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 sum of the roots of the equation (x - p)(2x + 1) = 0 is 1, find the value of x
1\(\frac{1}{2}\)
\(\frac{1}{2}\)
-\(\frac{3}{2}\)
-1\(\frac{1}{2}\)
Correct answer is A
(x - p)(2x + 1) = 0
2x2 + x - 2px - p = 0
2x2 + x (1 - 2p) - p = 0
2x2 - (2p - 1)x - p = 0
divide through by 2
x2 - \(\frac{(2p - 1)}{2}\)x - \(\frac{p}{2}\) = 0
compare to x2 - (sum of roots)x + product of roots = 0
sum of roots = \(\frac{2p - 1}{2}\)
But sum of roots = 1
Given; \(\frac{2p - 1}{2}\) = 1
2p - 1 = 2 x 1
2p - 1 = 2
2p = 2 + 1 = 3
p = \(\frac{3}{2}\)
p = 1\(\frac{1}{2}\)
2
3
4
5
Correct answer is A
5 1 6 2seven
-2 6 4 4seven
--------
2 2 1 5
--------
the missing number is 2
What is the value of x when y = 5? y = \(\frac{1}{2}\) x + 1
8
9
10
11
Correct answer is A
when y = 5; x = ?; y = \(\frac{1}{2}\)x + 1
5 = \(\frac{1}{2}\)x + 1
5 - 1 = \(\frac{1}{2}\)x
4 = \(\frac{1}{2}\)x
x = 4 x 2
x = 8
y = 2x
y = x + 1
y = x
y = \(\frac{1}{2}x + 1\)
Correct answer is D
y = mx + c; when x = 0; y = 1
1 = m(0) + c; 1 = 0 + c; c = 1
when x = 2; y = 2
2 = m(2) + c; 2 = 2m + c; but c = 1
2 = 2m + 1
2 - 1 = 2m
2m = 1
m = \(\frac{1}{2}\)
y = \(\frac{1}{2}\)x + 1
18
20
22
28
Correct answer is B
mean age = \(\frac{\text{sum of ages}}{\text{no. of men}}\)
50 = \9\frac{sum}{R}\)
sum = 50R.....(1)
Sum of ages of the men that left = 55 + 63 = 188
remaining sum = 50R - 118
remaining no. of men = R - 2
now mean age = 50 - 1 = 49 years
49 = \(\frac{50R - 118}{R - 2}\)
49(R - 2) = 50R - 118
49R - 50R = -188 - 98
-R = -20
R = 20