JAMB Mathematics Past Questions & Answers - Page 118

586.

The bar chart shows the distribution of marks scored by 60 pupils in a test in which the maximum score was 10. If the pass mark was 5, what percentage of the pupils failed the test?

A.

59.4%

B.

50.0%

C.

41.7%

D.

25.0%

Correct answer is C

\(\begin{array}{c|c} x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline f & 1 & 9 & 4 & 7 & 10 & 8 & 7 & 9 & 8 & 2 & 1\end{array}\)
no pupils who failed the test = 1 + 3 + 4 + 7 + 10

= 25

5 of pupils who fail = \(\frac{25}{60}\) x 100%

= 41.70%

587.

The pie chart shows the monthly expenditure pf a public servant. The monthly expenditure on housing is twice that of school fees. How much does the worker spend on housing if his monthly income is N7200?

A.

N1000

B.

N2000

C.

N3000

D.

N4000

Correct answer is B

Let the monthly expenditure on school fees be x let the monthly expenditure on housing be 2x; angle of housing and school fee in a pie chart = 360o - (120o + 90o)

= 360o - 210o = 150o

Angle of housing in a pie-chart = \(\frac{2}{3}\) x 150 = 100

\(\frac{100}{360}\) x 7200 = N2,000

588.

A chord of a circle radius \(\sqrt{3cm}\) subtends an angle of 60° on the circumference of he circle. Find the length of the chord

A.

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

B.

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

C.

3

D.

\(\sqrt{3}\)cm

Correct answer is D

Length of chord = \(2r \times \sin(\frac{\theta}{2})\)

= \( 2 \times \sqrt{3} \times \sin(\frac{60}{2})\)

= \(2 \times \sqrt{3} \times \frac{1}{2}\)

= \(\sqrt{3}\) cm.

589.

In the diagram, QTR is a straight line and < PQT = 30o. find the sin of < PTR

A.

\(\frac{8}{15}\)

B.

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

C.

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

D.

\(\frac{15}{16}\)

Correct answer is C

\(\frac{10}{\sin 30^o} = \frac{15}{\sin x} = \frac{10}{0.5} = \frac{15}{\sin x}\)

\(\frac{15}{20} = \sin x\)

sin x = \(\frac{15}{20} = \frac{3}{4}\)

N.B x = < PRQ

590.

TQ is tangent to circle XYTR, < YXT = 32o, RTQ = 40o. find < YTR

A.

108o

B.

121o

C.

140o

D.

148o

Correct answer is A

< TWR = < QTR = 40o (alternate segment)

< TWR = < TXR = 40o(Angles in the same segments)
< YXR = 40o + 32o = 72o

< YXR + < YTR = 180o(Supplementary)

72o + < YTR = 180o

< YTR = 180o - 72o

= 108o