Assume that the distance that a car runs on one litre of petrol varies inversely as the square of the speed at which it is driven. It gives a run of 25 km per litre at speed of 30 km/h. At what speed should it be driven to get a run of 36 km/L?

Options

A. 150 km/h
B. 225 km/h
C. 36 km/h
D. 25 km/h

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship
The distance that a car runs on one litre of petrol (let's denote this as \( x \)) varies inversely as the square of the speed (denote speed as \( v \)). This can be expressed mathematically as:
\[ x = \frac{k}{v^2} \]
where \( k \) is a constant.

### Step 2: Find the constant \( k \)
We know that when the speed \( v_1 = 30 \) km/h, the distance \( x_1 = 25 \) km per litre. We can substitute these values into the equation to find \( k \):
\[ 25 = \frac{k}{30^2} \]
Calculating \( 30^2 \):
\[ 30^2 = 900 \]
Now substituting back:
\[ 25 = \frac{k}{900} \]
To find \( k \), we rearrange the equation:
\[ k = 25 \times 900 = 22500 \]

### Step 3: Set up the equation for the new distance
Now we need to find the speed \( v_2 \) when the distance \( x_2 = 36 \) km per litre. We use the same relationship:
\[ 36 = \frac{22500}{v_2^2} \]

### Step 4: Solve for \( v_2^2 \)
Rearranging the equation to solve for \( v_2^2 \):
\[ v_2^2 = \frac{22500}{36} \]
Calculating \( \frac{22500}{36} \):
\[ v_2^2 = 625 \]

### Step 5: Find \( v_2 \)
Taking the square root of both sides to find \( v_2 \):
\[ v_2 = \sqrt{625} = 25 \text{ km/h} \]

### Conclusion
The speed at which the car should be driven to get a run of 36 km/L is **25 km/h**.

---