(^2\) - 3a - 27 = 02a\(^2\) - 9a + 6a - 27 = 0 a(2a - 9) + 3(2a - 9) = 0 (a + 3)(2a - 9) = 0 a = -3 or a = \(\frac{9}{2}\) |
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24. |
A sector of a circle of radius 7cm has an area of 44cm2. Calculate the angle of the sector correct to the nearest degree [Take π = 22/7] A. 6o B. 26o C. 52o D. 103o E. 206o Detailed Solutionπr2 = 360o44cm2 = θ; ∴ θ = 44πr2 x 360o = 102.9o = 103o |
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25. |
Calculate the surface area of a sphere of radius 7cm [Take π = 22/7] A. 86cm2 B. 154cm2 C. 616cm2 D. 1434cm2 E. 4312cm2 Detailed SolutionSurface area of a sphere = \(4\pi r^2\)= \(4 \times \frac{22}{7} \times 7 \times 7\) = \(616 cm^2\) |
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26. |
If the radius of the parallel of latitude 30°N is equal to the radius of the parallel of latitude θ°S, what is the value of θ? A. 75o B. 60o C. 45o D. 30o E. 0o Detailed Solution\(\theta_{1} = \theta_{2} = 30°\) |
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27. |
A cylindrical container closed at both ends, has a radius of 7cm and height 5cm [Take π = 22/7] A. 35cm2 B. 154cm2 C. 220cm2 D. 528cm2 E. 770cm2 Detailed SolutionTSA of a cylinder = \(2\pi r^2 + 2\pi rh\)= \(2\pi r (r + h)\) = \(2 \times \frac{22}{7} \times 7 (7 + 5)\) = \(44 \times 12\) = \(528 cm^2\) |
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28. |
A cylindrical container closed at both ends, has a radius of 7cm and height 5cm [Take π = 22/7]. What is the volume of the container? A. 35cm3 B. 154cm3 C. 220cm3 D. 528cm3 E. 770cm3 Detailed SolutionVolume = \(\pi r^2 h\)= \(\frac{22}{7} \times 7^2 \times 5\) = \(770 cm^3\) |
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29. |
In the diagram above, |PQ| = |PR| = |RS| and ∠RPS = 32°. Find the value of ∠QPR A. 72o B. 64o C. 52o D. 32o E. 26o Detailed SolutionFrom the figure, < PSR = 32° (base angles of an isos. triangle)\(\therefore\) < PRS = 180° - (32° + 32°) = 116° (sum of angles in a triangle) < QRP = 180° - 116° = 64° (angle on a straight line) < PQR = 64° (base angles of an isos. triangle) < QPR = 180° - (64° + 64°) = 52° |
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30. |
In the diagram above, WXYZ is a rhombus and ∠WYX = 20°. What is the value of ∠XZY A. 20o B. 30o C. 45o D. 60o E. 70o Detailed SolutionDiagonals bisect at 90°; < YXZ = 90° - 20° = 70°But ZY = XY (sides of a rhombus) \(\therefore\) < XYZ = 70° (base angle of an isos. triangle) |
21. |
Solve the equation 7y\(^2\) = 3y A. y = 3 or 7 B. y = 0 or 7 C. y = 0 or 3/7 D. y = 0 or 9 E. y = 0 or 10 Detailed Solution7y\(^2\) = 3y7y\(^2\) - 3y = 0 y(7y - 3) = 0 y = 0 or y = \(\frac{3}{7}\) |
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22. |
Find the value of m which makes x\(^2\) + 8 + m a perfect square A. 2 B. 4 C. 8 D. 12 E. 16 Detailed Solutionx\(^2\) + 8 + mTo make it a perfect square, we add \((\frac{b}{2})^2\) = \((\frac{8}{2})^2\) = 16 |
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23. |
Solve the equation 2a\(^2\) - 3a - 27 = 0 A. 3/2, 9 B. -2/3, 9 C. 3, 9/2 D. -3, -9/2 E. -3, 9/2 Detailed Solution2a\(^2\) - 3a - 27 = 02a\(^2\) - 9a + 6a - 27 = 0 a(2a - 9) + 3(2a - 9) = 0 (a + 3)(2a - 9) = 0 a = -3 or a = \(\frac{9}{2}\) |
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24. |
A sector of a circle of radius 7cm has an area of 44cm2. Calculate the angle of the sector correct to the nearest degree [Take π = 22/7] A. 6o B. 26o C. 52o D. 103o E. 206o Detailed Solutionπr2 = 360o44cm2 = θ; ∴ θ = 44πr2 x 360o = 102.9o = 103o |
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25. |
Calculate the surface area of a sphere of radius 7cm [Take π = 22/7] A. 86cm2 B. 154cm2 C. 616cm2 D. 1434cm2 E. 4312cm2 Detailed SolutionSurface area of a sphere = \(4\pi r^2\)= \(4 \times \frac{22}{7} \times 7 \times 7\) = \(616 cm^2\) |
26. |
If the radius of the parallel of latitude 30°N is equal to the radius of the parallel of latitude θ°S, what is the value of θ? A. 75o B. 60o C. 45o D. 30o E. 0o Detailed Solution\(\theta_{1} = \theta_{2} = 30°\) |
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27. |
A cylindrical container closed at both ends, has a radius of 7cm and height 5cm [Take π = 22/7] A. 35cm2 B. 154cm2 C. 220cm2 D. 528cm2 E. 770cm2 Detailed SolutionTSA of a cylinder = \(2\pi r^2 + 2\pi rh\)= \(2\pi r (r + h)\) = \(2 \times \frac{22}{7} \times 7 (7 + 5)\) = \(44 \times 12\) = \(528 cm^2\) |
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28. |
A cylindrical container closed at both ends, has a radius of 7cm and height 5cm [Take π = 22/7]. What is the volume of the container? A. 35cm3 B. 154cm3 C. 220cm3 D. 528cm3 E. 770cm3 Detailed SolutionVolume = \(\pi r^2 h\)= \(\frac{22}{7} \times 7^2 \times 5\) = \(770 cm^3\) |
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29. |
In the diagram above, |PQ| = |PR| = |RS| and ∠RPS = 32°. Find the value of ∠QPR A. 72o B. 64o C. 52o D. 32o E. 26o Detailed SolutionFrom the figure, < PSR = 32° (base angles of an isos. triangle)\(\therefore\) < PRS = 180° - (32° + 32°) = 116° (sum of angles in a triangle) < QRP = 180° - 116° = 64° (angle on a straight line) < PQR = 64° (base angles of an isos. triangle) < QPR = 180° - (64° + 64°) = 52° |
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30. |
In the diagram above, WXYZ is a rhombus and ∠WYX = 20°. What is the value of ∠XZY A. 20o B. 30o C. 45o D. 60o E. 70o Detailed SolutionDiagonals bisect at 90°; < YXZ = 90° - 20° = 70°But ZY = XY (sides of a rhombus) \(\therefore\) < XYZ = 70° (base angle of an isos. triangle) |