During routine workshop maintenance, 80 liters of used engine oil are collected. If each disposal drum has a marked capacity of 25 liters and regulations require drums to be filled to only 80 percent for safe transport, how many drums are required?
4 drums
First find the usable volume of one drum, which is only 80% of its 25 litre capacity: \(25\times\frac{80}{100}=20\) litres per drum.
Now divide the total oil by the usable volume per drum: \(\frac{80}{20}=4\) drums.
Since 4 drums hold exactly \(4\times20=80\) litres, all the oil fits with no overflow.
Hence, the answer is 4 drums.
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Important Questions from Simplification
Simplify the following expression.
\(\left(\frac{7}{16} \div \frac{1}{2}\:of\: \frac{1}{5}\right)\times \frac{4}{5}-\frac{1}{3}\times\frac{5}{8}\div \frac{1}{2}+\frac{3}{4}\)
The value of \(\left( {2\frac{6}{7}of4\frac{1}{5} \div \frac{2}{3}} \right) \times 5\frac{1}{9} \div \left( {\frac{3}{4} \times 2\frac{2}{3}of\frac{1}{2} \div \frac{1}{4}} \right)\) is:
The value of \(\left[ {\frac{4}{7}\rm \;of\;2\frac{4}{5} \times 1\frac{2}{3} - \left( {3\frac{1}{2} - 2\frac{1}{6}} \right)} \right] \div \left( {3\frac{1}{5} \div 4\frac{1}{2}\;\rm of\;\;5\frac{1}{3}} \right)\) is:
The value of \(\frac{{0.0203 \times 2.92}}{{0.7 \times 0.0365 \times 2.9}} \div \frac{{{{\left( {12.12} \right)}^2} - {{\left( {8.12} \right)}^2}}}{{{{\left( {0.25} \right)}^2} + \left( {0.25} \right)\left( {19.99} \right)}}\) is:
The value of 4 ÷ 12 of [3 ÷ 4 of {(4 - 2) × 6 ÷ 2}] - 2 × 6 ÷ 8 + 3 is:
