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.
18
20
30
38
Correct answer is D
Let n(M \(\cup\) N \) = x
Then 20 - x + x + 30
- x = n(M \(\cup\) N)
50 - x = 40
50 - 40 = x
10 = x
x = 10
Hence, n(M \(\cup\N)' = 8 + (20 - 10) + (30 + 10)
= 8 + 10 + 20
= 38
42cm
48cm
52cm
60cm
Correct answer is C
Let the length of a side of the rhombus be n
Then, n\(^2\) = 5\(^2\) + 12\(^2\)
= 25 + 144 = 169
n = \(\sqrt{169}\)
= 13cm
Hence, perimeter of rhombus = 4n = 4 x 13
= 52cm
Given that Y is 20cm on a bearing of 300\(^o\) from x, how far south of y is x?
10cm
15cm
25cm
30cm
Correct answer is A
In \(\bigtriangleup\)YSC, sin 30\(^o\) = \(\frac{YS}{20}\)
|YS| = 20 sin 30\(^o\)
= 20 x 0.5
10m
If tan x = \(\frac{4}{3}\), 0\(^o\) < x < 90\(^o\), find the value of sin x - cos x
\(\frac{1}{10}\)
\(\frac{1}{5}\)
\(\frac{5}{12}\)
1\(\frac{2}{5}\)
Correct answer is B
From the diagram,
h\(^2\) = 4\(^2\) + 3\(^2\) (pythagoras')
h\(^2\) = 16 + 9 = 25
h = \(\sqrt{25}\) = 5
Hence, sin x - cos x
= \(\frac{4}{5} - \frac{3}{5}\)
= \(\frac{1}{5}\)
170\(^o\)
192\(^o\)
177\(^o\)
182\(^o\)
Correct answer is B
Length of arc, L = 21.4 - 2 x 4.2cm
= 21.4 - 8.4
= 13cm
But L = \(\frac{\theta}{360^o}\) x 2\(\pi r\)
i.e 13 = \(\frac{\theta}{360^o}\) x 2 x \(\frac{22}{7}\) x 4.2
= 13 x 360\(^o\) x 7
= \(\theta\) x 2 x 22 x 4.2
\(\theta\) = \(\frac{13 \times 360^o \times 7}{44 \times 4.2}\)
= \(\approx\) 177.27\(^o\)
\(\approx\) 177\(^o\) (to the nearest degree)