Answer:
Let's denote the capacity of the drum as \( C \) liters.
Given:
- The drum is \( \frac{3}{4} \) full initially, so the amount of oil initially present \( = \frac{3}{4} \times C \).
- After drawing 30 liters of oil, the drum is \( \frac{7}{12} \) full, so the remaining amount of oil \( = \frac{7}{12} \times C \).
According to the problem:
\[ \frac{3}{4} \times C - 30 = \frac{7}{12} \times C \]
To solve for \( C \), let's eliminate the fractions by finding a common denominator, which is 12 in this case.
Convert \( \frac{3}{4} \) to have a denominator of 12:
\[ \frac{3}{4} = \frac{3 \times 3}{4 \times 3} = \frac{9}{12} \]
Now rewrite the equation:
\[ \frac{9}{12} \times C - 30 = \frac{7}{12} \times C \]
To eliminate the fractions, multiply through by 12:
\[ 9C - 360 = 7C \]
Subtract \( 7C \) from both sides:
\[ 9C - 7C = 360 \]
\[ 2C = 360 \]
Divide both sides by 2:
\[ C = \frac{360}{2} \]
\[ C = 180 \]
Therefore, the capacity of the drum is \( \boxed{180} \) liters.
Answer:
A drum of kerosene oil is 3 /4 Full. When 30 Liters of oil is drawn from it, it is 7 / 12 . full. Find the capacity of the drum= 34/3 liters
