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.
If the exterior angles of a pentagon are x°, (x + 5)°, (x + 10)°, (x + 15)° and (x + 20)°, find x
118o
72o
62o
36o
Correct answer is C
The sum of the exterior angles of a regular polygon = 360°.
\(\therefore x + (x + 5) + (x + 10) + (x + 15) + (x + 20) = 360°\)
\(5x + 50 = 360° \implies 5x = 360° - 50° = 310°\)
\(x = \frac{310°}{5} = 62°\)
8\(\sqrt{3}\)
\(\frac{16}{\sqrt{3}}\)
\(\sqrt{3}\)
\(\frac{10}{\sqrt{3}}\)
Correct answer is A
\(\frac{x}{4}\) = \(\frac{\tan 60}{1}\)
x = 4 tan60
= 4\(\sqrt{3}\)
BD = 2x
= 8\(\sqrt{3}\)
The area of a square is 144 sq cm. Find the length of its diagonal
11\(\sqrt{3cm}\)
12cm
12\(\sqrt{2cm}\)
13cm
Correct answer is C
BD = \(\sqrt{x^2 + x^2}\)
= \(\sqrt{12^2 + 12^2}\)
= \(\sqrt{144 + 144}\)
= 2(144)
= 12\(\sqrt{2cm}\)
2480
1240
620
124
Correct answer is B
Given the first and last term of an A.P, the sum of the terms is given by
\(S_{n} = \frac{n}{2} [a + l]\)
where a = first term; l = last term and n = number of terms.
\(\therefore S_{20} = \frac{20}{2} [7 + 117]\)
= \(10 (124)\)
= 1240
At what value of x is the function y = x2 - 2x - 3 minimum?
1
-1
-4
4
Correct answer is A
For y = ax2 - bx + c for minimum y
\(\frac{dy}{dx}\) = 2x - 2
= \(\frac{dy}{dx}\) = 0
∴ 2x - 2 = 0
x = 1