Mathematics (Core)
Paper 1 | Objectives | 46 Questions
WASSCE/WAEC MAY/JUNE
Year: 2003
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper 1 | Objectives | 46 Questions
WASSCE/WAEC MAY/JUNE
Year: 2003
Level: SHS
Time:
Type: Question Paper
Answers provided
No description provided
This paper is yet to be rated
| # | Question | Ans |
|---|---|---|
| 1. |
Simplify \(\left(\frac{3}{4} - \frac{1}{3}\right)\times 4\frac{1}{3}\div 3\frac{1}{4}\) A. \(\frac{5}{9}\) B. \(\frac{5}{16}\) C. \(\frac{8}{13}\) D. \(\frac{16}{5}\) Detailed Solution\((\frac{3}{4} - \frac{1}{3}) \times 4\frac{1}{3} \div 3\frac{1}{4}\)= \((\frac{9 - 4}{12}) \times \frac{13}{3} \div \frac{13}{4}\) = \(\frac{5}{12} \times \frac{13}{3} \times \frac{4}{13}\) = \(\frac{5}{9}\) |
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| 2. |
 The table below gives the distribution of marks obtained by a number of pupils in a class test.The mode of the distribution is A. 3 B. 5 C. 8 D. 18 |
A |
| 3. |
 The table below gives the distribution of marks obtained by a number of pupils in a class test.Find the median of the distribution A. 1 B. 2 C. 3 D. 4 |
C |
| 4. |
 The table below gives the distribution of marks obtained by a number of pupils in a class test.How many pupils scored at least 2 marks? A. 12 B. 23 C. 37 D. 49 Detailed Solution= 12 + 18 + 11 + 8= 49 students |
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| 5. |
 Interprete the inequality represented on the number line A. -4< X ≤ 5 B. -4 ≤ x< 5 C. -4 D. -4≤x≤5 |
A |
| 6. |
A student bought 3 notebooks and 1 pen for N35. After misplacing these items, she again bought 2 notebooks and 2 pens, all of the same type for A. N5.00 B. N7.50 C. N10.00 D. N15.00 Detailed SolutionRepresent notebook with xRepresent pen with y ∴ 3x+y=35 ---- (i) 2x+2y=30 ---- (ii) Multiply eqn (i) by 2 and eqn (ii) by 1 6x+2y=70 ---- (iii) 2x+2y=30 ---- (iv) Subtract eqn (iv) from eqn (iii) 4x = 40 X = 10 Substitute x = 10 in eqn (i) 3x+y=35 3(10)+y=35 Y = 5 ∴ The cost of pen is N5.00 |
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| 7. |
Simplify \(\frac{4}{a-3}-\frac{1}{a+3}\) A. \(\frac{3a+11}{(a-3)(a-2)}\) B. \(\frac{3a+11}{(a+3)(a-2)}\) C. \(\frac{3a+11}{(a+3)(a+2)}\) D. \(\frac{3a+11}{(a-3)(a+2)}\) Detailed Solution\(\frac{4}{a-3}-\frac{1}{a+3}\\= \frac{4(a+2)-1(a-3)}{(a-3)(a+2)}\\ =\frac{3a+11}{(a-3)(a+2)} \) |
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| 8. |
Simplify \(\frac{4\sqrt{18}}{\sqrt{8}}\) A. 2 B. 3 C. 6 D. 12 Detailed Solution\(\frac{4\sqrt{18}}{\sqrt{8}}\\= \frac{4\sqrt{9\times 2}}{\sqrt{4 \times 2}} = \frac{12\sqrt{2}}{2\sqrt{2}}=\frac{6\sqrt{2}}{\sqrt{2}}\\ Rationalising \frac{6\sqrt{2}}{\sqrt{2}}\\ =\frac{6\sqrt{2}}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = 6\) |
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| 9. |
 In the diagram, PQ/RT /RS/ =/QS/, ∠QST= 70o and ∠PQW = 55o A. 235o B. 290o C. 305o D. 325o Detailed SolutionReflex ∠WQS = 360o - (55o + 70o)=360o - 125o = 235o |
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| 10. |
Three men, Bedu, Bakre and Kofi shared' N500 in the ratio 3:2: x respectively. If Bedu’s share is N150, find the value of x. A. 1 B. 4 C. 5 D. 6 Detailed Solution\(5+x\rightarrow 500\\3 \rightarrow 150\\ ∴ 3 \times 500 = 150 \times (5+x)\\ 1500 = 750 + 150x\\ x=\frac{750}{150}=5\) |
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