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 \(P = \sqrt{QR\left(1+\frac{3t}{R}\right)}\), make R the subject of the formula.
\(R = \frac{3Qt}{P^2 - Q}\)
\(R = \frac{P^2 – 3t}{Q+1}\)
\(R = \frac{P^2 + 3t}{Q - 1}\)
\(R = \frac{P^2-3Qt}{Q}\)
Correct answer is D
No explanation has been provided for this answer.
2√3
4√3
6√3
12√3
Correct answer is B
\(sin \theta = \frac{opp}{hyp}\\
sin 60^o = \frac{|YZ|}{|XZ|}=\frac{6}{P}\\
P sin 60^o = 6\\
P = \frac{6}{sin60^o}\\
=\frac{6}{\sqrt{\frac{3}{2}}}=4\sqrt{3}\)
I only
II only
III only
I and III only
Correct answer is B
Using the formula, \((n - 2) \times 180°\) to get the sum of the interior angles. Then we can have
\((n - 2) \times 180° = 108n\) ... (1)
\((n - 2) \times 180° = 116n\) ... (2)
\((n - 2) \times 180° = 120n\) ... (3)
Solving the above given equations, where n is not a positive integer then that angle is not the interior for a regular polygon.
(1): \(180n - 360 = 108n \implies 72n = 360\)
\(n = 5\) (regular pentagon)
(2): \(180n - 360 = 116n \implies 64n = 360\)
\(n = 5.625\)
(3): \(180n - 360 = 120n \implies 60n = 360\)
\(n = 6\) (regular hexagon)
Hence, 116° is not an angle of a regular polygon.
If \(\frac{3^{(1-n)}}{9^{-2n}}=\frac{1}{9}\) find n
\(-\frac{3}{2}\)
\(\frac{1}{3}\)
-1
-3
Correct answer is C
\(\frac{3^{(1-n)}}{9^{-2n}}=\frac{1}{9}\\
3^{1-n}\times 3^{-2(-2n)} = 3^{-2}\\
1-n-2(-2n)= -2\\
1-n+4n=-2\\
n=-1\)
0.01014
0.01021
0.01015
0.01016
Correct answer is A
option A CANNOT BE BECAUSE THE LAST NUMBER BEFORE 1 CAN ONLY BE ROUNDED DOWN TO ZERO.