A train covers certain distance between two places at a uniform speed. If the train moved 10 km/h faster, it would take 2 h less, and if the train were slower by 10 km/h, it would take 3 h more than the scheduled time. Find the distance covered by the train.

Options

A. 300 km
B. 600 km
C. 800 km
D. 1200 km

Answer

To solve the problem step by step, let's denote the following variables:

- Let \( D \) be the distance covered by the train (in kilometers).
- Let \( S \) be the usual speed of the train (in km/h).
- The time taken at the usual speed is \( \frac{D}{S} \) hours.

### Step 1: Set up the equations based on the conditions given.

1. **Condition 1**: If the train moves 10 km/h faster, its speed becomes \( S + 10 \) km/h. The time taken will be \( \frac{D}{S + 10} \) hours. According to the problem, this time is 2 hours less than the usual time:
\[ \frac{D}{S + 10} = \frac{D}{S} - 2 \]

2. **Condition 2**: If the train moves 10 km/h slower, its speed becomes \( S - 10 \) km/h. The time taken will be \( \frac{D}{S - 10} \) hours. This time is 3 hours more than the usual time:
\[ \frac{D}{S - 10} = \frac{D}{S} + 3 \]

### Step 2: Rearranging the equations

From the first condition:
\[ \frac{D}{S + 10} + 2 = \frac{D}{S} \]
Multiplying through by \( S(S + 10) \) to eliminate the denominators:
\[ D \cdot S + 2S(S + 10) = D(S + 10) \]
Expanding and simplifying:
\[ DS + 2S^2 + 20S = DS + 10D \]
Cancelling \( DS \) from both sides:
\[ 2S^2 + 20S = 10D \quad \text{(Equation 1)} \]

From the second condition:
\[ \frac{D}{S - 10} - 3 = \frac{D}{S} \]
Multiplying through by \( S(S - 10) \):
\[ D \cdot S - 3S(S - 10) = D(S - 10) \]
Expanding and simplifying:
\[ DS - 3S^2 + 30S = DS - 10D \]
Cancelling \( DS \) from both sides:
\[ -3S^2 + 30S = -10D \]
Multiplying through by -1:
\[ 3S^2 - 30S = 10D \quad \text{(Equation 2)} \]

### Step 3: Equate the two equations for \( D \)

From Equation 1:
\[ D = \frac{2S^2 + 20S}{10} = \frac{S^2 + 10S}{5} \]

From Equation 2:
\[ D = \frac{3S^2 - 30S}{10} = \frac{3(S^2 - 10S)}{10} \]

Setting the two expressions for \( D \) equal to each other:
\[ \frac{S^2 + 10S}{5} = \frac{3(S^2 - 10S)}{10} \]

### Step 4: Cross-multiply and simplify

Cross-multiplying gives:
\[ 10(S^2 + 10S) = 15(S^2 - 10S) \]
Expanding both sides:
\[ 10S^2 + 100S = 15S^2 - 150S \]
Rearranging gives:
\[ 5S^2 - 250S = 0 \]
Factoring out \( S \):
\[ S(5S - 250) = 0 \]
Thus, \( S = 0 \) or \( S = 50 \).

Since speed cannot be zero, we have:
\[ S = 50 \text{ km/h} \]

### Step 5: Calculate the distance \( D \)

Substituting \( S = 50 \) back into either equation for \( D \):
\[ D = \frac{2(50^2) + 20(50)}{10} = \frac{2(2500) + 1000}{10} = \frac{5000 + 1000}{10} = \frac{6000}{10} = 600 \text{ km} \]

### Final Answer
The distance covered by the train is **600 km**.