A car can travel `12(1)/(2)` km in 1 litre of petro how much distances can it travel in `42(3)/(5)` litres of petrol

Answer

To solve the problem step by step, we need to find out how far a car can travel with `42(3)/(5)` litres of petrol, given that it can travel `12(1)/(2)` km with 1 litre.

### Step 1: Convert the mixed fraction `12(1)/(2)` to an improper fraction.
To convert a mixed fraction to an improper fraction, we use the formula:
\[ \text{Improper Fraction} = \left(\text{Whole Number} \times \text{Denominator} + \text{Numerator}\right) / \text{Denominator} \]
For `12(1)/(2)`, we have:
\[ 12(1)/(2) = \frac{12 \times 2 + 1}{2} = \frac{24 + 1}{2} = \frac{25}{2} \]

### Step 2: Convert the mixed fraction `42(3)/(5)` to an improper fraction.
Using the same method:
\[ 42(3)/(5) = \frac{42 \times 5 + 3}{5} = \frac{210 + 3}{5} = \frac{213}{5} \]

### Step 3: Multiply the distance per litre by the total litres of petrol.
Now we need to multiply the distance the car can travel per litre by the total litres:
\[ \text{Distance} = \left(\frac{25}{2}\right) \times \left(\frac{213}{5}\right) \]

### Step 4: Simplify the multiplication.
We can multiply the numerators and the denominators:
\[ \text{Distance} = \frac{25 \times 213}{2 \times 5} = \frac{5325}{10} \]

### Step 5: Convert the improper fraction to a mixed fraction.
To convert `5325/10` to a mixed fraction, we divide the numerator by the denominator:
\[ 5325 \div 10 = 532 \quad \text{(whole number)} \]
The remainder is:
\[ 5325 - (532 \times 10) = 5325 - 5320 = 5 \]
Thus, we can write it as:
\[ 532 \frac{5}{10} \]
Now, simplify `5/10` to `1/2`:
\[ 532 \frac{1}{2} \]

### Final Answer:
The car can travel **`532(1)/(2)` km** with `42(3)/(5)` litres of petrol.

---