Year :
2004
Title :
Mathematics (Core)
Exam :
WASSCE/WAEC MAY/JUNE
Paper 1 | Objectives
11 - 20 of 50 Questions
| # | Question | Ans |
|---|---|---|
| 11. |
Find the area of a rectangle of length 4cm and whose diagonal is 6cm, (Leave your answer in surd form) A. 8√3cm2 B. 12√3cm2 C. 16√2cm2 D. 16√3cm2 Detailed Solution
|BC| = 8cm In right-angled BAC; 8\(^2\) = 4\(^2\) + |AC|\(^2\) |AC|\(^2\) = 8\(^2\) - 4\(^2\) |AC\\(^2\) = 64 - 16 → 48 |AC| = \(\sqrt{48}\)cm → \(4\sqrt{3}cm\) The area of rectangle = L x B = |AB| x |AC| = \((4 \times 4\sqrt{3}cm^2\) =16\sqrt{3}cm^2\) |
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| 12. |
Given that x + y = 7 and 3x-y = 5, evaluate \(\frac{y}{2}-3\). A. -1 B. 1 C. 3 D. 4 Detailed Solution\(x + y = 7 ------ I\\3x – y = 5 ------ II\) Add I and II; \(4x = 12 => x = \frac{12}{4} = 3\\ Y = 7 – 3 = 4, evaluate \hspace{1mm} \frac{y}{2} – 3\\ \frac{4}{2} – 3 => 2-3 = -1\) |
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| 13. |
 In the diagram, POQ is the diameter of the circle centre O. Calculate ∠QRS A. 35o B. 70o C. 100o D. 125o Detailed Solution∠PSQ = 90° (angle in a semi-circle)(when SR is joined to SP) ∠SPQ = 180 – (90+35) = 180 – 125 = 55° ∠QRS + ∠SPQ = 180° (opposite angles in a cyclic quad is supplementary) ∠QRS = 180° - 55° = 125° |
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| 14. |
If \(\left(\frac{1}{4}\right)^{(2-y)} = 1\), find y. A. -2 B. \(-\frac{1}{2}\). C. \(\frac{1}{2}\) D. 2 Detailed Solution\(2^{-2(2-y)}-x=2^{0}; -4 + 2y = 0\\2y = 4; y = 2\) |
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| 15. |
Calculate the total surface area of a cupboard which measures 12cm by 10cm by 8cm A. 1920cm2 B. 592cm2 C. 296cm2 D. 148cm2 Detailed SolutionTotal surfaced area of a cupboard = lb + lh + bhl = length h = height b = base l = 12cm h = 10cm b = 8cm T.S.A \(= (12 \times 8 \times 12 \times 10 + 8 \times 10)cm^\\ = (96 + 120 + 80)cm^2 = 296cm^2\) |
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| 16. |
If \(\frac{x}{a+1}+\frac{y}{b}\) 1. Make y the subject of the relation A. \(\frac{b(a-x+1)}{a+1}\) B. \(\frac{a+1}{b(a-x+1)}\) C. \(\frac{a(b-x+1)}{b+1}\) D. \(\frac{b}{a(b-x+1)}\) Detailed SolutionIf \(\frac{bx+y(a+1)}{b(a+1)}=1\\bx + ya + y = ba + b; y(a+1) = ba+b-bx\\ y(a+1)=b(a+1-x); y = \frac{b(1+a-x}{a+1}\) |
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| 17. |
Lf \(log_q p= r\), express p in terms of q and r A. p =qr B. p =rq C. \(p = \frac{r}{q}\) D. p=qr Detailed SolutionIf \(log_q p\) = r convert from logarithm to indicesP = q\(^r\) |
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| 18. |
Find the next two terms of the sequence A. 61,110 B. 67,116 C. 81,140 D. 91, 140 Detailed Solution\(1 + 2^2 = 5 + 3^2 = 14 + 4^2 = 30 + 5^2 = 55\)\(55 + 6^2 = 91 + 7^2 = 140\). 91, 140 are the next two terms. |
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| 19. |
Each interior angle of a regular polygon is 108°. How many sides has it? A. 5 B. 7 C. 9 D. 14 Detailed SolutionThe exterior angle of the polygon = 180 – 108 = 72°. The sum of exterior angle of the regular polygon = 360°The number of sides \(=\frac{360}{72}=5\) |
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| 20. |
Solve the equation 10-3x-x2 = 0 A. x=2 or-5 B. x= -2 or 5 C. x = 1 or 10 D. x = 2 or 5 Detailed Solution10 – 3x - x2 = 0; 10 – 5x + 2x - x2 = 05(2 – x) + x(2 – x) = 0; (2 – x)(5 + x) = 0 2 – x = 0 or 5 + x = 0; x = 2 or x = -5 |
| 11. |
Find the area of a rectangle of length 4cm and whose diagonal is 6cm, (Leave your answer in surd form) A. 8√3cm2 B. 12√3cm2 C. 16√2cm2 D. 16√3cm2 Detailed Solution
|BC| = 8cm In right-angled BAC; 8\(^2\) = 4\(^2\) + |AC|\(^2\) |AC|\(^2\) = 8\(^2\) - 4\(^2\) |AC\\(^2\) = 64 - 16 → 48 |AC| = \(\sqrt{48}\)cm → \(4\sqrt{3}cm\) The area of rectangle = L x B = |AB| x |AC| = \((4 \times 4\sqrt{3}cm^2\) =16\sqrt{3}cm^2\) |
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| 12. |
Given that x + y = 7 and 3x-y = 5, evaluate \(\frac{y}{2}-3\). A. -1 B. 1 C. 3 D. 4 Detailed Solution\(x + y = 7 ------ I\\3x – y = 5 ------ II\) Add I and II; \(4x = 12 => x = \frac{12}{4} = 3\\ Y = 7 – 3 = 4, evaluate \hspace{1mm} \frac{y}{2} – 3\\ \frac{4}{2} – 3 => 2-3 = -1\) |
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| 13. |
In the diagram, POQ is the diameter of the circle centre O. Calculate ∠QRS A. 35o B. 70o C. 100o D. 125o Detailed Solution∠PSQ = 90° (angle in a semi-circle)(when SR is joined to SP) ∠SPQ = 180 – (90+35) = 180 – 125 = 55° ∠QRS + ∠SPQ = 180° (opposite angles in a cyclic quad is supplementary) ∠QRS = 180° - 55° = 125° |
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| 14. |
If \(\left(\frac{1}{4}\right)^{(2-y)} = 1\), find y. A. -2 B. \(-\frac{1}{2}\). C. \(\frac{1}{2}\) D. 2 Detailed Solution\(2^{-2(2-y)}-x=2^{0}; -4 + 2y = 0\\2y = 4; y = 2\) |
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| 15. |
Calculate the total surface area of a cupboard which measures 12cm by 10cm by 8cm A. 1920cm2 B. 592cm2 C. 296cm2 D. 148cm2 Detailed SolutionTotal surfaced area of a cupboard = lb + lh + bhl = length h = height b = base l = 12cm h = 10cm b = 8cm T.S.A \(= (12 \times 8 \times 12 \times 10 + 8 \times 10)cm^\\ = (96 + 120 + 80)cm^2 = 296cm^2\) |
| 16. |
If \(\frac{x}{a+1}+\frac{y}{b}\) 1. Make y the subject of the relation A. \(\frac{b(a-x+1)}{a+1}\) B. \(\frac{a+1}{b(a-x+1)}\) C. \(\frac{a(b-x+1)}{b+1}\) D. \(\frac{b}{a(b-x+1)}\) Detailed SolutionIf \(\frac{bx+y(a+1)}{b(a+1)}=1\\bx + ya + y = ba + b; y(a+1) = ba+b-bx\\ y(a+1)=b(a+1-x); y = \frac{b(1+a-x}{a+1}\) |
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| 17. |
Lf \(log_q p= r\), express p in terms of q and r A. p =qr B. p =rq C. \(p = \frac{r}{q}\) D. p=qr Detailed SolutionIf \(log_q p\) = r convert from logarithm to indicesP = q\(^r\) |
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| 18. |
Find the next two terms of the sequence A. 61,110 B. 67,116 C. 81,140 D. 91, 140 Detailed Solution\(1 + 2^2 = 5 + 3^2 = 14 + 4^2 = 30 + 5^2 = 55\)\(55 + 6^2 = 91 + 7^2 = 140\). 91, 140 are the next two terms. |
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| 19. |
Each interior angle of a regular polygon is 108°. How many sides has it? A. 5 B. 7 C. 9 D. 14 Detailed SolutionThe exterior angle of the polygon = 180 – 108 = 72°. The sum of exterior angle of the regular polygon = 360°The number of sides \(=\frac{360}{72}=5\) |
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| 20. |
Solve the equation 10-3x-x2 = 0 A. x=2 or-5 B. x= -2 or 5 C. x = 1 or 10 D. x = 2 or 5 Detailed Solution10 – 3x - x2 = 0; 10 – 5x + 2x - x2 = 05(2 – x) + x(2 – x) = 0; (2 – x)(5 + x) = 0 2 – x = 0 or 5 + x = 0; x = 2 or x = -5 |