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})