A drum of kerosene is $\frac{3}{4}$ full. When 30 litres of kerosene is drawn from it, it remains $\frac{7}{12}$ full. The capacity of the drum is

### Concept & Formula This is a classic algebraic volume problem based on fractional differences. When a specific quantity is removed, the fractional fullness of the container decreases. By equating the difference between the initial and final fractions to the absolute volume removed, we can find the total capacity. ### Step-by-Step Solution * **Given:** Initial fractional volume = $\frac{3}{4}$ Final fractional volume = $\frac{7}{12}$ Volume removed = 30 litres. Let total capacity be $x$. * **Calculation:** Write the equation representing the removal of kerosene: $$ \frac{3}{4}x - 30 = \frac{7}{12}x $$ * Rearrange the equation to group the $x$ terms on one side: $$ \frac{3}{4}x - \frac{7}{12}x = 30 $$ * Find a common denominator (which is 12) to evaluate the left side: $$ \frac{9x}{12} - \frac{7x}{12} = 30 $$ $$ \frac{2x}{12} = 30 $$ $$ \frac{1}{6}x = 30 $$ * Multiply by 6 to isolate $x$: $$ x = 180 $$ ### Exam Strategy & Shortcut Use the **Fractional Equivalence Method**. Calculate the difference in fractions first mentally: $\frac{3}{4} - \frac{7}{12} = \frac{9}{12} - \frac{7}{12} = \frac{2}{12} = \frac{1}{6}$. You now know that $\frac{1}{6}$ of the drum equals 30 litres. To find the whole drum (1 full unit), simply multiply $30 \times 6 = 180$. This skips writing out full algebraic steps and saves time. ### Common Pitfall Students often struggle with finding the common denominator quickly, leading to calculation errors. Another mistake is misreading "drawn from it" as "added to it," which would lead to incorrect addition instead of subtraction. ### Final Answer Therefore, the correct answer is **180 litres**.