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.
Correct 241.34(3 x 10\(^{-3}\))\(^2\) to 4 significant figures
0.0014
0.001448
0.0022
0.002172
Correct answer is D
first work out the expression and then correct the answer to 4 s.f = 241.34..............(A)
(3 x 10-\(^3\))\(^2\)............(B)
= 3\(^2\)x\(^2\)
= \(\frac{1}{10^3}\) x \(\frac{1}{10^3}\)
(Note that x\(^2\) = \(\frac{1}{x^3}\))
= 24.34 x 3\(^2\) x \(\frac{1}{10^6}\)
= \(\frac{2172.06}{10^6}\)
= 0.00217206
= 0.002172(4 s.f)
Find the derivative of the function y = 2x\(^2\)(2x - 1) at the point x = -1?
18
16
-4
-6
Correct answer is B
y = 2x\(^2\)(2x - 1)
y = 4x\(^3\) - 2x\(^2\)
dy/dx = 12x\(^2\) - 4x
at x = -1
dy/dx = 12(-1)\(^2\) - 4(-1)
= 12 + 4
= 16
\(\frac{-2}{7}\)
\(\frac{7}{6}\)
\(\frac{-6}{7}\)
2
Correct answer is D
Line: 2y+8x-17=0
recall y = mx + c
2y = -8x + 17
y = -4x + \(\frac{17}{2}\)
Slope m\(_1\) = 4
parallel lines: m\(_1\). m\(_2\) = -4
where Slope ( -4) = \(\frac{y_2 - y_1}{x_2 - x_1}\) at points (-1, -p) and (-2,2)
-4( \(x_2 - x_1\) ) = \(y_2 - y_1\)
-4 ( -2 - -1) = 2 - -p
p = 4 - 2 = 2
13 cm
4 cm
6 cm
7 cm
Correct answer is A
Area of Trapezium = 1/2(sum of parallel sides) * h
91 = \(\frac{1}{2}\) (5 + 9)h
cross multiply
91 = 7h
h = \(\frac{91}{7}\)
h = 13cm
Determine the maximum value of y=3x\(^2\) + 5x - 3
6
0
2
No correct option
Correct answer is D
y=3x\(^2\) + 5x - 3
dy/dx = 6x + 5
as dy/dx = 0
6x + 5 = 0
x = \(\frac{-5}{6}\)
Maximum value: 3 \( ^2{\frac{-5}{6}}\) + 5 \(\frac{-5}{6}\) - 3
3 \(\frac{75}{36}\) - \(\frac{25}{6}\) - 3
Using the L.C.M. 36
= \(\frac{25 - 50 - 36}{36}\)
= \(\frac{-61}{36}\)
No correct option