41 - 45 of 45 Questions
# | Question | Ans |
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41. |
MN is tangent to the given circle at M, MR and MQ are two chords. IF QNM is 60o and MNQ is 40o. Find RMQ A. 120o B. 110o C. 60o D. 20o Detailed SolutionMRQ = 60o(angle in the alternate segment are equal) MQN = 80o(angle sum of a \(\bigtriangleup\) = 180o) 60 = x = 80o(exterior angle = sum of opposite interior angles) x = 80o - 60o = 20o RMQ = 20o |
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42. |
In the diagram, HK is parallel to QR, PH = 4cm and HQ = 3cm. What is the ratio of KR:PR? A. 7:3 B. 3:7 C. 3:4 D. 4:3 Detailed SolutionIn \(\bigtriangleup\)PHK and \(\bigtriangleup\)PAR = \(\frac{PH}{PQ} = \frac{PK}{PR}\)\(\frac{4}{7} \times \frac{PR-KR}{PR}\) 4PR = 7(PR - KR) = 7PR - 7KR \(\frac{KR}{PR} = \frac{3}{7}\) KR:PR = 3:7 |
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43. |
If PQR is a straight line with QS = QR, calculate TPQ, If QT\\SR and TQS = 3yo A. 62o B. 56\(\frac{3}{2}\)o C. 20\(\frac{3}{2}\)o D. 18o Detailed SolutionSince QS = QRthen, angle SQR = angle SRQ 2 SQR = 180 - 56, SQR = \(\frac{124}{2}\) = 62o QTP = 62o QTP = 62o, corresponding angle 3y + 56 + 62 = 180 = 3y = 180 - 118 3y = 62 = 180 3y = 180 - 118 3y = 62 y = \(\frac{62}{3}\) = 20\(\frac{3}{2}\) |
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44. |
If PST is a straight line and PQ = QS = SR in the diagram, find y. A. 24o B. 48o C. 72o D. 84o Detailed Solution< PSQ = < SQR = < SRQ = 24o< QSR = 180o - 48o = 132o < PSQ + < QSR + y + 180 (angle on a straight lines) 24 + 132 + y = 180o = 156o + y = 180 y = 180o - 156o = 24o |
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45. |
In a class of 30 students, the marks scored in an examination are displayed in the histogram. What percentage of the student scored more than 40%? A. 14% B. 40% C. 40\(\frac{3}{4}\)% D. 53\(\frac{1}{3}\)% Detailed SolutionThis histogram is transferred into this frequency table\(\begin{array}{c|c} Marks & 20 & 40 & 60 & 80 & 100 \\ \hline students & 9 & 7 & 6 & 6 & 2\end{array}\) Students who scored more than 40 = 6 + 6 + 2 = 14 i.e. \(\frac{14}{30}\) x 100% = 46\(\frac{3}{4}\)% |
41. |
MN is tangent to the given circle at M, MR and MQ are two chords. IF QNM is 60o and MNQ is 40o. Find RMQ A. 120o B. 110o C. 60o D. 20o Detailed SolutionMRQ = 60o(angle in the alternate segment are equal) MQN = 80o(angle sum of a \(\bigtriangleup\) = 180o) 60 = x = 80o(exterior angle = sum of opposite interior angles) x = 80o - 60o = 20o RMQ = 20o |
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42. |
In the diagram, HK is parallel to QR, PH = 4cm and HQ = 3cm. What is the ratio of KR:PR? A. 7:3 B. 3:7 C. 3:4 D. 4:3 Detailed SolutionIn \(\bigtriangleup\)PHK and \(\bigtriangleup\)PAR = \(\frac{PH}{PQ} = \frac{PK}{PR}\)\(\frac{4}{7} \times \frac{PR-KR}{PR}\) 4PR = 7(PR - KR) = 7PR - 7KR \(\frac{KR}{PR} = \frac{3}{7}\) KR:PR = 3:7 |
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43. |
If PQR is a straight line with QS = QR, calculate TPQ, If QT\\SR and TQS = 3yo A. 62o B. 56\(\frac{3}{2}\)o C. 20\(\frac{3}{2}\)o D. 18o Detailed SolutionSince QS = QRthen, angle SQR = angle SRQ 2 SQR = 180 - 56, SQR = \(\frac{124}{2}\) = 62o QTP = 62o QTP = 62o, corresponding angle 3y + 56 + 62 = 180 = 3y = 180 - 118 3y = 62 = 180 3y = 180 - 118 3y = 62 y = \(\frac{62}{3}\) = 20\(\frac{3}{2}\) |
44. |
If PST is a straight line and PQ = QS = SR in the diagram, find y. A. 24o B. 48o C. 72o D. 84o Detailed Solution< PSQ = < SQR = < SRQ = 24o< QSR = 180o - 48o = 132o < PSQ + < QSR + y + 180 (angle on a straight lines) 24 + 132 + y = 180o = 156o + y = 180 y = 180o - 156o = 24o |
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45. |
In a class of 30 students, the marks scored in an examination are displayed in the histogram. What percentage of the student scored more than 40%? A. 14% B. 40% C. 40\(\frac{3}{4}\)% D. 53\(\frac{1}{3}\)% Detailed SolutionThis histogram is transferred into this frequency table\(\begin{array}{c|c} Marks & 20 & 40 & 60 & 80 & 100 \\ \hline students & 9 & 7 & 6 & 6 & 2\end{array}\) Students who scored more than 40 = 6 + 6 + 2 = 14 i.e. \(\frac{14}{30}\) x 100% = 46\(\frac{3}{4}\)% |