Mathematics (Core)
Paper 1 | Objectives | 49 Questions
WASSCE/WAEC MAY/JUNE
Year: 2010
Level: SHS
Time:
Type: Question Paper
Answers provided
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Paper 1 | Objectives | 49 Questions
WASSCE/WAEC MAY/JUNE
Year: 2010
Level: SHS
Time:
Type: Question Paper
Answers provided
No description provided
This paper is yet to be rated
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| # | Question | Ans |
|---|---|---|
| 1. |
Simplify 0.000215 x 0.000028 and express your answer in standard form A. 6.03 x 109 B. 6.02 x 109 C. 6.03 x 10-9 D. 6.02 x 10-9
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Detailed Solution0.000215215 x 0.000028= 215 x 10-4 x 28 x 10-4 = 215 x 28 x 10-6-6 = 6020 x 10-12 = 6.020 x 103 x 10-12 = 6.02 x 103 - 12 = 6.02 x 10-9 |
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| 2. |
Factorize: x + y - ax = ay A. (x - y)(1 - a) B. (x + y)(1 + a) C. (x + y)(1 - a) D. (x - y)(1 + a)
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Detailed Solutionx + y - ax = ay= (x + y) -a(x + y) = (x + y)(1 - a) |
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| 3. |
A car uses one litre of petrol for every 14km. If one of petrol cost N63.00, how far can the car go with N900.00 worth of petrol? A. 420km B. 405km C. 210km D. 200km
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Detailed Solution1 litre = N63.00xlitres = N900.00 x x N63 = N900 x 1 litre x = \(\frac{900}{63} \times litre\) x = \(\frac{100}{7}\) litres Also, 1 litre = 14km \(\frac{100}{7}\) = y y x 1 litre = \(\frac{100litre}{7}\) x 14km y = 20km |
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| 4. |
Correct 0.002473 to 3 significant figure A. 0.002 B. 0.0024 C. 0.00247 D. 0.0025 |
C |
| 5. |
Simplify 1\(\frac{1}{2} + 2\frac{1}{3} \times \frac{3}{4} - \frac{1}{2}\) A. -2\(\frac{1}{3}\) B. -2\(\frac{1}{4}\) C. 2\(\frac{1}{8}\) D. 2\(\frac{3}{4}\)
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Detailed Solution1\(\frac{1}{2} + 2\frac{1}{3} \times \frac{3}{4} - \frac{1}{2}\)\(\frac{3}{2} + \frac{7}{3} \times \frac{3}{4} - \frac{1}{2} = \frac{3}{2} + \frac{7}{4} - \frac{1}{2}\) = \(\frac{6 + 7 - 2}{4} = \frac{11}{4}\) = 2\(\frac{3}{4}\) |
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| 6. |
The sum of 2 consecutive whole numbers is \(\frac{5}{6}\) of their product, find the numbers A. 3, 4 B. 1, 2 C. 2, 3 D. 0, 1
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Detailed SolutionLet the no. be x and x + 1x + (x + 1) = \(\frac{5}{6}\) of x(x + 1) 2x + 1 = \(\frac{5}{6}\) x(x + 1) 6(2x + 1) = 5x2 + 5x 12x + 6 = 5x2 + 5x 5x2 + 5x - 12x - 6 = 0 5x2 - 7x - 6 = 0 5x2 - 10x + 3x - 6 = 0 5x(x - 2) + 3(x - 2) = 0 (5x + 3)(x - 2) = 0 (5x + 3)(x - 2) = 0 (5x + 3) = 0 x - 2 = 0 for (5x + 3) = 0 5x = -3 |
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| 7. |
A casting is made up of Copper and Zinc. If 65% of the casting is Zinc and there are 147g of Copper. What is the mass of the casting? A. 320g B. 420g C. 520g D. 620g
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Detailed Solution%Zinc + %Copper = 100%65% + %Copper = 100% %Copper = 100% - 65% = 35% %Copper = \(\frac{\text{Mass of % Copper}}{\text{Mass of Casting}}\) x 100% Mass of Casting = \(\frac{147}{35}\) x 100% = 420g |
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| 8. |
Given that P = {x : 1 \(\leq x \leq 6\)}, and Q = {x : 2 \(\leq x \leq 10\)} where x is an integer. Find n(P \(\cap\) Q) A. 5 B. 6 C. 8 D. 10
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Detailed SolutionP = {1, 2, 3, 4, 5, 6}; Q = {3, 4, 5, 6, 7, 8, 9}P \(\cap\) Q) = {2, 3, 4, 5, 6} n(P \(\cap\) Q) = 5 |
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| 9. |
The sum of 6 and one-third of x is one more than twice x, find x A. x = 7 B. x = 5 C. x = 3 D. x = 2
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Detailed Solution6 = \(\frac{1}{3}x\) = 1 + 2x6 = 1 = \(\frac{2x}{1} - \frac{x}{3}\) 5 = \(\frac{6x - x}{3} = \frac{5x}{3}\) 5 = \(\frac{5x}{3}\) 3 x 5 = 5x 15 = 5x x = \(\frac{15}{5}\) = 3 |
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| 10. |
Given that = {x: -2 < x \(\leq\) 9}, where x is an integer what is n(T)? A. 9 B. 10 C. 11 D. 12
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Detailed SolutionT = {-1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}n(T) = 11 |
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