1 - 10 of 35 Questions
# | Question | Ans |
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1. |
If s = \(\sqrt{(\frac{a^2}{x^2} - \frac{b^2}{y^2})}\)what does y equal? A. \(\sqrt{\frac{(b^2 - a^2)}{(s^2 - x^2)}}\) B. \(\sqrt{\frac{(b^2 - a^2)}{(s^2 \times 2)}}\) C. \(\sqrt{\frac{x^2 - a^2 - b^2}{s}}\) D. \(\frac{x^2 - a^2 - b^2}{s}\) E. \(\sqrt{\frac{(b^2 x^2)}{(a^2 - s^2 x^2)}}\) Detailed Solutions = \(\sqrt{(\frac{a^2}{x^2} - \frac{b^2}{y^2})}\)s\(^2\) = \(\frac{a^2}{x^2} - \frac{b^2}{y^2}\) \(\frac{b^2}{y^2}\) = \(\frac{a^2}{x^2}\) - s\(^2\) \(\frac{b^2}{y^2}\) = \(\frac{a^2 - s^2 x^2}{x^2}\) \(\frac{1}{y^2}\) = \((\frac{a^2 - s^2 x^2}{x^2}) \times \frac{1}{b^2}\) \(\frac{1}{y^2}\) = \(\frac{a^2 - s^2 x^2}{b^2 x^2}\) \(\therefore\) y\(^2\) = \(\frac{b^2 x^2}{a^2 - s^2 x^2}\) y = \(\sqrt{\frac{b^2 x^2}{a^2 - s^2 x^2}}\) |
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2. |
Mr Aborowa bought a car for N200,000.00 and later sold it for N170,000.00. What is the percentage loss? A. 117.60% B. 117.60% C. 117.60% D. 17.65% Detailed SolutionUsing these:Loss = C.p - S.p : Percentage Loss = ( Loss X 100) ÷ C.p = %17.65 |
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3. |
Find the 19th term of the A.P. \(\frac{5}{6}\), \(\frac{8}{6}\), \(\frac{11}{6}\)................. A. 7\(\frac{1}{2}\) B. 9 C. 9\(\frac{1}{2}\) D. 9\(\frac{5}{6}\) E. 10 Detailed Solutionfirst term (a) = \(\frac{5}{6}\)common difference = \(\frac{8}{6}\) - \(\frac{5}{6}\) → \(\frac{3}{6}\) or \(\frac{1}{2}\) A.P formula → T\(_n\) = a + (n - 1)d T\(_n\) = \(\frac{5}{6}\) + (19 - 1)\(\frac{1}{2}\) T\(_n\) = \(\frac{5}{6}\) + 9 → 9\(\frac{5}{6}\) |
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4. |
Simplify 27\(^{-\frac{1}{3}}\) \(\times\) 64\(^{-\frac{1}{3}}\) \(\times\) 4\(^{\frac{1}{3}}\) A. 48 B. 12 C. \(\frac{4}{3}\) D. \(\frac{1}{12}\) E. \(\frac{1}{48}\) |
D |
5. |
The bearing of a point Y from point x is 150°. What is the bearing of X from Y? A. 330° B. 240° C. 150° D. 120° E. 30° |
A |
6. |
If the mean of 3, 5, 8, k, 14 and 17 is 11, what is the value of k A. 58 B. 38 C. 19 D. 11 E. 967 Detailed SolutionTo calculate Mean:add up all the numbers ⇒ 3+5+8+k+14+17 = 47 + k then divide by how many numbers there are ⇒ 6 The Mean is already given as 11 11 = (47 + k) / 6 Cross multiply to solve for k : k = (6 X 11) - 47 k = 19 |
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7. |
From the kip of a storey-buliding, the angle of depression of a mango on a tree 20 m away from the base of the storey-building is 68°. if the mango is 4.5 m above the ground, what is the height of the storey-building to the nearest metre? A. 55m B. 54m C. 50m D. 49m E. 45m |
B |
8. |
The mean age of 12 boys involved survey is 19 years, 3 months. lf the-age of one of the boys is 22 years, what is the mean age of the other-boys? A. 10.5 years B. 19.0 years C. 35.4 years D. 264.0 years E. 423.5 years |
B |
9. |
If cos x = - \(\frac{5}{13}\) where 180° < X < 270°, what is the value of tan x -sin x ? A. \(\frac{111}{13}\) B. \(\frac{321}{65}\) C. -\(\frac{216}{65}\) D. \(\frac{112}{13}\) E. \(\frac{131}{65}\) |
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10. |
Given that (3 - y) ÷ 2x = (6y + 7) ÷ (4x - 5), find the value of x when y = 2 A. \(\frac{64}{5}\) B. 6 C. \(\frac{5}{34}\) D. \(\frac{-5}{34}\) E. \(\frac{-64}{5}\) |
D |
1. |
If s = \(\sqrt{(\frac{a^2}{x^2} - \frac{b^2}{y^2})}\)what does y equal? A. \(\sqrt{\frac{(b^2 - a^2)}{(s^2 - x^2)}}\) B. \(\sqrt{\frac{(b^2 - a^2)}{(s^2 \times 2)}}\) C. \(\sqrt{\frac{x^2 - a^2 - b^2}{s}}\) D. \(\frac{x^2 - a^2 - b^2}{s}\) E. \(\sqrt{\frac{(b^2 x^2)}{(a^2 - s^2 x^2)}}\) Detailed Solutions = \(\sqrt{(\frac{a^2}{x^2} - \frac{b^2}{y^2})}\)s\(^2\) = \(\frac{a^2}{x^2} - \frac{b^2}{y^2}\) \(\frac{b^2}{y^2}\) = \(\frac{a^2}{x^2}\) - s\(^2\) \(\frac{b^2}{y^2}\) = \(\frac{a^2 - s^2 x^2}{x^2}\) \(\frac{1}{y^2}\) = \((\frac{a^2 - s^2 x^2}{x^2}) \times \frac{1}{b^2}\) \(\frac{1}{y^2}\) = \(\frac{a^2 - s^2 x^2}{b^2 x^2}\) \(\therefore\) y\(^2\) = \(\frac{b^2 x^2}{a^2 - s^2 x^2}\) y = \(\sqrt{\frac{b^2 x^2}{a^2 - s^2 x^2}}\) |
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2. |
Mr Aborowa bought a car for N200,000.00 and later sold it for N170,000.00. What is the percentage loss? A. 117.60% B. 117.60% C. 117.60% D. 17.65% Detailed SolutionUsing these:Loss = C.p - S.p : Percentage Loss = ( Loss X 100) ÷ C.p = %17.65 |
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3. |
Find the 19th term of the A.P. \(\frac{5}{6}\), \(\frac{8}{6}\), \(\frac{11}{6}\)................. A. 7\(\frac{1}{2}\) B. 9 C. 9\(\frac{1}{2}\) D. 9\(\frac{5}{6}\) E. 10 Detailed Solutionfirst term (a) = \(\frac{5}{6}\)common difference = \(\frac{8}{6}\) - \(\frac{5}{6}\) → \(\frac{3}{6}\) or \(\frac{1}{2}\) A.P formula → T\(_n\) = a + (n - 1)d T\(_n\) = \(\frac{5}{6}\) + (19 - 1)\(\frac{1}{2}\) T\(_n\) = \(\frac{5}{6}\) + 9 → 9\(\frac{5}{6}\) |
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4. |
Simplify 27\(^{-\frac{1}{3}}\) \(\times\) 64\(^{-\frac{1}{3}}\) \(\times\) 4\(^{\frac{1}{3}}\) A. 48 B. 12 C. \(\frac{4}{3}\) D. \(\frac{1}{12}\) E. \(\frac{1}{48}\) |
D |
5. |
The bearing of a point Y from point x is 150°. What is the bearing of X from Y? A. 330° B. 240° C. 150° D. 120° E. 30° |
A |
6. |
If the mean of 3, 5, 8, k, 14 and 17 is 11, what is the value of k A. 58 B. 38 C. 19 D. 11 E. 967 Detailed SolutionTo calculate Mean:add up all the numbers ⇒ 3+5+8+k+14+17 = 47 + k then divide by how many numbers there are ⇒ 6 The Mean is already given as 11 11 = (47 + k) / 6 Cross multiply to solve for k : k = (6 X 11) - 47 k = 19 |
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7. |
From the kip of a storey-buliding, the angle of depression of a mango on a tree 20 m away from the base of the storey-building is 68°. if the mango is 4.5 m above the ground, what is the height of the storey-building to the nearest metre? A. 55m B. 54m C. 50m D. 49m E. 45m |
B |
8. |
The mean age of 12 boys involved survey is 19 years, 3 months. lf the-age of one of the boys is 22 years, what is the mean age of the other-boys? A. 10.5 years B. 19.0 years C. 35.4 years D. 264.0 years E. 423.5 years |
B |
9. |
If cos x = - \(\frac{5}{13}\) where 180° < X < 270°, what is the value of tan x -sin x ? A. \(\frac{111}{13}\) B. \(\frac{321}{65}\) C. -\(\frac{216}{65}\) D. \(\frac{112}{13}\) E. \(\frac{131}{65}\) |
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10. |
Given that (3 - y) ÷ 2x = (6y + 7) ÷ (4x - 5), find the value of x when y = 2 A. \(\frac{64}{5}\) B. 6 C. \(\frac{5}{34}\) D. \(\frac{-5}{34}\) E. \(\frac{-64}{5}\) |
D |