The difference between the time taken by two trains to travel a distance of 300 km is 2 hours 30 minutes. If the difference between their speeds is 6 kmph, find the speed of slower train?(in kmph)

Options

A. A)25
B. B)20
C. C)18
D. D)24

Answer

To solve the problem step by step, we can follow these steps:

### Step 1: Understand the given information
We know that:
- The distance traveled by both trains is 300 km.
- The difference in time taken by the two trains is 2 hours and 30 minutes.
- The difference in their speeds is 6 km/h.

### Step 2: Convert the time difference into hours
The time difference of 2 hours and 30 minutes can be converted into hours:
\[ 2 \text{ hours } 30 \text{ minutes} = 2 + \frac{30}{60} = 2.5 \text{ hours} \]

### Step 3: Set up the equations
Let the speed of the slower train be \( V \) km/h. Then, the speed of the faster train will be \( V + 6 \) km/h.

The time taken by the slower train to cover 300 km is:
\[ \text{Time}_{\text{slower}} = \frac{300}{V} \]

The time taken by the faster train to cover 300 km is:
\[ \text{Time}_{\text{faster}} = \frac{300}{V + 6} \]

### Step 4: Set up the equation based on the time difference
According to the problem, the difference in time taken by the two trains is 2.5 hours:
\[ \frac{300}{V} - \frac{300}{V + 6} = 2.5 \]

### Step 5: Solve the equation
To solve the equation, we can multiply through by \( V(V + 6) \) to eliminate the denominators:
\[ 300(V + 6) - 300V = 2.5V(V + 6) \]
This simplifies to:
\[ 300 \times 6 = 2.5V^2 + 15V \]
\[ 1800 = 2.5V^2 + 15V \]

### Step 6: Rearrange the equation
Rearranging gives us:
\[ 2.5V^2 + 15V - 1800 = 0 \]

### Step 7: Simplify the equation
To make calculations easier, we can multiply the entire equation by 2 to eliminate the decimal:
\[ 5V^2 + 30V - 3600 = 0 \]

### Step 8: Factor the quadratic equation
Now we need to factor the quadratic equation. We can look for two numbers that multiply to \( -3600 \) and add to \( 30 \). These numbers are \( 60 \) and \( -60 \):
\[ (5V - 60)(V + 60) = 0 \]

### Step 9: Solve for \( V \)
Setting each factor to zero gives:
1. \( 5V - 60 = 0 \) → \( V = 12 \) (not valid since we need a positive speed)
2. \( V + 60 = 0 \) → \( V = -60 \) (not valid)

### Step 10: Use the quadratic formula if necessary
If factoring does not yield valid results, we can use the quadratic formula:
\[ V = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 5, b = 30, c = -3600 \):
\[ V = \frac{-30 \pm \sqrt{30^2 - 4 \times 5 \times (-3600)}}{2 \times 5} \]
Calculating the discriminant:
\[ 30^2 + 72000 = 900 + 72000 = 72900 \]
Taking the square root:
\[ \sqrt{72900} = 270 \]
Now substituting back:
\[ V = \frac{-30 \pm 270}{10} \]
Calculating the two possible values:
1. \( V = \frac{240}{10} = 24 \) (valid)
2. \( V = \frac{-300}{10} = -30 \) (not valid)

Thus, the speed of the slower train is:
\[ \text{Speed of slower train} = 24 \text{ km/h} \]

### Final Answer
The speed of the slower train is **24 km/h**.

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