Pabitra had `19(2)/(7)l` of petrol in the fuel tank of his vehicle at the start of a journey. During the journey, `13(1)/(2)l` of petrol was consumed. What fraction of the initial petrol was left in the fuel tank after the journey?
(a)`(11)/(40)`
(b)`(3)/(10)`
(c)`(3)/(5)`
(d)`(7)/(10)`

Options

A. `(11)/(40)`
B. `(3)/(10)`
C. `(3)/(5)`
D. `(7)/(10)`

Answer

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

### Step 1: Convert mixed numbers to improper fractions
Pabitra had `19(2)/(7)l` of petrol. We can convert this mixed number to an improper fraction.

\[ 19 \frac{2}{7} = \frac{19 \times 7 + 2}{7} = \frac{133 + 2}{7} = \frac{135}{7} \]

### Step 2: Convert the consumed petrol to an improper fraction
Next, we need to convert `13(1)/(2)l` of petrol consumed into an improper fraction.

\[ 13 \frac{1}{2} = \frac{13 \times 2 + 1}{2} = \frac{26 + 1}{2} = \frac{27}{2} \]

### Step 3: Calculate the remaining petrol
Now, we will calculate how much petrol is left after the journey by subtracting the consumed petrol from the initial petrol.

\[ \text{Remaining petrol} = \text{Initial petrol} - \text{Consumed petrol} = \frac{135}{7} - \frac{27}{2} \]

To perform this subtraction, we need a common denominator. The least common multiple of 7 and 2 is 14.

Convert both fractions:

\[ \frac{135}{7} = \frac{135 \times 2}{7 \times 2} = \frac{270}{14} \]
\[ \frac{27}{2} = \frac{27 \times 7}{2 \times 7} = \frac{189}{14} \]

Now subtract:

\[ \text{Remaining petrol} = \frac{270}{14} - \frac{189}{14} = \frac{270 - 189}{14} = \frac{81}{14} \]

### Step 4: Find the fraction of initial petrol left
Now we need to find the fraction of the initial petrol that is left in the tank.

\[ \text{Fraction left} = \frac{\text{Remaining petrol}}{\text{Initial petrol}} = \frac{\frac{81}{14}}{\frac{135}{7}} \]

To divide fractions, we multiply by the reciprocal:

\[ \text{Fraction left} = \frac{81}{14} \times \frac{7}{135} = \frac{81 \times 7}{14 \times 135} \]

### Step 5: Simplify the fraction
Now we simplify:

\[ \frac{81 \times 7}{14 \times 135} = \frac{567}{1890} \]

Now, we can simplify this fraction. The greatest common divisor (GCD) of 567 and 1890 is 189.

\[ \frac{567 \div 189}{1890 \div 189} = \frac{3}{10} \]

### Final Answer
The fraction of the initial petrol that was left in the fuel tank after the journey is:

\[ \frac{3}{10} \]

### Conclusion
Thus, the correct option is (b) \( \frac{3}{10} \).

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