31 - 40 of 48 Questions
# | Question | Ans |
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31. |
In a sample survey of a University Community, the following table shows the percentage distribution of the number of members per house hold A. 4 B. 3 C. 5 D. 4.5 E. None Detailed Solution\(\begin{array}{c|c} x & f \\\hline 1 & 3\\ 2 & 13\\3 & 15\\ 4 & 28\\ 5 & 21\\ 6 & 10\\ 7 & 7\\ 8 & 4\end{array}\)Median is half of total frequency = 50th term 4 falls in the range = 4 |
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32. |
On a square paper of length 2.524375cm is inscribed square diagram of length 0.524375cm. Find the area of the paper not covered by the diagram. correct to 3 significant figures. A. 6.00cm2 B. 6.10cm2 C. 6cmv D. 6.09cm2 E. 4.00cm2 Detailed SolutionArea of the paper = area of square = L x B or S2where s = S x k Area of the paper = (2.524)2 area of the diagram = (0.524)2 area not covered = (2.524)2 - (0.524)2 = 6.370576 - 0.274576 = 6.096 = 6.10cm2 (2 s.f) |
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33. |
Simplify \(\sqrt[3]{\frac{27a^{-9}}{8}}\). A. \(\frac{9a^2}{2a^3}\) B. \(\frac{3}{2a^3}\) C. \(\frac{2}{3a^2}\) D. \(\frac{3a^2}{2}\) Detailed Solution\(\sqrt[3]{\frac{27a ^{-9}}{8}}\) = \(\frac{3a^{-3}}{2}\)= \(\frac{3}{2}\) x \(\frac{1}{a^3}\) = \(\frac{3}{2a^3}\) |
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34. |
PQR is the diagram of a semicircle RSP with centre at Q and radius of length 3.5cm. If QPT = 60o. Find the perimeter of the figure PTRS. \(\pi\) = \(\frac{22}{7}\) A. 25cm B. 18cm C. 36cm D. 29cm E. 25.5cm Detailed SolutionCircumference of PRS = \(\frac{\pi}{2}\) = \(\frac{22}{7}\) x \(\frac{7}{1}\) x \(\frac{1}{2}\)= 11cm Side PT = 7cm, Side TR = 7cm Perimeter(PTRS) = 11cm + 7cm + 7cm = 25cm |
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35. |
In a triangle PQT, QR = \(\sqrt{3}cm\), PR = 3cm, PQ = \(2\sqrt{3}\)cm and PQR = 30o. Find angles P and R A. P = 60o and R = 90o B. P = 30o and R = 120o C. P = 90o and R = 60o D. P = 60o and R 60o E. P = 45o and R = 105o Detailed SolutionBy using cosine formula, p2 = Q2 + R2 - 2QR cos pCos P = \(\frac{Q^2 + R^2 - p^2}{2 QR}\) = \(\frac{(3)^2 + 2(\sqrt{3})^2 - 3^2}{2\sqrt{3}}\) = \(\frac{3 + 12 - 9}{12}\) = \(\frac{6}{12}\) = \(\frac{1}{2}\) = 0.5 Cos P = 0.5 p = cos-1 0.5 = 60o = < P = 60o If < P = 60o and < Q = 30 < R = 180o - 90o angle P = 60o and angle R is 90o |
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36. |
Find the mean of the following 24.57, 25.63, 24.32, 26.01, 25.77 A. 25.12 B. 25.30 C. 25.26 D. 25.50 E. 25.75 Detailed Solution\(\frac{24.57 + 25.63 + 24.32 + 26.01 + 25.77}{5}\)mean = \(\frac{126.3}{5}\) = 25.26 |
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37. |
A man drove for 4 hours at a certain speed, he then doubled his speed and drove for another 3 hours. Although he covered 600 kilometers. At what speed did he drive for the last 3 hours? A. 120km/hr B. 60km/hr C. 670km/hr D. 40km/hr Detailed SolutionSpeed = \(\frac{distance}{time}\)let x represent the speed, d represent distance x = \(\frac{d}{4}\) d = 4x 2x = \(\frac{600 - d}{3}\) 6x = 600 - d 6x = 600 - 4x 10x = 600 x = \(\frac{600}{10}\) = 60km/hr |
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38. |
In a class of 60 pupils, the statistical distribution of the numbers of pupils offering Biology, History, French, Geography and Additional mathematics is as shown in the pie chart. How many pupils offer Additional Mathematics? A. 15 B. 10 C. 18 D. 12 E. 20 Detailed Solution(2x - 24)° + (3x - 18)° + (2x + 12)° + (x + 12)° + x° = 360°9x = 360° + 18° x = \(\frac{378}{9}\) = 42°, if x = 42°, then add maths = \(\frac{2x - 42}{360}\) x 60 = \(\frac{2 \times 42 - 24}{360}\) x 60 = \(\frac{84 - 24}{6}\) = 10 |
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39. |
In the figure, find PRQ A. 66\(\frac{1}{2}\)o B. 62\(\frac{1}{2}\)o C. 125o D. 105o E. 65o Detailed SolutionAngle subtended at any part of the circumference of the circle \(\frac{125^o}{2}\) at centre = 360o - 235o = 125o\(\bar{PQR}\) = \(\frac{125}{2}\) = 62\(\frac{1}{2}\)o |
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40. |
Which of the following equations represents the graph? A. y = 1 + 2x + 3x2 B. y = 1 - 2x + 3x2 C. y = 1 + 2x - 3x2 D. y = 1 - 2x - 3x2 E. y = 3x2 + 2x - 1 Detailed SolutionThe roots of the function are 1 and \(\frac{1}{3}\)sum of roots = -1 + \(\frac{1}{3}\) = -\(\frac{2}{3}\) product of roots = -1 x \(\frac{1}{3}\) = -\(\frac{1}{3}\) x2 - (sum of roots)x + (product of roots) = 0 x2 + (-\(\frac{2}{3}\))x - (-\(\frac{1}{3}\)) = 0 x2 + \(\frac{2x}{3}\) - \(\frac{1}{3}\) = 0 3x2 + 2x - 1 = 0 |
31. |
In a sample survey of a University Community, the following table shows the percentage distribution of the number of members per house hold A. 4 B. 3 C. 5 D. 4.5 E. None Detailed Solution\(\begin{array}{c|c} x & f \\\hline 1 & 3\\ 2 & 13\\3 & 15\\ 4 & 28\\ 5 & 21\\ 6 & 10\\ 7 & 7\\ 8 & 4\end{array}\)Median is half of total frequency = 50th term 4 falls in the range = 4 |
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32. |
On a square paper of length 2.524375cm is inscribed square diagram of length 0.524375cm. Find the area of the paper not covered by the diagram. correct to 3 significant figures. A. 6.00cm2 B. 6.10cm2 C. 6cmv D. 6.09cm2 E. 4.00cm2 Detailed SolutionArea of the paper = area of square = L x B or S2where s = S x k Area of the paper = (2.524)2 area of the diagram = (0.524)2 area not covered = (2.524)2 - (0.524)2 = 6.370576 - 0.274576 = 6.096 = 6.10cm2 (2 s.f) |
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33. |
Simplify \(\sqrt[3]{\frac{27a^{-9}}{8}}\). A. \(\frac{9a^2}{2a^3}\) B. \(\frac{3}{2a^3}\) C. \(\frac{2}{3a^2}\) D. \(\frac{3a^2}{2}\) Detailed Solution\(\sqrt[3]{\frac{27a ^{-9}}{8}}\) = \(\frac{3a^{-3}}{2}\)= \(\frac{3}{2}\) x \(\frac{1}{a^3}\) = \(\frac{3}{2a^3}\) |
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34. |
PQR is the diagram of a semicircle RSP with centre at Q and radius of length 3.5cm. If QPT = 60o. Find the perimeter of the figure PTRS. \(\pi\) = \(\frac{22}{7}\) A. 25cm B. 18cm C. 36cm D. 29cm E. 25.5cm Detailed SolutionCircumference of PRS = \(\frac{\pi}{2}\) = \(\frac{22}{7}\) x \(\frac{7}{1}\) x \(\frac{1}{2}\)= 11cm Side PT = 7cm, Side TR = 7cm Perimeter(PTRS) = 11cm + 7cm + 7cm = 25cm |
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35. |
In a triangle PQT, QR = \(\sqrt{3}cm\), PR = 3cm, PQ = \(2\sqrt{3}\)cm and PQR = 30o. Find angles P and R A. P = 60o and R = 90o B. P = 30o and R = 120o C. P = 90o and R = 60o D. P = 60o and R 60o E. P = 45o and R = 105o Detailed SolutionBy using cosine formula, p2 = Q2 + R2 - 2QR cos pCos P = \(\frac{Q^2 + R^2 - p^2}{2 QR}\) = \(\frac{(3)^2 + 2(\sqrt{3})^2 - 3^2}{2\sqrt{3}}\) = \(\frac{3 + 12 - 9}{12}\) = \(\frac{6}{12}\) = \(\frac{1}{2}\) = 0.5 Cos P = 0.5 p = cos-1 0.5 = 60o = < P = 60o If < P = 60o and < Q = 30 < R = 180o - 90o angle P = 60o and angle R is 90o |
36. |
Find the mean of the following 24.57, 25.63, 24.32, 26.01, 25.77 A. 25.12 B. 25.30 C. 25.26 D. 25.50 E. 25.75 Detailed Solution\(\frac{24.57 + 25.63 + 24.32 + 26.01 + 25.77}{5}\)mean = \(\frac{126.3}{5}\) = 25.26 |
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37. |
A man drove for 4 hours at a certain speed, he then doubled his speed and drove for another 3 hours. Although he covered 600 kilometers. At what speed did he drive for the last 3 hours? A. 120km/hr B. 60km/hr C. 670km/hr D. 40km/hr Detailed SolutionSpeed = \(\frac{distance}{time}\)let x represent the speed, d represent distance x = \(\frac{d}{4}\) d = 4x 2x = \(\frac{600 - d}{3}\) 6x = 600 - d 6x = 600 - 4x 10x = 600 x = \(\frac{600}{10}\) = 60km/hr |
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38. |
In a class of 60 pupils, the statistical distribution of the numbers of pupils offering Biology, History, French, Geography and Additional mathematics is as shown in the pie chart. How many pupils offer Additional Mathematics? A. 15 B. 10 C. 18 D. 12 E. 20 Detailed Solution(2x - 24)° + (3x - 18)° + (2x + 12)° + (x + 12)° + x° = 360°9x = 360° + 18° x = \(\frac{378}{9}\) = 42°, if x = 42°, then add maths = \(\frac{2x - 42}{360}\) x 60 = \(\frac{2 \times 42 - 24}{360}\) x 60 = \(\frac{84 - 24}{6}\) = 10 |
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39. |
In the figure, find PRQ A. 66\(\frac{1}{2}\)o B. 62\(\frac{1}{2}\)o C. 125o D. 105o E. 65o Detailed SolutionAngle subtended at any part of the circumference of the circle \(\frac{125^o}{2}\) at centre = 360o - 235o = 125o\(\bar{PQR}\) = \(\frac{125}{2}\) = 62\(\frac{1}{2}\)o |
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40. |
Which of the following equations represents the graph? A. y = 1 + 2x + 3x2 B. y = 1 - 2x + 3x2 C. y = 1 + 2x - 3x2 D. y = 1 - 2x - 3x2 E. y = 3x2 + 2x - 1 Detailed SolutionThe roots of the function are 1 and \(\frac{1}{3}\)sum of roots = -1 + \(\frac{1}{3}\) = -\(\frac{2}{3}\) product of roots = -1 x \(\frac{1}{3}\) = -\(\frac{1}{3}\) x2 - (sum of roots)x + (product of roots) = 0 x2 + (-\(\frac{2}{3}\))x - (-\(\frac{1}{3}\)) = 0 x2 + \(\frac{2x}{3}\) - \(\frac{1}{3}\) = 0 3x2 + 2x - 1 = 0 |