Mathematics questions and answers

Mathematics Questions and Answers

How good are you with figures and formulas? Find out with these Mathematics past questions and answers. This Test is useful for both job aptitude test candidates and students preparing for JAMB, WAEC, NECO or Post UTME.

1,636.

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.

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

D.

53\(\frac{1}{3}\)%

Correct answer is C

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

1,637.

If PST is a straight line and PQ = QS = SR in the diagram, find y.

A.

24o

B.

48o

C.

72o

D.

84o

Correct answer is A

< 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

1,638.

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

Correct answer is C

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\(\frac{3}{2}\)

1,639.

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

Correct answer is B

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

1,640.

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

Correct answer is D

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