A drum full of rice weighs `40(1)/(6)kg`. If the empty drum weighs `13(3)/(4)kg`, find the weight of rice in the drum.

Answer

To find the weight of rice in the drum, we will follow these steps:

### Step 1: Convert mixed numbers to improper fractions
The weight of the full drum is given as \( 40 \frac{1}{6} \) kg. We convert this mixed number to an improper fraction.

\[ 40 \frac{1}{6} = \frac{40 \times 6 + 1}{6} = \frac{240 + 1}{6} = \frac{241}{6} \text{ kg} \]

### Step 2: Convert the weight of the empty drum to an improper fraction
The weight of the empty drum is given as \( 13 \frac{3}{4} \) kg. We convert this mixed number to an improper fraction.

\[ 13 \frac{3}{4} = \frac{13 \times 4 + 3}{4} = \frac{52 + 3}{4} = \frac{55}{4} \text{ kg} \]

### Step 3: Set up the equation to find the weight of rice
Let \( W \) be the weight of rice in the drum. We can express this as:

\[ W = \text{Weight of full drum} - \text{Weight of empty drum} \]

Substituting the values we found:

\[ W = \frac{241}{6} - \frac{55}{4} \]

### Step 4: Find a common denominator
To subtract these fractions, we need a common denominator. The least common multiple (LCM) of 6 and 4 is 12.

### Step 5: Convert the fractions to have the common denominator
Convert \( \frac{241}{6} \) and \( \frac{55}{4} \) to have a denominator of 12:

\[ \frac{241}{6} = \frac{241 \times 2}{6 \times 2} = \frac{482}{12} \]

\[ \frac{55}{4} = \frac{55 \times 3}{4 \times 3} = \frac{165}{12} \]

### Step 6: Subtract the fractions
Now we can subtract the two fractions:

\[ W = \frac{482}{12} - \frac{165}{12} = \frac{482 - 165}{12} = \frac{317}{12} \text{ kg} \]

### Conclusion
The weight of rice in the drum is \( \frac{317}{12} \) kg.

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