41 - 49 of 49 Questions
# | Question | Ans |
---|---|---|
41. |
If x and y represent the mean and the median respectively of the following set of numbers 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, find the \(\frac{x}{y}\) correct to one decimal place A. 18 B. 15 C. 1.0 D. 5 Detailed SolutionMean \(\bar{x}\) = \(\frac{156}{10}\) = 15.6Median = \(\bar{y}\) = \(\frac{15 + 16}{2}\) \(\frac{31}{2}\) = 15.5 \(\frac{x}{y}\) = \(\frac{15.6}{15.5}\) = 1.0065 1.0(1 d.p) |
|
42. |
If two dice are thrown together, what is the probability of obtaining at least a score of 10? A. \(\frac{1}{6}\) B. \(\frac{1}{12}\) C. \(\frac{5}{6}\) D. \(\frac{11}{12}\) Detailed SolutionThe total sample space when two dice are thrown together is 6 x 6 = 36\(\begin{array}{c|c} & 1 & 2 & 3 & 4 & 5 & 6\\ 1. & 1.1 & 1.2 & 1.3 & 1.4 & 1.5 & 1.6\\ 2 & 2.1 & 2.2 & 2.3 & 2.4 & 2.5 & 2.6\\3 & 3.1 & 3.2 & 3.3 & 3.4 & 3.5 & 3.6\\ 4 & 4.1 & 4.2 & 4.3 & 4.4 & 4.5 & 4.6\\ 5 & 5.1 & 5.2 & 5.3 & 5.4 & 5.5 & 5.6\\6 & 6.1 & 6.2 & 6.3 & 6.4 & 6.5 & 6.6 \end{array}\) At least 10 means 10 and above P(at least 10) = \(\frac{6}{36}\) = \(\frac{1}{6}\) |
|
43. |
In the figure, PQRS is a circle. If chords QR and RS are equal, calculate the value of x A. 80o B. 60o C. 45o D. 40o Detailed SolutionSRT is a straight line, where QRT = 120SRQ = 180o - 120o = 60o - (angle on a straight line) also angle QRS = 180o - 100o (angle on a straight line) . In angles where QR = SR and angle SRQ = 60o x = 100 - 60 = 40o |
|
44. |
In the figure, PQ is a parallel to ST and QRS = 40o. Find the value of x A. 55o B. 60o C. 65o D. 75o Detailed SolutionFrom the figure, 3x + x - 40o = 180o4x = 180o + 40o 4x = 220o x = \(\frac{220}{4}\) = 55o |
|
45. |
In the figure, PS = 7cm and RY = 9cm. IF the area of parallelogram PQRS is 56cm2. Find the area of trapezium PQTS A. 56cm2 B. 112cm2 C. 120cm2 D. 176cm2 Detailed SolutionFrom the figure, PS = QR = YT = 7cmArea of parallelogram PQRS = 56cm 56 = base x height, where base = 7 7 x h = 56cm, h = \(\frac{56}{7}\) = 8cm Area of trapezium \(\frac{1}{2}\) (sum of two sides)x height where two sides are QT and PS but QT = QR + RY + YT = 7 +9 + 7 = 23cm Area of trapezium PQTS = \(\frac{1}{2}\)(23 + 7) x 8 \(\frac{1}{2}\) x 30 x 8 = 120cmsq |
|
46. |
In the figure, STQ = SRP, PT = TQ = 6cm and QS = 5cm. Find SR A. \(\frac{47}{5}\) B. 5 C. \(\frac{32}{5}\) D. \(\frac{22}{5}\) Detailed SolutionFrom similar triangle, \(\frac{QS}{QP} = \frac{TQ}{QR} = \frac{5}{12} = \frac{6}{QR}\)QT = \(\frac{6 \times 12}{5} = \frac{72}{5} = SR = QR - QS\) = \(\frac{72}{5} - 5 = \frac{72 - 25}{5}\) = \(\frac{47}{5}\) |
|
47. |
In the figure, PS = RS = QS and QRS = 50°. Find QPR A. 25O B. 40O C. 50O D. 65O Detailed SolutionIn the figure PS = RS = QS, they will have equal base QR = RPIn angle SQR, angle S = 50O In angle QRP, 65 + 65 = 130O Since RQP = angle RPQ = \(\frac{180 - 130}{2}\) = \(\frac{50}{2} = 25^o\) QPR = 25° |
|
48. |
In the figure, XR and YQ are tangents to the circle YZXP if ZXR = 45° and YZX = 55°, Find ZYQ A. 135O B. 125O C. 100O D. 90O Detailed Solution< RXZ = < ZYX = 45O(Alternate segment)< ZYQ = 90 + 45 = 135° |
|
49. |
In the figure, a solid consists of a hemisphere surmounted by a right circular cone, with radius 3.0cm and height 6.0cm. Find the volume of the solid A. 36\(\pi\)cm 3 B. 54\(\pi\)cm 2 C. 18\(\pi\)cm 2 D. 108\(\pi\)cm 2 Detailed SolutionThe volume of the solid = vol. of cone + vol. of hemispherevolume of cone = \(\frac{1}{3} \pi r^2 h\) = \(\frac{1 \pi}{3} \times (3)^2 x 6 = 18 \pi cm^2\) vol. of hemisphere = \(\frac{4 \pi r^3}{6} = \frac{2 \pi r^3}{3}\) = \(\frac{2 \pi}{3} \times (3)^3 = 18\pi cm^3\) vol. of solid = 18\(\pi\) + 18\(\pi\) = 36\(\pi\)cm3 |
41. |
If x and y represent the mean and the median respectively of the following set of numbers 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, find the \(\frac{x}{y}\) correct to one decimal place A. 18 B. 15 C. 1.0 D. 5 Detailed SolutionMean \(\bar{x}\) = \(\frac{156}{10}\) = 15.6Median = \(\bar{y}\) = \(\frac{15 + 16}{2}\) \(\frac{31}{2}\) = 15.5 \(\frac{x}{y}\) = \(\frac{15.6}{15.5}\) = 1.0065 1.0(1 d.p) |
|
42. |
If two dice are thrown together, what is the probability of obtaining at least a score of 10? A. \(\frac{1}{6}\) B. \(\frac{1}{12}\) C. \(\frac{5}{6}\) D. \(\frac{11}{12}\) Detailed SolutionThe total sample space when two dice are thrown together is 6 x 6 = 36\(\begin{array}{c|c} & 1 & 2 & 3 & 4 & 5 & 6\\ 1. & 1.1 & 1.2 & 1.3 & 1.4 & 1.5 & 1.6\\ 2 & 2.1 & 2.2 & 2.3 & 2.4 & 2.5 & 2.6\\3 & 3.1 & 3.2 & 3.3 & 3.4 & 3.5 & 3.6\\ 4 & 4.1 & 4.2 & 4.3 & 4.4 & 4.5 & 4.6\\ 5 & 5.1 & 5.2 & 5.3 & 5.4 & 5.5 & 5.6\\6 & 6.1 & 6.2 & 6.3 & 6.4 & 6.5 & 6.6 \end{array}\) At least 10 means 10 and above P(at least 10) = \(\frac{6}{36}\) = \(\frac{1}{6}\) |
|
43. |
In the figure, PQRS is a circle. If chords QR and RS are equal, calculate the value of x A. 80o B. 60o C. 45o D. 40o Detailed SolutionSRT is a straight line, where QRT = 120SRQ = 180o - 120o = 60o - (angle on a straight line) also angle QRS = 180o - 100o (angle on a straight line) . In angles where QR = SR and angle SRQ = 60o x = 100 - 60 = 40o |
|
44. |
In the figure, PQ is a parallel to ST and QRS = 40o. Find the value of x A. 55o B. 60o C. 65o D. 75o Detailed SolutionFrom the figure, 3x + x - 40o = 180o4x = 180o + 40o 4x = 220o x = \(\frac{220}{4}\) = 55o |
|
45. |
In the figure, PS = 7cm and RY = 9cm. IF the area of parallelogram PQRS is 56cm2. Find the area of trapezium PQTS A. 56cm2 B. 112cm2 C. 120cm2 D. 176cm2 Detailed SolutionFrom the figure, PS = QR = YT = 7cmArea of parallelogram PQRS = 56cm 56 = base x height, where base = 7 7 x h = 56cm, h = \(\frac{56}{7}\) = 8cm Area of trapezium \(\frac{1}{2}\) (sum of two sides)x height where two sides are QT and PS but QT = QR + RY + YT = 7 +9 + 7 = 23cm Area of trapezium PQTS = \(\frac{1}{2}\)(23 + 7) x 8 \(\frac{1}{2}\) x 30 x 8 = 120cmsq |
46. |
In the figure, STQ = SRP, PT = TQ = 6cm and QS = 5cm. Find SR A. \(\frac{47}{5}\) B. 5 C. \(\frac{32}{5}\) D. \(\frac{22}{5}\) Detailed SolutionFrom similar triangle, \(\frac{QS}{QP} = \frac{TQ}{QR} = \frac{5}{12} = \frac{6}{QR}\)QT = \(\frac{6 \times 12}{5} = \frac{72}{5} = SR = QR - QS\) = \(\frac{72}{5} - 5 = \frac{72 - 25}{5}\) = \(\frac{47}{5}\) |
|
47. |
In the figure, PS = RS = QS and QRS = 50°. Find QPR A. 25O B. 40O C. 50O D. 65O Detailed SolutionIn the figure PS = RS = QS, they will have equal base QR = RPIn angle SQR, angle S = 50O In angle QRP, 65 + 65 = 130O Since RQP = angle RPQ = \(\frac{180 - 130}{2}\) = \(\frac{50}{2} = 25^o\) QPR = 25° |
|
48. |
In the figure, XR and YQ are tangents to the circle YZXP if ZXR = 45° and YZX = 55°, Find ZYQ A. 135O B. 125O C. 100O D. 90O Detailed Solution< RXZ = < ZYX = 45O(Alternate segment)< ZYQ = 90 + 45 = 135° |
|
49. |
In the figure, a solid consists of a hemisphere surmounted by a right circular cone, with radius 3.0cm and height 6.0cm. Find the volume of the solid A. 36\(\pi\)cm 3 B. 54\(\pi\)cm 2 C. 18\(\pi\)cm 2 D. 108\(\pi\)cm 2 Detailed SolutionThe volume of the solid = vol. of cone + vol. of hemispherevolume of cone = \(\frac{1}{3} \pi r^2 h\) = \(\frac{1 \pi}{3} \times (3)^2 x 6 = 18 \pi cm^2\) vol. of hemisphere = \(\frac{4 \pi r^3}{6} = \frac{2 \pi r^3}{3}\) = \(\frac{2 \pi}{3} \times (3)^3 = 18\pi cm^3\) vol. of solid = 18\(\pi\) + 18\(\pi\) = 36\(\pi\)cm3 |