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 k in the equation \(\frac{1}{8}^{k+2}\) = 1
2
-4
-2
4
Correct answer is C
\((8^{-1})^{k+2}\) = \((8^{0})\)
base 8 cancel out on both sides
-1(k+2) = 0
-k -2 = 0
: k = -2
For what value of x is \(\frac{4 - 2x}{x + 1}\) undefined.
2
-1
1
-2
Correct answer is B
A rational expression is undefined when the denominator is equal to zero.
when x = -1
The denominator in this equation : x + 1
--> -1 + 1 = 0
This expression is undefined when x = -1
The equation of a line is given as 3 x - 5y = 7. Find its gradient (slope)
\(\frac{5}{3}\).
\(\frac{3}{5}\).
\(\frac{-3}{5}\).
\(\frac{-5}{3}\).
Correct answer is B
the form y=mx+c
where m is the gradient and c is the y-intercept.
the equation to gives 5y=3x - 7.
divide both side by 5 the coefficient of y → \(\frac{3}{5}\)x - \(\frac{7}{5}\)
comparing this with the general equation y=mx+c,
you can see that m= the gradient= \(\frac{3}{5}\).
A man will be (x+10)years old in 8years time. If 2years ago he was 63 years., find the value of x
55
63
57
67
Correct answer is B
A man will be (x+10) years old in 8years time.
As at today, he is x + 2 years of age.
The man was 63 years old 2 years ago, so he is 63+2=65 now.
8 years from now, he will be 65+8=73.
He will be (x+10) years old when he is 73. So
x+10=73
x=73-10=63
A fair die is tossed twice what is the probability of get a sum of at least 10.
\(\frac{5}{36}\)
\(\frac{2}{3}\)
\(\frac{5}{18}\)
\(\frac{1}{6}\)
Correct answer is D
\(\begin{array}{c|c}
& 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
1 & 1,1 & 1,2 & 1,3 & 1,4 & 1,5 & 1,6 \\ \hline 2 & 2,1 & 2,2 & 2,3 & 2,4 & 2,5 & 2,6 \\ \hline 3 & 3,1 & 3,2 & 3,3 & 3,4 & 3,5 & 3,6 \\ \hline 4 & 4,1 & 4,2 & 4,3 & 4,4 & 4,5 & 4,6 \\ 5 & 5,1 & 5,2 & 5,3 & 5,4 & 5,5 & 5,6 \\ \hline 6 & 6,1 & 6,2 & 6,3 & 6,4 & 6,5 & 6,6\end{array}\)
From the table above, event space, n(E) = 6
sample space, n(S) = 36
Hence, probability sum of scores is at least 10, is;
\(\frac{n(E)}{n(S)}\)
= \(\frac{6}{36}\)
= \(\frac{1}{6}\)