21 - 30 of 48 Questions
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21. |
\(\begin{array}{c|c} \oplus mod 10 & 2 & 4 & 6 & 8 \\ \hline 2 & 4 & 8 & 2 & 6 \\4 & 8 & 6 & 4 & 2\\ 4 & 8 & 6 & 4 & 2\\ 6 & 2 & 4 & 6 & 8\\ 8 & 6 & 2 & 8 & 4\end{array}\) A. 2 B. 4 C. 6 D. 8 Detailed SolutionThe inverse of 2 is 6 since 2 x 6 = 12; under mod 1012 = 2 which is also the value required |
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22. |
Solve for x and y \(\begin{pmatrix} 1 & 1 \\ 3 & y \end{pmatrix}\)\(\begin{pmatrix} x \\ 1 \end{pmatrix}\) = \(\begin{pmatrix} 4 \\ 1\end{pmatrix}\) A. x = 3, y = 8 B. x = 3, y = -8 C. x = -8, y = 3 D. x = 8, y = -3 Detailed Solution\(\begin{pmatrix} 1 & 1 \\ 3 & y \end{pmatrix}\)\(\begin{pmatrix} x \\ 1 \end{pmatrix}\) = \(\begin{pmatrix} 4 \\ 1\end{pmatrix}\) = x + 1 = 4x = 4 - 1 = 3 3x + y =1 3(3) = y = 1 = 9 + y = 1 y = 1 - 9 = -8 |
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23. |
The determination of the matrix \(\begin{pmatrix} 1 & 3 & 3 \\ 4 & 5 & 6\\ 2 & 0 & 1 \end{pmatrix}\) is A. -67 B. -57 C. -3 D. 3 Detailed Solution4 x 2 = -8 upward arrows = +ve2 x 5 x 3 = 30 0 x 6 = \(\frac{0}{22}\) downward arrows = -ve -1(1 x 5x - 1) = 5 - (2 x 6 x 2) = -24 = -(4 x 0 x 1) = \(\frac{0}{-19}\) therefore 22 - 19 = 3 |
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24. |
\(\begin{array}{c|c} \text{Age in years} & 10 & 11 & 12 \\ \hline \text{Number of pupils} & 6 & 27 & 7\end{array}\) A. \(\frac{27}{40}\) B. \(\frac{17}{20}\) C. \(\frac{33}{40}\) D. \(\frac{3}{20}\) Detailed SolutionTotal number of pupils = 7 + 27 + 6 = 40n(at least 11 years old) = 27 + 7 = 34. P(picking a pupil who is at least 11 years old) = \(\frac{34}{40} = \frac{17}{20}\). |
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25. |
If three angles of a quadrilateral are (3y - x - z)o, 3xo, (2z - 2y - x)o find the fourth angle in terms of x, y and z A. (360 - x- y - z)o B. (360 + x + y - z)o C. (180 - x + y + z)o D. (180 - x + y + z)o Detailed SolutionThe sum of angles of a quadrilateral is 360o∴ (3y - x - z)o + 3xo + (2z - 2y - x)o + po = 360o Where P is the fourth angle 3y - x - z + 3x + 2z - 2y - x + p = 360o p = 360 - (x + y + z) ∴ p = (360 - x - y - z)o |
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26. |
An open rectangular box is made of wood 2cm thick. If the internal dimensions of the box are 50cm long, 36cm wide and 20cm deep, the box volume of wood in the box is A. 11 520cm3 B. 36 000cm3 C. 38 200cm3 D. 47 520cm3 Detailed SolutionInternal dimension are 50cm, 36cm and 20cminternal volume = 50 x 36 x 20cm3 1000 x 36cm3 = 36000cm3 External dimension are 54cm x 40cm x 22cm = 2160cm2 x 22cm = 47520cm3 Volume of wood = Ext. volume - Int. volume = 47,520cm3 - 36,000cm3 = 11,520cm3 |
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27. |
Calculate the perimeter, in cm, of a sector of a circle of radius 8cm and angle 45o A. 2\(\pi\) B. 8 + 2\(\pi\) C. 16 + 2\(\pi\) D. 16 + 16\(\pi\) Detailed SolutionPerimeter = OP + OQ + PQ= 8 + 8 + PQ length PQ = \(\frac{\theta}{360 \times 2\pi r}\) = \(\frac{45}{360}\) x 2 x \(\pi\) x 8 = 2\(\pi\) Perimeter of sector 2r + L Where l = length of arc and r = radius ∴ P = 2(8) + 2\(\pi\) = 16 + 2\(\pi\) |
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28. |
What is the locus of a point P which moves on one side of a straight line XY, so that the angle XPY is always equal to 90o? A. the perpendicular bisector of XY B. a right-angled triangle C. a circle D. a semi circle Detailed SolutionSince XY is a fixed line andXPY = 90o P is on one side of XY P1P2P3......Pn are all possible cases where XPY = 90o the only possible tendency is a semicircle because angles in semicircle equals 90o |
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29. |
If M(4, q) is the mid-point of the line joining L(p, -2) and N(q, p). Find the values of p and q A. P = 2, q = 4 B. p = 3, q = 1 C. p = 5, q = 3 D. p = 6, q = 2 Detailed Solution\(\frac{p + q}{2}\) = 4p + q = 8 ....(i) \(\frac{p - 2}{2}\) = q p - 2q 2 ....(ii) q = 2, p = 6 3q = 6 |
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30. |
The angle of depression of a boat from the top of a cliff 10m high is 30. How far is the boat from the foot of the cliff? A. \(\frac{5√3m}{3}\) B. 5√3m C. 10√3m D. \(\frac{10√3m}{3}\) Detailed SolutionRQP = 90oPQS = 30o but RQS = 90o = \(\frac{RS}{10}\) RS = 10 tan 60o RS = 10 tan 60o RS = 10√3m |
21. |
\(\begin{array}{c|c} \oplus mod 10 & 2 & 4 & 6 & 8 \\ \hline 2 & 4 & 8 & 2 & 6 \\4 & 8 & 6 & 4 & 2\\ 4 & 8 & 6 & 4 & 2\\ 6 & 2 & 4 & 6 & 8\\ 8 & 6 & 2 & 8 & 4\end{array}\) A. 2 B. 4 C. 6 D. 8 Detailed SolutionThe inverse of 2 is 6 since 2 x 6 = 12; under mod 1012 = 2 which is also the value required |
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22. |
Solve for x and y \(\begin{pmatrix} 1 & 1 \\ 3 & y \end{pmatrix}\)\(\begin{pmatrix} x \\ 1 \end{pmatrix}\) = \(\begin{pmatrix} 4 \\ 1\end{pmatrix}\) A. x = 3, y = 8 B. x = 3, y = -8 C. x = -8, y = 3 D. x = 8, y = -3 Detailed Solution\(\begin{pmatrix} 1 & 1 \\ 3 & y \end{pmatrix}\)\(\begin{pmatrix} x \\ 1 \end{pmatrix}\) = \(\begin{pmatrix} 4 \\ 1\end{pmatrix}\) = x + 1 = 4x = 4 - 1 = 3 3x + y =1 3(3) = y = 1 = 9 + y = 1 y = 1 - 9 = -8 |
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23. |
The determination of the matrix \(\begin{pmatrix} 1 & 3 & 3 \\ 4 & 5 & 6\\ 2 & 0 & 1 \end{pmatrix}\) is A. -67 B. -57 C. -3 D. 3 Detailed Solution4 x 2 = -8 upward arrows = +ve2 x 5 x 3 = 30 0 x 6 = \(\frac{0}{22}\) downward arrows = -ve -1(1 x 5x - 1) = 5 - (2 x 6 x 2) = -24 = -(4 x 0 x 1) = \(\frac{0}{-19}\) therefore 22 - 19 = 3 |
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24. |
\(\begin{array}{c|c} \text{Age in years} & 10 & 11 & 12 \\ \hline \text{Number of pupils} & 6 & 27 & 7\end{array}\) A. \(\frac{27}{40}\) B. \(\frac{17}{20}\) C. \(\frac{33}{40}\) D. \(\frac{3}{20}\) Detailed SolutionTotal number of pupils = 7 + 27 + 6 = 40n(at least 11 years old) = 27 + 7 = 34. P(picking a pupil who is at least 11 years old) = \(\frac{34}{40} = \frac{17}{20}\). |
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25. |
If three angles of a quadrilateral are (3y - x - z)o, 3xo, (2z - 2y - x)o find the fourth angle in terms of x, y and z A. (360 - x- y - z)o B. (360 + x + y - z)o C. (180 - x + y + z)o D. (180 - x + y + z)o Detailed SolutionThe sum of angles of a quadrilateral is 360o∴ (3y - x - z)o + 3xo + (2z - 2y - x)o + po = 360o Where P is the fourth angle 3y - x - z + 3x + 2z - 2y - x + p = 360o p = 360 - (x + y + z) ∴ p = (360 - x - y - z)o |
26. |
An open rectangular box is made of wood 2cm thick. If the internal dimensions of the box are 50cm long, 36cm wide and 20cm deep, the box volume of wood in the box is A. 11 520cm3 B. 36 000cm3 C. 38 200cm3 D. 47 520cm3 Detailed SolutionInternal dimension are 50cm, 36cm and 20cminternal volume = 50 x 36 x 20cm3 1000 x 36cm3 = 36000cm3 External dimension are 54cm x 40cm x 22cm = 2160cm2 x 22cm = 47520cm3 Volume of wood = Ext. volume - Int. volume = 47,520cm3 - 36,000cm3 = 11,520cm3 |
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27. |
Calculate the perimeter, in cm, of a sector of a circle of radius 8cm and angle 45o A. 2\(\pi\) B. 8 + 2\(\pi\) C. 16 + 2\(\pi\) D. 16 + 16\(\pi\) Detailed SolutionPerimeter = OP + OQ + PQ= 8 + 8 + PQ length PQ = \(\frac{\theta}{360 \times 2\pi r}\) = \(\frac{45}{360}\) x 2 x \(\pi\) x 8 = 2\(\pi\) Perimeter of sector 2r + L Where l = length of arc and r = radius ∴ P = 2(8) + 2\(\pi\) = 16 + 2\(\pi\) |
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28. |
What is the locus of a point P which moves on one side of a straight line XY, so that the angle XPY is always equal to 90o? A. the perpendicular bisector of XY B. a right-angled triangle C. a circle D. a semi circle Detailed SolutionSince XY is a fixed line andXPY = 90o P is on one side of XY P1P2P3......Pn are all possible cases where XPY = 90o the only possible tendency is a semicircle because angles in semicircle equals 90o |
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29. |
If M(4, q) is the mid-point of the line joining L(p, -2) and N(q, p). Find the values of p and q A. P = 2, q = 4 B. p = 3, q = 1 C. p = 5, q = 3 D. p = 6, q = 2 Detailed Solution\(\frac{p + q}{2}\) = 4p + q = 8 ....(i) \(\frac{p - 2}{2}\) = q p - 2q 2 ....(ii) q = 2, p = 6 3q = 6 |
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30. |
The angle of depression of a boat from the top of a cliff 10m high is 30. How far is the boat from the foot of the cliff? A. \(\frac{5√3m}{3}\) B. 5√3m C. 10√3m D. \(\frac{10√3m}{3}\) Detailed SolutionRQP = 90oPQS = 30o but RQS = 90o = \(\frac{RS}{10}\) RS = 10 tan 60o RS = 10 tan 60o RS = 10√3m |