A man covers a certain distance on a toy train. If the train moved 4 km/h faster, it would take 30 minutes less. If it moved 2 km/h slower, it would have taken 20 minutes more. Find the distance.

Options

A. 60 km
B. 58 km
C. 55 km
D. 50 km

Answer

To solve the problem step by step, we will use the information provided in the question about the speed of the train and the time taken to cover the distance.

### Step 1: Define Variables
Let:
- \( D \) = Distance covered by the train (in km)
- \( S \) = Original speed of the train (in km/h)

### Step 2: Set Up the Equations
1. **When the speed is increased by 4 km/h:**
- New speed = \( S + 4 \) km/h
- Time taken at new speed = \( \frac{D}{S + 4} \)
- Original time taken = \( \frac{D}{S} \)
- According to the problem, the new time is 30 minutes (or \( \frac{1}{2} \) hours) less than the original time:
\[ \frac{D}{S} - \frac{D}{S + 4} = \frac{1}{2} \]

2. **When the speed is decreased by 2 km/h:**
- New speed = \( S - 2 \) km/h
- Time taken at new speed = \( \frac{D}{S - 2} \)
- According to the problem, the new time is 20 minutes (or \( \frac{1}{3} \) hours) more than the original time:
\[ \frac{D}{S - 2} - \frac{D}{S} = \frac{1}{3} \]

### Step 3: Solve the First Equation
From the first equation:
\[ \frac{D}{S} - \frac{D}{S + 4} = \frac{1}{2} \]
Multiply through by \( S(S + 4) \):
\[ D(S + 4) - DS = \frac{1}{2} S(S + 4) \]
This simplifies to:
\[ 4D = \frac{1}{2} S(S + 4) \]
\[ 8D = S(S + 4) \quad \text{(Equation 1)} \]

### Step 4: Solve the Second Equation
From the second equation:
\[ \frac{D}{S - 2} - \frac{D}{S} = \frac{1}{3} \]
Multiply through by \( S(S - 2) \):
\[ D(S) - D(S - 2) = \frac{1}{3} S(S - 2) \]
This simplifies to:
\[ 2D = \frac{1}{3} S(S - 2) \]
\[ 6D = S(S - 2) \quad \text{(Equation 2)} \]

### Step 5: Equate the Two Equations
From Equation 1 and Equation 2:
1. \( 8D = S(S + 4) \)
2. \( 6D = S(S - 2) \)

Now, we can express \( D \) from both equations:
From Equation 1:
\[ D = \frac{S(S + 4)}{8} \]
From Equation 2:
\[ D = \frac{S(S - 2)}{6} \]

### Step 6: Set the Two Expressions for D Equal
\[ \frac{S(S + 4)}{8} = \frac{S(S - 2)}{6} \]
Cross-multiply to eliminate the fractions:
\[ 6S(S + 4) = 8S(S - 2) \]
Expanding both sides:
\[ 6S^2 + 24S = 8S^2 - 16S \]
Rearranging gives:
\[ 2S^2 - 40S = 0 \]
Factoring out \( 2S \):
\[ 2S(S - 20) = 0 \]
Thus, \( S = 0 \) or \( S = 20 \). Since speed cannot be zero, we have:
\[ S = 20 \text{ km/h} \]

### Step 7: Find the Distance
Now substitute \( S \) back into either equation for \( D \). Using Equation 1:
\[ D = \frac{20(20 + 4)}{8} = \frac{20 \times 24}{8} = 60 \text{ km} \]

### Final Answer
The distance covered by the man on the toy train is **60 km**.