A train runs 30% faster than a car both start at the same time from A and reach at B at the same time because there was a halt of 9 minutes for train. Find out the speed of car if the distance between A and B is 78 km.

Answer

To solve the problem, we need to find the speed of the car given that the train runs 30% faster than the car and both reach their destination at the same time, despite the train halting for 9 minutes.

### Step-by-Step Solution:

1. **Define Variables:**
Let the speed of the car be \( x \) km/h. Then, the speed of the train will be \( 1.3x \) km/h (since the train is 30% faster than the car).

2. **Calculate Time Taken by Each Vehicle:**
The distance between A and B is 78 km.
- Time taken by the car to travel from A to B:
\[ \text{Time}_{\text{car}} = \frac{78}{x} \text{ hours} \]
- Time taken by the train to travel from A to B (including the halt):
\[ \text{Time}_{\text{train}} = \frac{78}{1.3x} + \frac{9}{60} \text{ hours} \]
(The halt of 9 minutes is converted to hours by dividing by 60.)

3. **Set Up the Equation:**
Since both the car and the train reach B at the same time, we can set their times equal to each other:
\[ \frac{78}{x} = \frac{78}{1.3x} + \frac{9}{60} \]

4. **Clear the Fractions:**
To eliminate the fractions, multiply through by \( 60x \):
\[ 60 \cdot 78 = 60 \cdot \frac{78}{1.3} + 9x \]
Simplifying gives:
\[ 4680 = \frac{4680}{1.3} + 9x \]

5. **Calculate \( \frac{4680}{1.3} \):**
\[ \frac{4680}{1.3} = 3600 \]
Now substitute back into the equation:
\[ 4680 = 3600 + 9x \]

6. **Solve for \( x \):**
Rearranging gives:
\[ 4680 - 3600 = 9x \]
\[ 1080 = 9x \]
\[ x = \frac{1080}{9} = 120 \text{ km/h} \]

7. **Conclusion:**
The speed of the car is \( 120 \) km/h.