JAMB Mathematics Past Questions & Answers - Page 258

2log \(\frac{2}{5}\) - log\(\frac{72}{125}\) + log 9

[\(\frac{2}{5}\))² x 9] = log \(\frac{4}{25}\) x \(\frac{9}{1}\) x \(\frac{125}{72}\)

= log \(\frac{72}{125}\)

= log \(\frac{5}{2}\)

= log \(\frac{10}{4}\)

= log 10 - log 4

= log₁₀ 10 - log₁₀ 22

= 1 - 2 log2

\(\frac{27}{5^1}\)(3)-3 x \(\frac{(1)^{-1}}{5}\) = \(\frac{27}{5}\) x \(\frac{1}{3^3}\) x \(\frac{1}{\frac{1}{5}}\)

= \(\frac{27}{5}\) x \(\frac{1}{27}\) x \(\frac{1}{5}\)

=  \(\frac{1}{25}\)

% of water in the mixture

= \(\frac{\text{Total Amount of water}}{\text{Total quantity of spirit}}\) x \(\frac{100}{1}\)

\(\frac{3(\frac{20}{100}) + 5 (\frac{15}{100})}{3 + 5}\) x \(\frac{100}{1}\)

= \(\frac{\frac{6}{10} + \frac{75}{100}}{8}\) x \(\frac{100}{1}\)

= \(\frac{0.6 + 0.75}{8}\) x \(\frac{100}{1}\)


= \(\frac{1.35}{8}\) x \(\frac{100}{1}\)

= \(\frac{33.75}{2}\)

= 16.875

= 16\(\frac{7}{8}\)

I = \(\frac{PRT}{100}\)

= \(\frac{10 \times 2 \times 4}{100}\)

= \(\frac{4}{5}\)

= 0.8

Total amount = N10.80

He pays N8.00

Remainder = 10.80 - 8.00

= N2.80

P produces 20,000 bags

P reduces production by 5%

i.e. reduction = \(\frac{5}{100}\) x \(\frac{20000}{1}\)

= 1000

Q produces 15000 bags

Q increases production by 5%

i.e. increase = \(\frac{5}{100}\) x \(\frac{15000}{1}\)

= 750

Effective loss by the two = 1000 - 750

= 250 bags