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.

826.

The population of students in a school is 810. If this is represented on a pie chart, calculate the sectoral angle for a class of 72 students

A.

32o

B.

45o

C.

60o

D.

75o

Correct answer is A

In a school with students' population 810, the sectoral angle for a class of 72 students is

= \(\frac{72}{810}\) x 360º

= 32º

827.

The angle of elevation of an aircraft from a point K on the horizontal ground 30\(\alpha\). If the aircraft is 800m above the ground, how far is it from K?

A.

400.00m

B.

692.82m

C.

923.76m

D.

1,600.99m

Correct answer is D

In \(\bigtriangleup\)KBC, sin 30 = \(\frac{800}{IKCI}\)

IKCL = \(\frac{800}{sin30^o}\)

= \(\frac{800}{0.5}\)

= 1600m

828.

PQRT is square. If x is the midpoint of PQ, Calculate correct to the nearest degree, LPXS

A.

53o

B.

55o

C.

63o

D.

65o

Correct answer is C

In the diagram given,

tan\(\alpha\) = \(\frac{1}{0.5}\) = 2

\(\alpha\) = tan - 1(2) = 63.43o

= 63o

829.

Given that tan x = \(\frac{2}{3}\), where 0o d" x d" 90o, Find the value of 2sinx.

A.

\(\frac{2\sqrt{13}}{13}\)

B.

\(\frac{3\sqrt{13}}{13}\)

C.

\(\frac{4\sqrt{13}}{13}\)

D.

\(\frac{6\sqrt{13}}{13}\)

Correct answer is C

tan x = \(\frac{2}{3}\)(given), is illustrated in a right-angled \(\Delta\)

thus m2 = 22 + 32

= 4 + 9 = 13

m = \(\sqrt{13}\)

Hence, 2sin x = 2 x \(\frac{2}{m}\)

2 x\(\frac{2}{\sqrt{13}}\)

= \(\frac{4}{\sqrt{13}}\)

= \(\frac{4}{\sqrt{13}} = \frac{\sqrt{13}}{\sqrt{13}}\)

= \(\frac{4\sqrt{13}}{13}\)

830.

\(\begin{array}{c|c}
x & 0 & 1\frac{1}{4} & 2 & 4\\
\hline
y & 3 & 5\frac{1}{2} & &
\end{array}\)
The table given shows some values for a linear graph. Find the gradient of the line

A.

1

B.

2

C.

3

D.

4

Correct answer is B

Ler: (x1, y1) = (0, 3)

(x2, y2) = (\(\frac{5}{4}, \frac{11}{2}\))

Using gradient, m = \(\frac{y_2 - y_2}{x_2 - x_1}\)

= \(\frac{\frac{11}{2} - 3}{\frac{5}{4} - 0}\)

= \(\frac{11 - 6}{2} + \frac{5}{4}\)

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

= \(\frac{5}{2} \times \frac{4}{5}\)

= 2