Year :
2021
Title :
Mathematics (Core)
Exam :
WASSCE/WAEC MAY/JUNE
Paper 1 | Objectives
11 - 20 of 49 Questions
| # | Question | Ans |
|---|---|---|
| 11. |
The circumference of a circular track is 9km. A cyclist rides round it a number of times and stops after covering a distance of 302km. How far is the cyclist from the starting point? A. 5km B. 6km C. 7km D. 3km Detailed SolutionCircumference of the circular track = 9kmDistance covered = 302km Number of complete circles or revolutions from the starting point = 302/9 =33 circles and additional 5km. So, the distance of the cyclist from the starting point would be 5km |
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| 12. |
Simplify 2√7- 14/√7+7/21 A. \(\frac{√21}{21}\) B. 7\(\frac{√21}{21}\) C. \(\frac{√21}{7}\) D. 3√21 Detailed Solution2√7- 14/√7 +7/√21L.C.M is √21 \(\frac{√21 * 2√7 - √3 *14 + 7}{√21}\) =\(\frac{2 * 7√3 - 14√3 + 7}{√21}\) = \(\frac{14√3 - 14√3 + 7}{√21}\) =\(\frac{7}{√21}\) or \(\frac{7 \times √21}{√21 \times √21}\) = \(\frac{7 \times √21}{21}\) or \(\frac{√21}{3}\) |
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| 13. |
If 4x+2y=16 and 6x-2y=4 , find the value of (y-x). A. 8 B. 2 C. 4 D. 6 Detailed SolutionUsing elimination method:4x+2y=16 * 6 --> 24x +12y=96 ... eqn iii 6x-2y=4 * 4 --> 24x - 8y = 16 ... eqn iv Subtract eqn iv from iii 20y = 80 y = 4 Subst. y for 4 in 4x + 2y = 16 --> 4x + 2(4) = 16 --> 4x = 16 - 8 --> 4x = 8 --> x = 2 : The value of (y-x) is 4 - 2 = 2 |
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| 14. |
 In the diagram, ∠ABC and ∠BCD are right angles, ∠BAD = t and ∠EDF = 70°. Find the value of t. A. 70° B. 165° C. 140° D. 110° Detailed SolutionVertically opp angles are equal ∠CDE = 70°Sum of interior angles in a quadilateral = 360° (90 + 90 + 70 + t)° = 360° t° = 360° - 250° t° = 110° |
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| 15. |
The sum of the interior angles of a regular polygon with k sides is (3k-10) right angles. Find the size of the exterior angle? A. 60° B. 40° C. 90° D. 120° Detailed SolutionIn a polygon with n sides, the sum of the angles =(n - 2)180, where n = the number of sides. In our problem, n = k. So, we have: (3k - 10)90 = (k - 2)180 270k - 900 = 180k - 360 Simplifying: 90k = 540 k = 6. So, we have a regular hexagon. Now, each exterior angle = 360/n We have: 360/6 = 60 Conclusion: Each exterior angle is 60 degrees. |
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| 16. |
Make u the subject in x =\(\frac{2u-3}{3u + 2}\) A. u = \(\frac{2x + 3}{3x - 2}\) B. u = \(\frac{2x - 3}{3x - 2}\) C. u = \(\frac{2x + 3}{2 - 3x}\) D. u = \(\frac{2x + 3}{3x + 2}\) Detailed Solutionx =\(\frac{2u-3}{3u + 2}\)cross multiply x(3u + 2) = 2u - 3 3ux + 2x = 2u - 3 collect like terms of u 2x + 3 = 2u - 3ux \(\frac{2x + 3}{2 - 3x}\) = u |
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| 17. |
A trader paid import duty of 38 kobo in the naira on the cost of an engine. If a total of #22,800.00 was paid as import duty, calculate the cost of the engine. A. #60,000.00 B. #120,000.00 C. #24,000.00 D. #18,000.00 |
A |
| 18. |
The height of an equilateral triangle of side is 10 3√ cm. calculate its perimeter. A. 20cm B. 60cm C. 40cm D. 30cm Detailed SolutionHeight of an equilateral triangle, h = a\(\frac{√3}{2}\), where a is the side of the equilateral triangle.10√3 = a\(\frac{√3}{2}\) cross multiply--> 2 * 10√3 = a√3 √3 strikes √3 on both sides 20 = a The perimeter of an equilateral triangle is: P = 3a P = 3 * 20 = 60cm P = 30√3 |
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| 19. |
In △LMN, |LM| = 6cm, ∠LNM = x and sin x = sin x = \(\frac{3}{5}\).Find the area of △LMN A. 60cm\(^2\) B. 48cm\(^2\) C. 24cm\(^2\) D. 30cm\(^2\) |
C |
| 20. |
 Consider the statements:P = All students offering Literature(L) also offer History(H);Q = Students offering History(H) do not offer Geography(G).Which of the Venn diagram correctly illustrate the two statements? A. A B. B C. C D. D |
C |
| 11. |
The circumference of a circular track is 9km. A cyclist rides round it a number of times and stops after covering a distance of 302km. How far is the cyclist from the starting point? A. 5km B. 6km C. 7km D. 3km Detailed SolutionCircumference of the circular track = 9kmDistance covered = 302km Number of complete circles or revolutions from the starting point = 302/9 =33 circles and additional 5km. So, the distance of the cyclist from the starting point would be 5km |
|
| 12. |
Simplify 2√7- 14/√7+7/21 A. \(\frac{√21}{21}\) B. 7\(\frac{√21}{21}\) C. \(\frac{√21}{7}\) D. 3√21 Detailed Solution2√7- 14/√7 +7/√21L.C.M is √21 \(\frac{√21 * 2√7 - √3 *14 + 7}{√21}\) =\(\frac{2 * 7√3 - 14√3 + 7}{√21}\) = \(\frac{14√3 - 14√3 + 7}{√21}\) =\(\frac{7}{√21}\) or \(\frac{7 \times √21}{√21 \times √21}\) = \(\frac{7 \times √21}{21}\) or \(\frac{√21}{3}\) |
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| 13. |
If 4x+2y=16 and 6x-2y=4 , find the value of (y-x). A. 8 B. 2 C. 4 D. 6 Detailed SolutionUsing elimination method:4x+2y=16 * 6 --> 24x +12y=96 ... eqn iii 6x-2y=4 * 4 --> 24x - 8y = 16 ... eqn iv Subtract eqn iv from iii 20y = 80 y = 4 Subst. y for 4 in 4x + 2y = 16 --> 4x + 2(4) = 16 --> 4x = 16 - 8 --> 4x = 8 --> x = 2 : The value of (y-x) is 4 - 2 = 2 |
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| 14. |
In the diagram, ∠ABC and ∠BCD are right angles, ∠BAD = t and ∠EDF = 70°. Find the value of t. A. 70° B. 165° C. 140° D. 110° Detailed SolutionVertically opp angles are equal ∠CDE = 70°Sum of interior angles in a quadilateral = 360° (90 + 90 + 70 + t)° = 360° t° = 360° - 250° t° = 110° |
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| 15. |
The sum of the interior angles of a regular polygon with k sides is (3k-10) right angles. Find the size of the exterior angle? A. 60° B. 40° C. 90° D. 120° Detailed SolutionIn a polygon with n sides, the sum of the angles =(n - 2)180, where n = the number of sides. In our problem, n = k. So, we have: (3k - 10)90 = (k - 2)180 270k - 900 = 180k - 360 Simplifying: 90k = 540 k = 6. So, we have a regular hexagon. Now, each exterior angle = 360/n We have: 360/6 = 60 Conclusion: Each exterior angle is 60 degrees. |
| 16. |
Make u the subject in x =\(\frac{2u-3}{3u + 2}\) A. u = \(\frac{2x + 3}{3x - 2}\) B. u = \(\frac{2x - 3}{3x - 2}\) C. u = \(\frac{2x + 3}{2 - 3x}\) D. u = \(\frac{2x + 3}{3x + 2}\) Detailed Solutionx =\(\frac{2u-3}{3u + 2}\)cross multiply x(3u + 2) = 2u - 3 3ux + 2x = 2u - 3 collect like terms of u 2x + 3 = 2u - 3ux \(\frac{2x + 3}{2 - 3x}\) = u |
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| 17. |
A trader paid import duty of 38 kobo in the naira on the cost of an engine. If a total of #22,800.00 was paid as import duty, calculate the cost of the engine. A. #60,000.00 B. #120,000.00 C. #24,000.00 D. #18,000.00 |
A |
| 18. |
The height of an equilateral triangle of side is 10 3√ cm. calculate its perimeter. A. 20cm B. 60cm C. 40cm D. 30cm Detailed SolutionHeight of an equilateral triangle, h = a\(\frac{√3}{2}\), where a is the side of the equilateral triangle.10√3 = a\(\frac{√3}{2}\) cross multiply--> 2 * 10√3 = a√3 √3 strikes √3 on both sides 20 = a The perimeter of an equilateral triangle is: P = 3a P = 3 * 20 = 60cm P = 30√3 |
|
| 19. |
In △LMN, |LM| = 6cm, ∠LNM = x and sin x = sin x = \(\frac{3}{5}\).Find the area of △LMN A. 60cm\(^2\) B. 48cm\(^2\) C. 24cm\(^2\) D. 30cm\(^2\) |
C |
| 20. |
Consider the statements:P = All students offering Literature(L) also offer History(H);Q = Students offering History(H) do not offer Geography(G).Which of the Venn diagram correctly illustrate the two statements? A. A B. B C. C D. D |
C |