JAMB - Mathematics (1991 - No. 7)
If 3 gallons of spirit containing 20% water are added to 5 gallons of another spirit containing 15% water, what percentage of the mixture is water?
2\(\frac{4}{5}\)%
16\(\frac{7}{8}\)%
18\(\frac{1}{8}\)%
18\(\frac{7}{8}\)%
Explanation
% 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\(\frac{7}{8}\)
= \(\frac{\text{Total Amount of water}}{\text{Total quantity of spirit}}\) 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\(\frac{7}{8}\)
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