21 - 30 of 49 Questions
# | Question | Ans |
---|---|---|
21. |
\(\begin{array}{c|c} A. 13 B. 14 C. 15 D. 16 Detailed SolutionMedian = \(\frac{N}{2}\)th age= \(\frac{50}{2}\)th age = 25th age = 14 years |
|
22. |
In the diagram, \(\bar{YW}\) is a tangent to the circle at X, |UV| = |VX| and < VXW = 50o. Find the value of < UXY. A. 70o B. 80o C. 105o D. 110o Detailed SolutionIn the diagram above, x1 = 50(angles in alternate segment)x1 = x2(base angles of isos. \(\Delta\)) < UXY + x2 + 50o = 180o(sum of angles on a straight line) < UXY + 50o + 50o = 180p < UXY + 100o = 180o < UXY = 180o - 100o = 80o |
|
23. |
In the diagram, \(\bar{PF}\), \(\bar{QT}\), \(\bar{RG}\) intersect at S and PG||RG. If < SPQ = 113o and < RSt = 220, find < PSQ A. 22o B. 45o C. 67o D. 89o Detailed SolutionIn In the diagram given, \(\alpha\) = 22o (vertically opp. angles), \(\alpha\) = \(\beta\) (alternate angles)< PSQ + 133o + \(\beta\) = 180o (sum of angles of a \(\Delta\)) < PSQ + 133o + 22o = 180o < PSQ = 180o - (133 + 22)o 45o |
|
24. |
In the diagram, O is the centre of the circle, < XOZ = (10cm)o and < XWZ = mo. Calculate the value of m. A. 30 B. 36 C. 40 D. 72 Detailed Solution\(\alpha\) + 10mo = 360o (angle at circumference) \(\alpha\) + 10mo = 360o(angles round a point) 2mo + 10mo = 360o 12mo = 360o |
|
25. |
Kweku walked 8m up to slope and was 3m above the ground. If he walks 12m further up the slope, how far above the ground will he be? A. 4.5m B. 6.0m C. 7.5m D. 9.0m Detailed Solution\(\frac{8}{3} = \frac{20}{h}\) h = \(\frac{3 \times 20}{8}\) = 7.5m |
|
26. |
In the diagram, TS is a tangent to the circle at S. |PR| and < PQR = 177o. Calculate < PST. A. 54o B. 44o C. 34o D. 27o Detailed Solutionx1 = x2 (base angles of isos. \(\Delta\)) x1 + x2 + \(\alpha\) = 180o (sum of angles of a \(\Delta\) 63o + 63o + \(\alpha\) = 180o< |
|
27. |
In the diagram, PR||SV||WY|, TX||QY|, < PQT = 48o and < TXW = 60o.Find < TQU. A. 120o B. 108o C. 72o D. 60o Detailed SolutionIn the diagram, < TUQ + 60o(corresp. angles)< QTU = 48o (alternate angles) < QU + 60o + 48o = 180o(sum of angles of a \(\Delta\)) < TQU = 180o - 108o = 72o |
|
28. |
In the diagram, TX is perpendicular to UW, |UX| = 1cm and |TX| = |WX| = \(\sqrt{3}\)cm. Find UTW A. 135o B. 105o C. 75o D. 60o Detailed SolutionIn \(\Delta\) UXT, tan\(\alpha\) = \(\frac{1}{\sqrt{3}}\)\(\alpha\) = tan-1(\(\frac{1}{\sqrt{3}}\)) = 30o In \(\Delta\)WXT, tan\(\beta\) \(\frac{\sqrt{3}}{\sqrt{3}}\) = 1 \(\beta\) = tan-1(1) = 45o Hence, < UTW = \(\alpha\) + \(\beta\) = 30o + 45o = 75o |
|
29. |
The figure is a pie chart which represents the expenditure of a family in a year. If the total income of the family was Le 10,800,000.00, how much was spent on food? A. Le 2,250,000.00 B. Le 22,700,000.00 C. Le 3,600,000.00 D. Le 4,500,000.00 Detailed Solutionsectoral angle representing food= 360o - (80 + 70 + 90)o = 120o Amount spent on food = \(\frac{\tect{sectoral angle}}{360^o}\) x Le 10,800,000 = Le 3,600,000 |
|
30. |
A fair die is thrown two times. What is the probability that the sum of the scores is at least 10? A. \(\frac{5}{36}\) B. \(\frac{1}{6}\) C. \(\frac{5}{18}\) D. \(\frac{2}{3}\) Detailed Solution\(\begin{array}{c|c}& 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & 1,1 & 1,2 & 1,3 & 1,4 & 1,5 & 1,6 \\ \hline 2 & 2,1 & 2,2 & 2,3 & 2,4 & 2,5 & 2,6 \\ \hline 3 & 3,1 & 3,2 & 3,3 & 3,4 & 3,5 & 3,6 \\ \hline 4 & 4,1 & 4,2 & 4,3 & 4,4 & 4,5 & 4,6 \\ 5 & 5,1 & 5,2 & 5,3 & 5,4 & 5,5 & 5,6 \\ \hline 6 & 6,1 & 6,2 & 6,3 & 6,4 & 6,5 & 6,6\end{array}\) From the table |
21. |
\(\begin{array}{c|c} A. 13 B. 14 C. 15 D. 16 Detailed SolutionMedian = \(\frac{N}{2}\)th age= \(\frac{50}{2}\)th age = 25th age = 14 years |
|
22. |
In the diagram, \(\bar{YW}\) is a tangent to the circle at X, |UV| = |VX| and < VXW = 50o. Find the value of < UXY. A. 70o B. 80o C. 105o D. 110o Detailed SolutionIn the diagram above, x1 = 50(angles in alternate segment)x1 = x2(base angles of isos. \(\Delta\)) < UXY + x2 + 50o = 180o(sum of angles on a straight line) < UXY + 50o + 50o = 180p < UXY + 100o = 180o < UXY = 180o - 100o = 80o |
|
23. |
In the diagram, \(\bar{PF}\), \(\bar{QT}\), \(\bar{RG}\) intersect at S and PG||RG. If < SPQ = 113o and < RSt = 220, find < PSQ A. 22o B. 45o C. 67o D. 89o Detailed SolutionIn In the diagram given, \(\alpha\) = 22o (vertically opp. angles), \(\alpha\) = \(\beta\) (alternate angles)< PSQ + 133o + \(\beta\) = 180o (sum of angles of a \(\Delta\)) < PSQ + 133o + 22o = 180o < PSQ = 180o - (133 + 22)o 45o |
|
24. |
In the diagram, O is the centre of the circle, < XOZ = (10cm)o and < XWZ = mo. Calculate the value of m. A. 30 B. 36 C. 40 D. 72 Detailed Solution\(\alpha\) + 10mo = 360o (angle at circumference) \(\alpha\) + 10mo = 360o(angles round a point) 2mo + 10mo = 360o 12mo = 360o |
|
25. |
Kweku walked 8m up to slope and was 3m above the ground. If he walks 12m further up the slope, how far above the ground will he be? A. 4.5m B. 6.0m C. 7.5m D. 9.0m Detailed Solution\(\frac{8}{3} = \frac{20}{h}\) h = \(\frac{3 \times 20}{8}\) = 7.5m |
26. |
In the diagram, TS is a tangent to the circle at S. |PR| and < PQR = 177o. Calculate < PST. A. 54o B. 44o C. 34o D. 27o Detailed Solutionx1 = x2 (base angles of isos. \(\Delta\)) x1 + x2 + \(\alpha\) = 180o (sum of angles of a \(\Delta\) 63o + 63o + \(\alpha\) = 180o< |
|
27. |
In the diagram, PR||SV||WY|, TX||QY|, < PQT = 48o and < TXW = 60o.Find < TQU. A. 120o B. 108o C. 72o D. 60o Detailed SolutionIn the diagram, < TUQ + 60o(corresp. angles)< QTU = 48o (alternate angles) < QU + 60o + 48o = 180o(sum of angles of a \(\Delta\)) < TQU = 180o - 108o = 72o |
|
28. |
In the diagram, TX is perpendicular to UW, |UX| = 1cm and |TX| = |WX| = \(\sqrt{3}\)cm. Find UTW A. 135o B. 105o C. 75o D. 60o Detailed SolutionIn \(\Delta\) UXT, tan\(\alpha\) = \(\frac{1}{\sqrt{3}}\)\(\alpha\) = tan-1(\(\frac{1}{\sqrt{3}}\)) = 30o In \(\Delta\)WXT, tan\(\beta\) \(\frac{\sqrt{3}}{\sqrt{3}}\) = 1 \(\beta\) = tan-1(1) = 45o Hence, < UTW = \(\alpha\) + \(\beta\) = 30o + 45o = 75o |
|
29. |
The figure is a pie chart which represents the expenditure of a family in a year. If the total income of the family was Le 10,800,000.00, how much was spent on food? A. Le 2,250,000.00 B. Le 22,700,000.00 C. Le 3,600,000.00 D. Le 4,500,000.00 Detailed Solutionsectoral angle representing food= 360o - (80 + 70 + 90)o = 120o Amount spent on food = \(\frac{\tect{sectoral angle}}{360^o}\) x Le 10,800,000 = Le 3,600,000 |
|
30. |
A fair die is thrown two times. What is the probability that the sum of the scores is at least 10? A. \(\frac{5}{36}\) B. \(\frac{1}{6}\) C. \(\frac{5}{18}\) D. \(\frac{2}{3}\) Detailed Solution\(\begin{array}{c|c}& 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & 1,1 & 1,2 & 1,3 & 1,4 & 1,5 & 1,6 \\ \hline 2 & 2,1 & 2,2 & 2,3 & 2,4 & 2,5 & 2,6 \\ \hline 3 & 3,1 & 3,2 & 3,3 & 3,4 & 3,5 & 3,6 \\ \hline 4 & 4,1 & 4,2 & 4,3 & 4,4 & 4,5 & 4,6 \\ 5 & 5,1 & 5,2 & 5,3 & 5,4 & 5,5 & 5,6 \\ \hline 6 & 6,1 & 6,2 & 6,3 & 6,4 & 6,5 & 6,6\end{array}\) From the table |