1993 - WAEC Mathematics Past Questions and Answers - page 1
S = {1, 2, 3, 4, 5, 6}, T = {2,4,5,7} and R = {1,4, 5}, and (S∩T) ∪ R
S = {1, 2, 3, 4, 5, 6}; T = {2, 4, 5, 7}; R = {1, 4, 5}
(S∩T) ∪ R = {2, 4, 5} ∪ {1, 4, 5}
= {1, 2, 4, 5}
Simplify: \(\frac{3}{4} \div 1\frac{1}{4} \times (1\frac{1}{2} - \frac{2}{3})\)
\(\frac{3}{4} \div 1\frac{1}{4} \times (1\frac{1}{2} - \frac{2}{3})\)
\(\frac{3}{4} \div \frac{5}{4} \times (\frac{9 - 4}{6})\)
= \(\frac{3}{4} \times \frac{4}{5} \times \frac{5}{6}\)
= \(\frac{1}{2}\)
Solve the inequality: 3m + 3 > 9
Convert 89\(_{10}\) to a number in base two.
\(89_{10}\)
| 2 | 89 |
| 2 | 44 r 1 |
| 2 | 22 r 0 |
| 2 | 11 r 0 |
| 2 | 5 r 1 |
| 2 | 2 r 1 |
| 2 | 1 r 0 |
| 0 r 1 |
\(89_{10} = 1011001_{2}\)
The nth term of a sequence is given by (-1)\(^{n-2}\) x 2\(^{n+1}\). Find the sum of the second and third terms.
when n = 2
(-1)\(^{n-2}\) 2\(^{n+1}\) = 2
When n = 3
(-1)\(^{n-2}\) 2\(^{n+1}\) = -4
Sum = 2 - 4 = -2
Simplify: \(\frac{4^{-\frac{1}{2}} \times 16^{\frac{3}{4}}}{4^{\frac{1}{2}}}\)
\(\frac{4^{-\frac{1}{2}} \times 16^{\frac{3}{4}}}{4^{\frac{1}{2}}}\)
= \(\frac{16^{\frac{3}{4}}}{4^{\frac{1}{2}} \times 4^{\frac{1}{2}}}\)
= \(\frac{(2^4)^{\frac{3}{4}}}{4^{\frac{1}{2}} \times 4^{\frac{1}{2}}}\)
= \(\frac{2^3}{4}\)
= 2
Simplify: \(\frac{\log \sqrt{27}}{\log \sqrt{81}}\)
\(\frac{\log \sqrt{27}}{\log 81}\)
= \(\frac{\log \sqrt{3^3}}{\log 3^4}\)
= \(\frac{\log 3^{\frac{3}{2}}}{\log 3^4}\)
= \(\frac{\frac{3}{2} \log 3}{4 \log 3}\)
= \(\frac{\frac{3}{2}}{4}\)
= \(\frac{3}{8}\)
Factorize the expression 2s\(^2\) - 3st - 2t\(^2\).
2s\(^2\) - 3st - 2t\(^2\)
= 2s\(^2\) - 4st + st - 2t\(^2\)
= 2s(s - 2t) + t(s - 2t)
= (2s + t)(s - 2t)
Solve the equation x\(^2\) - 2x - 3 = 0
x\(^2\) - 2x - 3 = 0
x\(^2\) - 3x + x - 3 = 0
x(x - 3) + 1(x - 3) = 0
(x + 1)(x - 3) = 0
x = (-1, 3)