JAMB Mathematics Past Questions & Answers

201.

If $\sin x = \frac{4}{5}$ , find $\frac{1 + \cot^2 x}{\csc^2 x - 1}$ .

$\frac{13}{2}$

$\frac{25}{9}$

$\frac{3}{13}$

$\frac{4}{11}$

View answer & explanation

Correct answer is B

$\sin x = \frac{opp}{Hyp} = \frac{4}{5}$

5 $^2$ = 4 $^2$ + adj $^2$

adj $^2$ = 25 - 16 = 9

adj = $\sqrt{9}$ = 3

$\tan x = \frac{4}{3}$

$\cot x = \frac{1}{\frac{4}{3}} = \frac{3}{4}$

$\cot^2 x = (\frac{3}{4})^2 = \frac{9}{16}$

$\csc x = \frac{1}{\sin x}$

= $\frac{1}{\frac{4}{5}} = \frac{5}{4}$

$\csc^2 x = (\frac{5}{4})^2 = \frac{25}{16}$

$\therefore \frac{1 + \cot^2 x}{\csc^2 x - 1} = \frac{1 + \frac{9}{16}}{\frac{25}{16} - 1}$

= $\frac{25}{16} \div \frac{9}{16}$

= $\frac{25}{9}$

202.

Calculate the volume of the regular three dimensional figure drawn above, where < ABC = 90° (a right- angled triangle).

394 cm $^3$

425 cm $^3$

268 cm $^3$

540 cm $^3$

View answer & explanation

Correct answer is D

|AC| = |DF| = 13 cm

Using Pythagoras theorem,

|AC| $^2$ = |AB| $^2$ + |BC| $^2$

13 $^2$ = 12 $^2$ + |BC| $^2$

|BC| $^2$ = 169 - 144 = 25

|BC| = $\sqrt{25}$

= 5 cm

Volume of triangular prism = $\frac{1}{2} \times base \times length \times height$

= $\frac{1}{2} \times 5 \times 12 \times 18$

= 540 cm $^3$

203.

In the circle above, with centre O and radius 7 cm. Find the length of the arc AB, when < AOB = 57°

5.32 cm

4.39 cm

7.33 cm

6.97 cm

View answer & explanation

Correct answer is D

Length of arc = $\frac{\theta}{360°} \times 2 \pi r$

= $\frac{57}{360} \times 2 \times \frac{22}{7} \times 7$

= 6.97 cm

204.

The histogram above represents the number of candidates who did Further Mathematics examination in a school. How many candidates scored more than 40?

100

150

View answer & explanation

Correct answer is C

Number of students that scored above 40 = 55 + 45 + 30 + 15 + 5 = 150 students.

205.

Marks	1	2	3	4	5
Frequency	2y - 2	y - 1	3y - 4	3 - y	6 - 2y

The table above is the distribution of data with mean equals to 3. Find the value of y.

View answer & explanation

Correct answer is B

Marks (x)	1	2	3	4	5
Frequency (f)	2y - 2	y - 1	3y - 4	3 - y	6 - 2y	3y + 2
fx	2y - 2	2y - 2	9y - 12	12 - 4y	30 - 10y	26 - y

Mean = $\frac{\sum fx}{\sum f}$

$3 = \frac{26 - y}{3y + 2}$

$3(3y + 2) = 26 - y$

$9y + 6 = 26 - y$

$9y + y = 26 - 6$

$10y = 20 \implies y = 2$

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