Year :
2013
Title :
Mathematics (Core)
Exam :
WASSCE/WAEC MAY/JUNE
Paper 1 | Objectives
11 - 20 of 48 Questions
| # | Question | Ans |
|---|---|---|
| 11. |
If x = 64 and y = 27, evaluate: \(\frac{x^{\frac{1}{2}} - y^{\frac{1}{3}}}{y - x^{\frac{2}{3}}}\) A. 2\(\frac{1}{5}\) B. 1 C. \(\frac{5}{11}\) D. \(\frac{11}{43}\) Detailed Solution\(\frac{x^{\frac{1}{2}} - y^{\frac{1}{3}}}{y - x^{\frac{2}{3}}}\)substitute x = 64 and y = 27 \(\frac{64^{\frac{1}{2}} - 27^{\frac{1}{3}}}{27 - 64^{\frac{2}{3}}} = \frac{\sqrt{64} - 3\sqrt{27}}{27 - (3\sqrt{64})^2}\) = \(\frac{8 - 3}{27 - 16}\) = \(\frac{5}{11}\) |
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| 12. |
If \(\frac{1}{2}\)x + 2y = 3 and \(\frac{3}{2}\)x and \(\frac{3}{2}\)x - 2y = 1, find (x + y) A. 3 B. 2 C. 1 D. 5 Detailed Solution\(\frac{1}{2}\)x + 2y = 3......(i)(multiply by 2)\(\frac{3}{2}\)x - 2y = 1......(ii)(multiply by 2) x + 4y = 6......(iii) 3x - 4y = 2.....(iv) add (iii) and (iv) 4x = 8, x = \(\frac{8}{4}\) = 2 substitute x = 2 into equation (iii) x + 4y = 6 2 + 4y = 6 4y = 6 - 2 4y = 4 y = \(\frac{4}{4}\) = 1(x + y) 2 + 1 = 3 |
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| 13. |
Given that p\(\frac{1}{3}\) = \(\frac{3\sqrt{q}}{r}\), make q the subject of the equation A. q = p\(\sqrt{r}\) B. q = p3r C. q = pr3 D. q = pr\(\frac{1}{3}\) Detailed Solutionp\(\frac{1}{3}\) = \(\frac{3\sqrt{q}}{r}\)(cross multiply)3\(\sqrt{q}\) = r x 3\(\frac{\sqrt{q}}{r}\)(cross multiply) 3\(\sqrt{q}\) = r x 3\(\sqrt{p}\) cube root both side q = 3\(\sqrt{r}\) x p q = r\(\frac{1}{3}\)p = pr\(\frac{1}{3}\) |
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| 14. |
A chord is 2cm from the centre of a circle. If the radius of the circle is 5cm, find the length of the chord A. 2\(\sqrt{21}\)cm B. \(\sqrt{42}\)cm C. 2\(\sqrt{19}\)cm D. \(\sqrt{21}\)cm Detailed SolutionFrom \(\bigtriangleup\) OMQ find /MQ/ by Pythagoras OQ2 = OM2 + MQ252 = 22 + MQ2 25 = 4 + MQ2 25 - 4 = MQ2 21 - MQ2 MQ2 = 21 MQ2 = \(\sqrt{21}\) Length of chord = 2 x \(\sqrt{21}\) = 2\(\sqrt{21}\)cm |
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| 15. |
A cube and cuboid have the same base area. The volume of the cube is 64cm\(^3\) while that of the cuboid is 80cm\(^3\). Find the height of the cuboid A. 1cm B. 3cm C. 5cm D. 6cm Detailed SolutionVolume of a cube with side a cm = a\(^3\)a\(^3\) = 64 \(\implies\) a = 4cm Base area of a cube = a\(^2\) = 4\(^2\) = 16 cm\(^2\) \(\implies\) Base area of the cuboid = 16 cm\(^2\) Volume of cuboid = Base area x height 80 = 16 x h h = \(\frac{80}{16}\) = 5 cm |
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| 16. |
If sin x = \(\frac{5}{13}\) and 0o \(\leq\) x \(\leq\) 90o, find the value of (cos x - tan x) A. \(\frac{7}{13}\) B. \(\frac{12}{13}\) C. \(\frac{79}{156}\) D. \(\frac{209}{156}\) Detailed SolutionSin x = \(\frac{5}{13}\)0o \(\leq\) x \(\leq\) 90o, (cos x - tan x) AC2 = AB2 + BC2 132 = 52 + BC2 169 - 25 + BC2 169 - 25 = BC2 144 = BC2 Cos x = \(\frac{Adj}{Hyp}\) = \(\frac{12}{13}\) BC = \(\sqrt{144}\) BC = 12 tan x = \(\frac{opp}{adj} = \frac{5}{12}\) BC |
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| 17. |
An object is 6m away from the base of a mast. The angle of depression of the object from the top pf the mast is 50o, Find, correct to 2 decimal places, the height of the mast A. 8.60m B. 7.51m C. 7.15m D. 1.19m Detailed SolutionTan 50o = \(\frac{h}{6}\)h = 6 tan 50 = 6 x 1.1917 = 7.1505 = 7.15 |
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| 18. |
The bearing of Y from X is 060o and the bearing of Z from Y = 060o. Find the bearing of X from Z A. 300o B. 240o C. 180o D. 120o |
B |
| 19. |
Which of the following is not a probability of Mary scoring 85% in a mathematics test? A. 0.15 B. 0.57 C. 0.94 D. 1.01 Detailed SolutionThe probability not scoring 85% is = 1 - pro(scoring 85%)= 1 - \(\frac{85}{100}\) = 1 - 0.85 = 0.15 not scoring would be less than 1 |
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| 20. |
If 2 log x (3\(\frac{3}{8}\)) = 6, find the value of x A. \(\frac{3}{2}\) B. \(\frac{4}{3}\) C. \(\frac{2}{3}\) D. \(\frac{1}{2}\) Detailed Solution2 log x (3\(\frac{3}{8}\)) = 6(divide both sides by 2)\(\frac{2 log_x(3 \frac{3}{8})}{2} = \frac{6}{2}\) \(\log_x \frac{27}{8} = 3\) \(\frac{27}{8} = x^3\) \(x^3 = \frac{27}{8}\) x = \(\sqrt{\frac{27}{8}}\) = (\(\frac{27}{8}\))\(\frac{1}{3}\) = \(\frac{(3^3)^{\frac{1}{3}}}{(2^3)^{\frac{1}{3}}}\) = \(\frac{3}{2}\) |
| 11. |
If x = 64 and y = 27, evaluate: \(\frac{x^{\frac{1}{2}} - y^{\frac{1}{3}}}{y - x^{\frac{2}{3}}}\) A. 2\(\frac{1}{5}\) B. 1 C. \(\frac{5}{11}\) D. \(\frac{11}{43}\) Detailed Solution\(\frac{x^{\frac{1}{2}} - y^{\frac{1}{3}}}{y - x^{\frac{2}{3}}}\)substitute x = 64 and y = 27 \(\frac{64^{\frac{1}{2}} - 27^{\frac{1}{3}}}{27 - 64^{\frac{2}{3}}} = \frac{\sqrt{64} - 3\sqrt{27}}{27 - (3\sqrt{64})^2}\) = \(\frac{8 - 3}{27 - 16}\) = \(\frac{5}{11}\) |
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| 12. |
If \(\frac{1}{2}\)x + 2y = 3 and \(\frac{3}{2}\)x and \(\frac{3}{2}\)x - 2y = 1, find (x + y) A. 3 B. 2 C. 1 D. 5 Detailed Solution\(\frac{1}{2}\)x + 2y = 3......(i)(multiply by 2)\(\frac{3}{2}\)x - 2y = 1......(ii)(multiply by 2) x + 4y = 6......(iii) 3x - 4y = 2.....(iv) add (iii) and (iv) 4x = 8, x = \(\frac{8}{4}\) = 2 substitute x = 2 into equation (iii) x + 4y = 6 2 + 4y = 6 4y = 6 - 2 4y = 4 y = \(\frac{4}{4}\) = 1(x + y) 2 + 1 = 3 |
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| 13. |
Given that p\(\frac{1}{3}\) = \(\frac{3\sqrt{q}}{r}\), make q the subject of the equation A. q = p\(\sqrt{r}\) B. q = p3r C. q = pr3 D. q = pr\(\frac{1}{3}\) Detailed Solutionp\(\frac{1}{3}\) = \(\frac{3\sqrt{q}}{r}\)(cross multiply)3\(\sqrt{q}\) = r x 3\(\frac{\sqrt{q}}{r}\)(cross multiply) 3\(\sqrt{q}\) = r x 3\(\sqrt{p}\) cube root both side q = 3\(\sqrt{r}\) x p q = r\(\frac{1}{3}\)p = pr\(\frac{1}{3}\) |
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| 14. |
A chord is 2cm from the centre of a circle. If the radius of the circle is 5cm, find the length of the chord A. 2\(\sqrt{21}\)cm B. \(\sqrt{42}\)cm C. 2\(\sqrt{19}\)cm D. \(\sqrt{21}\)cm Detailed SolutionFrom \(\bigtriangleup\) OMQ find /MQ/ by Pythagoras OQ2 = OM2 + MQ252 = 22 + MQ2 25 = 4 + MQ2 25 - 4 = MQ2 21 - MQ2 MQ2 = 21 MQ2 = \(\sqrt{21}\) Length of chord = 2 x \(\sqrt{21}\) = 2\(\sqrt{21}\)cm |
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| 15. |
A cube and cuboid have the same base area. The volume of the cube is 64cm\(^3\) while that of the cuboid is 80cm\(^3\). Find the height of the cuboid A. 1cm B. 3cm C. 5cm D. 6cm Detailed SolutionVolume of a cube with side a cm = a\(^3\)a\(^3\) = 64 \(\implies\) a = 4cm Base area of a cube = a\(^2\) = 4\(^2\) = 16 cm\(^2\) \(\implies\) Base area of the cuboid = 16 cm\(^2\) Volume of cuboid = Base area x height 80 = 16 x h h = \(\frac{80}{16}\) = 5 cm |
| 16. |
If sin x = \(\frac{5}{13}\) and 0o \(\leq\) x \(\leq\) 90o, find the value of (cos x - tan x) A. \(\frac{7}{13}\) B. \(\frac{12}{13}\) C. \(\frac{79}{156}\) D. \(\frac{209}{156}\) Detailed SolutionSin x = \(\frac{5}{13}\)0o \(\leq\) x \(\leq\) 90o, (cos x - tan x) AC2 = AB2 + BC2 132 = 52 + BC2 169 - 25 + BC2 169 - 25 = BC2 144 = BC2 Cos x = \(\frac{Adj}{Hyp}\) = \(\frac{12}{13}\) BC = \(\sqrt{144}\) BC = 12 tan x = \(\frac{opp}{adj} = \frac{5}{12}\) BC |
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| 17. |
An object is 6m away from the base of a mast. The angle of depression of the object from the top pf the mast is 50o, Find, correct to 2 decimal places, the height of the mast A. 8.60m B. 7.51m C. 7.15m D. 1.19m Detailed SolutionTan 50o = \(\frac{h}{6}\)h = 6 tan 50 = 6 x 1.1917 = 7.1505 = 7.15 |
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| 18. |
The bearing of Y from X is 060o and the bearing of Z from Y = 060o. Find the bearing of X from Z A. 300o B. 240o C. 180o D. 120o |
B |
| 19. |
Which of the following is not a probability of Mary scoring 85% in a mathematics test? A. 0.15 B. 0.57 C. 0.94 D. 1.01 Detailed SolutionThe probability not scoring 85% is = 1 - pro(scoring 85%)= 1 - \(\frac{85}{100}\) = 1 - 0.85 = 0.15 not scoring would be less than 1 |
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| 20. |
If 2 log x (3\(\frac{3}{8}\)) = 6, find the value of x A. \(\frac{3}{2}\) B. \(\frac{4}{3}\) C. \(\frac{2}{3}\) D. \(\frac{1}{2}\) Detailed Solution2 log x (3\(\frac{3}{8}\)) = 6(divide both sides by 2)\(\frac{2 log_x(3 \frac{3}{8})}{2} = \frac{6}{2}\) \(\log_x \frac{27}{8} = 3\) \(\frac{27}{8} = x^3\) \(x^3 = \frac{27}{8}\) x = \(\sqrt{\frac{27}{8}}\) = (\(\frac{27}{8}\))\(\frac{1}{3}\) = \(\frac{(3^3)^{\frac{1}{3}}}{(2^3)^{\frac{1}{3}}}\) = \(\frac{3}{2}\) |