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JAMB Mathematics Past Questions & Answers - Page 108

536.

From the diagram given, find the bearing of R from S

226^o

224^o

136^o

134^o

View answer & explanation

Correct answer is B

The bearing of R from S = 180 + 44 = 224º, because

-Following the clockwise movement at 90º:

-NORTH -EAST, EAST-SOUTH, SOUTH - WEST & WEST - NORTH

- It is observed that R is situated between SOUTH - WEST poles on S

537.

In the diagram above, O is the centre of the circle, If < UOT = 70^o and < RST = 100^o. Calculate < RUO

20^o

25^o

50^o

80^o

View answer & explanation

Correct answer is B

< OUT = < OTU

< OUT = $\frac{180^o - 70^o}{2} = \frac{110^o}{2} = 55^o$

< RUT = 180^o - 100^o = 80^o

< RUO = < RUT - < OUT

= 80^o - 55^o

= 25^o

538.

in the diagram above, calculate the value of x.

60^o

100^o

120^o

140^o

View answer & explanation

Correct answer is C

No explanation has been provided for this answer.

539.

The diagram above is the graph of the function f(x). Determine the range of values of x for which f(x) $\geq$ 0.

x $\geq$ 2

0 $\geq$ x $\geq$ 2

-2 $\geq$ x $\geq$ 0, x $\leq$ 2

x $\geq$ -2, 0 $\geq$ x $\geq$ 2

View answer & explanation

Correct answer is C

No explanation has been provided for this answer.

540.

The graph above shows the cumulative frequency curve of the distribution of marks in a class test. What percentage of the students scored more than 20 marks?

68%

28%

17%

View answer & explanation

Correct answer is B

$\begin{array}{c|c} \text{class-marks(x)} & \text{(freq. (f)} & \text{cum. freq.}\\\hline 9 & &3.....3\\14 & & 6....3 + 6 = 9\\ 19 & & 9.....9 + 9 = 18\\ 24 & & 3.....18 + 3 = 21\\ 29 & & 2.....21 + 2 = 23\\ 34 & & 2.....23 + 2 = 25\\ & & 25\end{array}$
% of students scoring more than 20 marks

= $\frac{7}{25}$ x 100% = 28%