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.
Solve for x and y if x - y = 2 and x2 - y2 = 8
(-1, 3)
(3, 1)
(-3, 1)
(1, 3)
Correct answer is B
x - y = 2 ...........(1)
x2 - y2 = 8 ........... (2)
x - 2 = y ............ (3)
Put y = x -2 in (2)
x2 - (x - 2)2 = 8
x2 - (x2 - 4x + 4) = 8
x2 - x2 + 4x - 4 = 8
4x = 8 + 4 = 12
x = \(\frac{12}{4}\)
= 3
from (3), y = 3 - 2 = 1
therefore, x = 3, y = 1
If 9x2 + 6xy + 4y2 is a factor of 27x3 - 8y3, find the other factor.
2y + 3x
2y - 3x
3x + 2y
3x - 2y
Correct answer is D
27x3 - 8y3 = (3x - 2y)3
But 9x2 + 6xy + 4y2 = (3x +2y)2
So, 27x3 - 8y3 = (3x - 2y)(3x - 2y)2
Hence the other factor is 3x - 2y
Make Q the subject of formula if p = \(\frac{M}{5}\)(X + Q) + 1
\(\frac{5P - MX + 5}{M}\)
\(\frac{5P - MX - 5}{M}\)
\(\frac{5P + MX + 5}{M}\)
\(\frac{5P + MX - 5}{M}\)
Correct answer is B
p = \(\frac{M}{5}\)(X + Q) + 1
P - 1 = \(\frac{M}{5}\)(X + Q)
\(\frac{5}{M}\)(p - 1) = X + Q
\(\frac{5}{M}\)(p - 1)- x = Q
Q = \(\frac{5(p -1) - Mx}{M}\)
= \(\frac{5p - 5 - Mx}{M}\)
= \(\frac{5p - Mx - 5}{M}\)
Find the equation of a line parallel to y = -4x + 2 passing through (2,3)
y + 4x + 11 = 0
y - 4x - 11 = 0
y + 4x - 11 = 0
y - 4x + 11 = 0
Correct answer is C
By comparing y = mx + c
with y = -4x + 2,
the gradient of y = -4x + 2 is m1 = -4
let the gradient of the line parallel to the given line be m2,
then, m2 = m1 = -4
(condition for parallelism)
using, y - y1 = m2(x - x1)
Hence the equation of the parallel line is
y - 3 = -4(x-2)
y - 3 = -4 x + 8
y + 4x = 8 + 3
y + 4x = 11
y + 4x - 11 = 0
At what value of X does the function y = -3 - 2x + X2 attain a minimum value?
-1
14
4
1
Correct answer is D
Given that y = -3 - 2x + X2
then, \(\frac{dy}{dx}\) = -2 + 2x
At maximum value, \(\frac{dy}{dx}\) = O
therefore, -2 + 2x
2x = 2
x = 2/2 = 1