A distance of 600 km is to be covered in 2 parts in 1st Phase 120 km is travelled by train and rest by car and it took total of 8 hr, but if 200 km is covered by train and rest by car it takes 20 min more. Find the average speed of car and train ?

Options

A. a)80, 60 km/hr
B. b)60, 40 km/hr
C. c)80, 40 km/hr
D. d)60, 35 km/hr

Answer

To solve the problem step by step, we will define the variables and set up equations based on the information given.

### Step 1: Define Variables
Let:
- \( y \) = average speed of the train (in km/h)
- \( x \) = average speed of the car (in km/h)

### Step 2: Set Up the First Equation
In the first phase, 120 km is traveled by train and the remaining distance (600 km - 120 km = 480 km) is traveled by car. The total time taken is 8 hours.

Using the formula for time, we have:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \]
So, the equation for the first phase is:
\[ \frac{120}{y} + \frac{480}{x} = 8 \quad \text{(1)} \]

### Step 3: Set Up the Second Equation
In the second phase, 200 km is traveled by train and the remaining distance (600 km - 200 km = 400 km) is traveled by car. The total time taken is 20 minutes more than the first phase, which is \( 8 \text{ hours} + \frac{20}{60} \text{ hours} = \frac{25}{3} \text{ hours} \).

So, the equation for the second phase is:
\[ \frac{200}{y} + \frac{400}{x} = \frac{25}{3} \quad \text{(2)} \]

### Step 4: Solve the Equations
We now have two equations:
1. \( \frac{120}{y} + \frac{480}{x} = 8 \)
2. \( \frac{200}{y} + \frac{400}{x} = \frac{25}{3} \)

To eliminate one variable, we can express \( \frac{1}{y} \) and \( \frac{1}{x} \) in terms of a new variable. Let:
- \( u = \frac{1}{y} \)
- \( v = \frac{1}{x} \)

Rewriting the equations:
1. \( 120u + 480v = 8 \) (Multiply by 60 to eliminate fractions)
2. \( 200u + 400v = \frac{25}{3} \) (Multiply by 3 to eliminate fractions)

### Step 5: Multiply to Eliminate
Multiply the first equation by 20 and the second by 12 to make the coefficients of \( u \) the same:
1. \( 2400u + 9600v = 160 \)
2. \( 2400u + 4800v = 100 \)

### Step 6: Subtract the Equations
Now, subtract the second equation from the first:
\[ (2400u + 9600v) - (2400u + 4800v) = 160 - 100 \]
This simplifies to:
\[ 4800v = 60 \]
So,
\[ v = \frac{60}{4800} = \frac{1}{80} \]

### Step 7: Substitute Back to Find \( u \)
Now substitute \( v \) back into one of the original equations to find \( u \):
Using \( 120u + 480v = 8 \):
\[ 120u + 480 \cdot \frac{1}{80} = 8 \]
\[ 120u + 6 = 8 \]
\[ 120u = 2 \quad \Rightarrow \quad u = \frac{1}{60} \]

### Step 8: Find Speeds
Now we can find \( y \) and \( x \):
\[ y = \frac{1}{u} = 60 \text{ km/h (speed of the train)} \]
\[ x = \frac{1}{v} = 80 \text{ km/h (speed of the car)} \]

### Final Answer
The average speed of the train is 60 km/h and the average speed of the car is 80 km/h.

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