A train can travel 40% faster than a car. Both the train and the car start from point A at the same time and reach point , which is 70km away from point A at the same time. On the way, however, the train lost about 15 minutes while stopping a stations. The speed of the car in km/h is

Options

A. 120
B. 80
C. 90
D. 100

Answer

To solve the problem step by step, we will denote the speed of the car as \( S \) km/h. The speed of the train will then be \( 1.4S \) km/h since it is 40% faster than the car.

### Step 1: Establish the relationship between the speeds and the time taken.
Both the train and the car cover the same distance of 70 km, but the train stops for 15 minutes. We need to express the time taken by both vehicles in terms of their speeds.

- **Time taken by the car**:
\[ T_{car} = \frac{70}{S} \]

- **Time taken by the train** (including the 15 minutes lost):
\[ T_{train} = \frac{70}{1.4S} + \frac{15}{60} \]
(Note: 15 minutes is converted to hours by dividing by 60)

### Step 2: Set the times equal to each other.
Since both the train and the car reach the destination at the same time, we can set their times equal:
\[ \frac{70}{S} = \frac{70}{1.4S} + \frac{15}{60} \]

### Step 3: Simplify the equation.
To eliminate the fractions, we can multiply through by \( 1.4S \times 60 \) (the least common multiple of the denominators):
\[ 70 \times 60 \times 1.4 = 70 \times 60 + 15 \times 1.4S \]

This simplifies to:
\[ 5880 = 4200 + 21S \]

### Step 4: Solve for \( S \).
Rearranging the equation gives:
\[ 5880 - 4200 = 21S \]
\[ 1680 = 21S \]
\[ S = \frac{1680}{21} = 80 \text{ km/h} \]

### Conclusion:
The speed of the car is \( 80 \) km/h.