Year : 
1989
Title : 
Mathematics (Core)
Exam : 
JAMB Exam

Paper 1 | Objectives

41 - 45 of 45 Questions

# Question Ans
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 Solution

QMN = 60o

MRQ = 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
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 Solution

In \(\bigtriangleup\)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
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 Solution

Since QS = QR

then, 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
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 Solution

This 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 Solution

QMN = 60o

MRQ = 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
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 Solution

In \(\bigtriangleup\)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
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 Solution

Since QS = QR

then, 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
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 Solution

This 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\(\frac{3}{4}\)%