JAMB Mathematics Past Questions & Answers - Page 220

1,096.

Find the probability of selecting a figure which is a parallelogram from a square, a rectangle, a rhombus, a kite and a trapezium

A.

\(\frac{3}{5}\)

B.

\(\frac{2}{5}\)

C.

\(\frac{4}{5}\)

D.

\(\frac{1}{5}\)

Correct answer is A

A rectangle, a square and a rhombus are the only parallelogram

∴ Probability of a parallelogram = \(\frac{3}{5}\)

1,098.

The numbers 3, 2, 8, 5, 7, 12, 9 and 14 are the marks scored by a group of students in a class test. If P is scored by an group of students in a class test. If P is the mean and Q the median, the P + Q is

A.

18

B.

17\(\frac{1}{2}\)

C.

16

D.

15

Correct answer is D

Re-arranging them 2, 3, 5, 7, 8, 9, 12, 14

Mean = P and median = Q

P = \(\frac{2 + 2 + 5 + 7 + 8 + 9 + 12 + 14}{8}\)

\(\frac{60}{8}\) = 7.5

median (Q) = \(\frac{7 + 8}{2}\)

\(\frac{15}{2}\) = 7.5

P + Q = 7.5 + 7.5 = 15

1,099.

The goals scored by 40 football teams from three league divisions are recorded below
\(\begin{array}{c|c} \text{Number of goals} & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline frequency & 4 & 3 & 15 & 16 & 1 & 0 & 1\end{array}\)
What is the total number of goals scored by all the terms?

A.

21

B.

40

C.

91

D.

96

Correct answer is C

Let x rep. number of goals
\(\begin{array}{c|c} "f & " & frequency\\\hline x & f & fx\\ 0 & 4 & 0\\1 & 3 & 3\\2 & 15 & 30\\3 & 16 & 48\\4 & 1 & 4\\5 & 0 & 0 \\ 6 & 1 & 6\end{array}\)

\(\sum fx\) = 91

Total number of goals scored is \(\sum fx\)= 91

1,100.

Taking the period of day light on a certain day to be from 5.30a.m to 7.00p.m. Calculate the angle of a pie chart designed to show the periods of the day light and of darkness on that day

A.

187o30'

B.

172o30'

C.

135o225o

D.

202o30' 157o 30'

Correct answer is D

Period of day light is from 5.30 to 7.00p.m - 13hrs 30min. interval

24hrs make 1 day

∴ period of darkness = 24hrs - 13hrs 30mins = 10hrs 30mins.

Angle of sector of daylight = \(\frac{13 \times 60 + 30 \times 360}{(24 \times 60)}\)

= \(\frac{810}{1440}\) x 360o

= \(\frac{291600}{1440}\)

= 202.5o

202.5o = 202.30o

darkness period = 360o - 202o 30'

= 157o30'

= 202o30' 157o 30'