2011 - JAMB Mathematics Past Questions and Answers - page 1

2q3₅ = 77₈

2 x 5² + q x 5¹ + 3 x 5⁰ = 7 x 8¹ + 7 x 8⁰

2 x 25 + q x 5 + 3 x 1 = 7 x 8 + 7 x 1

50 + 5q + 3 = 56 + 7

5q = 63 - 53

q = \(\frac{10}{5}\)

q = 2

\(\frac{3\frac{2}{3} \times \frac{5}{6} \times \frac{2}{3}}{\frac{11}{15} \times \frac{3}{4} \times \frac{2}{27}}\)

\(\frac{\frac{11}{3} \times \frac{5}{6} \times \frac{2}{3}}{\frac{11}{15} \times \frac{3}{4} \times \frac{2}{27}}\)

\(\frac{110}{54} \div \frac{66}{1620}\)

50

S.I. = \(\frac{P \times R \times T}{100}\)

If T = 9 months, it is equivalent to \(\frac{9}{12}\) years

S.I. = \(\frac{5000 \times 4 \times 9}{100 \times 12}\)

S.I. = N150

M:N:Q == 5:4:3

i.e M = 5, N = 4, Q = 3

Substituting values into equation, we have...

\(\frac{2N - Q}{M}\)

= \(\frac{2(4) - 3}{5}\)

= \(\frac{8 - 3}{5}\)

= \(\frac{5}{5}\)

= 1

\((\frac{16}{81})^{\frac{1}{4}} \div (\frac{9}{16})^{-\frac{1}{2}}\)

\((\frac{16}{81})^{\frac{1}{4}} \div (\frac{16}{9})^{\frac{1}{2}}\)

\((\frac{2^4}{3^4})^{\frac{1}{4}} \div (\frac{4^2}{3^2})^{\frac{1}{2}}\)

\(\frac{2^{4 \times \frac{1}{4}}}{3^{4 \times \frac{1}{4}}} \div \frac{4^{2 \times \frac{1}{2}}}{3^{2 \times \frac{1}{2}}}\)

\(\frac{2}{3} \div \frac{4}{3}\)

\(\frac{2}{3} \times \frac{3}{4}\)

\(\frac{2}{4}\)

\(\frac{1}{2}\)

log\(_{3}^{18}\) + log\(_{3}^{3}\) - log\(_{3}^{x}\) = 3

log\(_{3}^{18}\) + log\(_{3}^{3}\) - log\(_{3}^{x}\) = 3log₃3

log\(_{3}^{18}\) + log\(_{3}^{3}\) - log\(_{3}^{x}\) = log₃3³

log₃(\(\frac{18 \times 3}{X}\)) = log₃3³

\(\frac{18 \times 3}{X}\) = 3³

18 x 3 = 27 x X

x = \(\frac{18 \times 3}{27}\)

= 2

\(\frac{2 - \sqrt5}{3 - \sqrt5}\) x \(\frac{3 + \sqrt5}{3 + \sqrt5}\)

\(\frac{(2 - \sqrt5)(3 + \sqrt5)}{(3 - \sqrt5)(3 + \sqrt5)}\) = \(\frac{6 +2\sqrt5 - 3\sqrt5 - \sqrt25}{9 + 3\sqrt5 - 3\sqrt5 - \sqrt25}\)

= \(\frac{6 - \sqrt5 - 5}{9 - 5}\)

= \(\frac{1 - \sqrt5}{4}\)

(\(\sqrt2 + \frac{1}{\sqrt3})(\sqrt2 - \frac{1}{\sqrt3}\))

\(\sqrt4 - \frac {\sqrt2}{\sqrt3} + \frac {\sqrt2}{\sqrt3} - \frac {1}{\sqrt9}\)

= 2 - \(\frac {1}{3}\)

= \(\frac {16 - 1}{3}\)

= \(\frac{5}{3}\)

The first poster has 7 ways to be arranges, the second poster can be arranged in 6 ways and the third poster in 5 ways.

= 7 x 6 x 5

= 210 ways

or \(\frac{7}{P_3}\) = \(\frac{7!}{(7 - 3)!}\) = \(\frac{7!}{4!}\)

= \(\frac{7 \times 6 \times 5 \times 4!}{4!}\)

= 210 ways

T = \(\frac{KR^2 + M}{3}\)

3T = KR² + M

KR² = 3T - M

R² = \(\frac{3T - M}{K}\)

R = \(\sqrt\frac{3T - M}{K}\)