11 - 20 of 46 Questions
# | Question | Ans |
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11. |
Simplify \(\left(1\frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2\) A. \(2\frac{1}{3}\) B. \(1\frac{1}{3}\) C. 1 D. \(\frac{3}{7}\) Detailed Solution\(\left(1\frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2=\left(\frac{5}{3}\right)^2 - \left(\frac{2}{3}\right)^2\)Difference of two squares \(\left(\frac{5}{3}-\frac{2}{3}\right)\left(\frac{5}{3}+\frac{2}{3}\right)=\left(\frac{3}{3}\right)\left(\frac{7}{3}\right)\\ \frac{7}{3}=2\frac{1}{3}\) |
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12. |
Find the value of x which satisfies the equation A. x =2 B. x=4 C. x=6 D. x = 14 Detailed Solution5(x - 7) = 7 - 2x5x - 35 = 7 - 2x 5x + 2x = 7 + 35 7x = 42 x = \(\frac{42}{7}\) = 6 |
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13. |
Evaluate \(\frac{log8}{log\left(\frac{1}{4}\right)}\) A. -2 B. \(\frac{-3}{2}\) C. \(\frac{1}{2}\) D. 4 Detailed Solution\(\frac{log8}{log\frac{1}{4}}=\frac{log2^3}{log2^{-2}}=\frac{3log2}{-2log2}=-\frac{3}{2}\) |
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14. |
A pyramid of volume 120cm3 has a rectangular base which measures 5cm by 6cm. Calculate the height of the pyramid A. 5cm B. 9cm C. 12cm D. 15cm Detailed Solution\(V= \frac{1}{2}\hspace{1mm}base\hspace{1mm}area\hspace{1mm}\times\hspace{1mm}height\\120=\frac{1}{3}\times 5 \times 6 \times h; h = \frac{120}{10}=12cm\) |
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15. |
PQRS is a cyclic quadrilateral. If ∠QPS = 75°, what is the size of ∠QRS? A. 180o B. 150o C. 105o D. 75o Detailed Solution< QPS = 75° < QRS = 180° - 75° = 105° (opposite angles of a cyclic quadrilateral) |
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16. |
In the diagram, ZM is a straight line. Calculate the value of x. A. 27o B. 30o C. 35o D. 37o Detailed SolutionThe sum of angles in a triangle is 180o2xo + (2x -21)o + ∠ZYP = 180o Also ∠ZYP = 180o - (3x+14)o ∴2xo + (2x-21)o + 180o - (3x+14)o = 180 x = 35o |
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17. |
Find the value of x in the equation 3x\(^2\) - 8x - 3 = 0 A. \(\frac{1}{3},-3\) B. \(-\frac{1}{3},-3\) C. \(-\frac{1}{3},3\) D. \(\frac{1}{3},3\) Detailed Solution3x\(^2\) - 8x - 3 = 03x\(^2\) - 9x + x - 3 = 0 3x(x - 3) + 1(x - 3) = 0 (3x + 1)(x - 3) = 0 3x = -1 \(\implies\) x = \(-\frac{1}{3}\) or x = 3. |
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18. |
In\( ∆ PQR, P\hat{Q}P = 84^°, |Q\hat{P}R |= 43^°\) and |PQ| = 5cm. Find /QR/ in cm, correct to 1 decimal place. A. 3.4 B. 4.3 C. 5.9 D. 6.2 Detailed Solution\(\frac{|QR|}{sinP}=\frac{|PQ|}{sinR}\\ \frac{x}{sin43^o}=\frac{5}{sin53^o}\\ x = \frac{5\times sin43^o}{sin53^o}\\ =\frac{5\times 0.6820}{0.7986}\\ =4.3\) |
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19. |
Given, that \(4P4_5 = 119_{10}\), find the value of P A. 1 B. 2 C. 3 D. 4 Detailed Solution\(4P4_5 = 119_{10}\)\(4 \times 5^2 + P \times 5^1 + 4 \times 5^0 = 119\) \(100 + 5P + 4 = 119\) \(5P = 119 - 104 = 15\) \(P = 3\) |
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20. |
A ladder, 6m long, leans against a vertical wall at an angle 53o to the horizontal. How high up the wall does the ladder reach? A. 3.611m B. 4.521m C. 4.792m D. 7.962m Detailed SolutionSin 53 = \(\frac {h}{6} \)h = 6 sin 53 = 6 x 0.7986 = 4.792m |
11. |
Simplify \(\left(1\frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2\) A. \(2\frac{1}{3}\) B. \(1\frac{1}{3}\) C. 1 D. \(\frac{3}{7}\) Detailed Solution\(\left(1\frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2=\left(\frac{5}{3}\right)^2 - \left(\frac{2}{3}\right)^2\)Difference of two squares \(\left(\frac{5}{3}-\frac{2}{3}\right)\left(\frac{5}{3}+\frac{2}{3}\right)=\left(\frac{3}{3}\right)\left(\frac{7}{3}\right)\\ \frac{7}{3}=2\frac{1}{3}\) |
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12. |
Find the value of x which satisfies the equation A. x =2 B. x=4 C. x=6 D. x = 14 Detailed Solution5(x - 7) = 7 - 2x5x - 35 = 7 - 2x 5x + 2x = 7 + 35 7x = 42 x = \(\frac{42}{7}\) = 6 |
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13. |
Evaluate \(\frac{log8}{log\left(\frac{1}{4}\right)}\) A. -2 B. \(\frac{-3}{2}\) C. \(\frac{1}{2}\) D. 4 Detailed Solution\(\frac{log8}{log\frac{1}{4}}=\frac{log2^3}{log2^{-2}}=\frac{3log2}{-2log2}=-\frac{3}{2}\) |
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14. |
A pyramid of volume 120cm3 has a rectangular base which measures 5cm by 6cm. Calculate the height of the pyramid A. 5cm B. 9cm C. 12cm D. 15cm Detailed Solution\(V= \frac{1}{2}\hspace{1mm}base\hspace{1mm}area\hspace{1mm}\times\hspace{1mm}height\\120=\frac{1}{3}\times 5 \times 6 \times h; h = \frac{120}{10}=12cm\) |
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15. |
PQRS is a cyclic quadrilateral. If ∠QPS = 75°, what is the size of ∠QRS? A. 180o B. 150o C. 105o D. 75o Detailed Solution< QPS = 75° < QRS = 180° - 75° = 105° (opposite angles of a cyclic quadrilateral) |
16. |
In the diagram, ZM is a straight line. Calculate the value of x. A. 27o B. 30o C. 35o D. 37o Detailed SolutionThe sum of angles in a triangle is 180o2xo + (2x -21)o + ∠ZYP = 180o Also ∠ZYP = 180o - (3x+14)o ∴2xo + (2x-21)o + 180o - (3x+14)o = 180 x = 35o |
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17. |
Find the value of x in the equation 3x\(^2\) - 8x - 3 = 0 A. \(\frac{1}{3},-3\) B. \(-\frac{1}{3},-3\) C. \(-\frac{1}{3},3\) D. \(\frac{1}{3},3\) Detailed Solution3x\(^2\) - 8x - 3 = 03x\(^2\) - 9x + x - 3 = 0 3x(x - 3) + 1(x - 3) = 0 (3x + 1)(x - 3) = 0 3x = -1 \(\implies\) x = \(-\frac{1}{3}\) or x = 3. |
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18. |
In\( ∆ PQR, P\hat{Q}P = 84^°, |Q\hat{P}R |= 43^°\) and |PQ| = 5cm. Find /QR/ in cm, correct to 1 decimal place. A. 3.4 B. 4.3 C. 5.9 D. 6.2 Detailed Solution\(\frac{|QR|}{sinP}=\frac{|PQ|}{sinR}\\ \frac{x}{sin43^o}=\frac{5}{sin53^o}\\ x = \frac{5\times sin43^o}{sin53^o}\\ =\frac{5\times 0.6820}{0.7986}\\ =4.3\) |
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19. |
Given, that \(4P4_5 = 119_{10}\), find the value of P A. 1 B. 2 C. 3 D. 4 Detailed Solution\(4P4_5 = 119_{10}\)\(4 \times 5^2 + P \times 5^1 + 4 \times 5^0 = 119\) \(100 + 5P + 4 = 119\) \(5P = 119 - 104 = 15\) \(P = 3\) |
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20. |
A ladder, 6m long, leans against a vertical wall at an angle 53o to the horizontal. How high up the wall does the ladder reach? A. 3.611m B. 4.521m C. 4.792m D. 7.962m Detailed SolutionSin 53 = \(\frac {h}{6} \)h = 6 sin 53 = 6 x 0.7986 = 4.792m |