A milk vendor was selling milk to customers from a vessel containing `12 (4)/(5)` litres of milk. He sold of ` 11 (1)/(2)` litres and then refilled the vessel with `9(4)/(15)` litres of milk again. How much milk is available with the vendor now?

Options

A. `=8 (17)/(30)` litres
B. `=11 (17)/(30)` litres
C. `=10 (17)/(30)` litres
D. `=10 (17)/(57)` litres

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Convert the mixed numbers to improper fractions
The initial amount of milk is given as \( 12 \frac{4}{5} \) liters. To convert this to an improper fraction:
\[ 12 \frac{4}{5} = \frac{12 \times 5 + 4}{5} = \frac{60 + 4}{5} = \frac{64}{5} \]

The amount of milk sold is \( 11 \frac{1}{2} \) liters. Converting this to an improper fraction:
\[ 11 \frac{1}{2} = \frac{11 \times 2 + 1}{2} = \frac{22 + 1}{2} = \frac{23}{2} \]

The amount of milk refilled is \( 9 \frac{4}{15} \) liters. Converting this to an improper fraction:
\[ 9 \frac{4}{15} = \frac{9 \times 15 + 4}{15} = \frac{135 + 4}{15} = \frac{139}{15} \]

### Step 2: Calculate the remaining milk after selling
Now we need to find the remaining milk after selling:
\[ \text{Remaining milk} = \text{Initial milk} - \text{Sold milk} = \frac{64}{5} - \frac{23}{2} \]

To subtract these fractions, we need a common denominator. The least common multiple (LCM) of 5 and 2 is 10.

Converting both fractions:
\[ \frac{64}{5} = \frac{64 \times 2}{5 \times 2} = \frac{128}{10} \]
\[ \frac{23}{2} = \frac{23 \times 5}{2 \times 5} = \frac{115}{10} \]

Now we can subtract:
\[ \text{Remaining milk} = \frac{128}{10} - \frac{115}{10} = \frac{128 - 115}{10} = \frac{13}{10} \]

### Step 3: Add the refilled milk
Now we add the refilled milk:
\[ \text{Total milk} = \text{Remaining milk} + \text{Refilled milk} = \frac{13}{10} + \frac{139}{15} \]

Again, we need a common denominator. The LCM of 10 and 15 is 30.

Converting both fractions:
\[ \frac{13}{10} = \frac{13 \times 3}{10 \times 3} = \frac{39}{30} \]
\[ \frac{139}{15} = \frac{139 \times 2}{15 \times 2} = \frac{278}{30} \]

Now we can add:
\[ \text{Total milk} = \frac{39}{30} + \frac{278}{30} = \frac{39 + 278}{30} = \frac{317}{30} \]

### Step 4: Convert the improper fraction back to a mixed number
To convert \( \frac{317}{30} \) back to a mixed number:
\[ 317 \div 30 = 10 \quad \text{(whole number part)} \]
\[ 317 - (30 \times 10) = 17 \quad \text{(remainder)} \]
Thus, we can write:
\[ \frac{317}{30} = 10 \frac{17}{30} \]

### Final Answer
The amount of milk available with the vendor now is:
\[ 10 \frac{17}{30} \text{ liters} \]

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