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.
N68.50
N63.00
N60.00
N52.00
Correct answer is B
C = a + kS. If C = 74, S = 20
C = 96, S = 30, C= ? S = 15
74 = a + 20k......(1)
96 = a + 30k......(2)
subtract equation (1) from (2)
96 = a + 30k
-
74 = a + 20k
--------------
22 = 10k
k = 2.2
find a
74 = a + 44
a = 30
C = 30 + 2.2S
when S = 15, C = 30 + 2.2 x 15
= 30 + 33
= N63
Make R the subject of the fomula S = \(\sqrt{\frac{2R + T}{2RT}}\)
R = \(\frac{T}{(TS^2 + 1)}\)
R = \(\frac{T}{2(TS^2 - 2)}\)
R = \(\frac{T}{2(TS^2 + 1)}\)
R = \(\frac{R}{2(TS^2 + 1)}\)
Correct answer is B
S = \(\sqrt{\frac{2R + T}{2RT}}\)
Squaring both sides,
\(S^{2} = \frac{2R + T}{2RT}\)
\(S^{2} (2RT) = 2R + T\)
\(2S^{2} RT - 2R = T\)
\(R = \frac{T}{2TS^{2} - 2}\)
= \(\frac{T}{2(TS^{2} - 1)}
What are the values of y which satisfy the equation \(9^{y} - 4 \times 3^{y} + 3 = 0\) ?
-1 and 0
-1 and 1
1 and 3
0 and 1
Correct answer is D
\(9^{y} - 4 \times 3^{y} + 3 = 0\)
\(\equiv (3^{2})^{y} - 4 \times 3^{y} + 3 = 0\)
\((3^{y})^{2} - 4 \times 3^{y} + 3 = 0\)
Let \(3^{y}\) be r. Then,
\(r^{2} - 4r + 3 = 0\)
Solving the equation,
\(r^{2} - 3r - r + 3 = 0\)
\(r(r - 3) - 1(r - 3) = 0\)
\((r - 3)(r - 1) = 0\)
\(\therefore \text{r = 3 or 1}\)
Recall, \(3^{y} = r\)
\(3^{y} = 3 = 3^{1} \text{ or } 3^{y} = 1 = 3^{0}\)
\(\implies \text{y = 1 or 0}\)
Find p in terms of q if \(\log_{3} p + 3\log_{3} q = 3\)
(\(\frac{3}{q}\))3
(\(\frac{q}{3}\))\(\frac{1}{3}\)
(\(\frac{q}{3}\))3
(\(\frac{3}{q}\))\(\frac{1}{3}\)
Correct answer is A
\(\log_{3} p + 3\log_{3} q = 3\)
\(\log_{3} p + \log_{3} q^{3} = 3\)
\(\implies \log_{3} (pq^{3}) = 3\)
\(pq^{3} = 3^{3} = 27\)
\(\therefore p = \frac{27}{q^{3}}\)
= \((\frac{3}{q})^{3}\)
Simplify \(4 - \frac{1}{2 - \sqrt{3}}\)
2\(\sqrt{3}\)
-2 - \(\sqrt{3}\)
-2 + \(\sqrt{3}\)
2 - \(\sqrt{3}\)
Correct answer is D
\(4 - \frac{1}{2 - \sqrt{3}} = \frac{4(2 - \sqrt{3}) - 1}{2 - \sqrt{3}}\)
= \(\frac{8 - 4\sqrt{3} - 1}{2 - \sqrt{3}}\)
= \(\frac{7 - 4\sqrt{3}}{2 - \sqrt{3}}\)
Rationalizing,
\((\frac{7 - 4\sqrt{3}}{2 - \sqrt{3}}) (\frac{2 + \sqrt{3}}{2 + \sqrt{3}})\)
= \(\frac{14 + 7\sqrt{3} - 8\sqrt{3} - 12}{4 + 2\sqrt{3} - 2\sqrt{3} - 3}\)
= \(\frac{2 - \sqrt{3}}{1} = 2 - \sqrt{3}\)