A car travels for 11 hours. For the first 100km the car travels with a certain speed, and then it increases its speed by 15 km h to cover the remaining 280 km. The time (in hours) it takes to travel the 100 km part is :

Options

A. 4
B. 7
C. 10
D. 11

Answer

To solve the problem step by step, let's break it down:

### Step 1: Define the Variables
Let the initial speed of the car be \( x \) km/h.

### Step 2: Write the Time Equation
The car travels 100 km at speed \( x \) and then travels 280 km at speed \( x + 15 \) km/h. The total time taken for the journey is 11 hours. We can express this as:
\[ \frac{100}{x} + \frac{280}{x + 15} = 11 \]

### Step 3: Clear the Fractions
To eliminate the fractions, we can multiply through by \( x(x + 15) \):
\[ 100(x + 15) + 280x = 11x(x + 15) \]

### Step 4: Expand and Rearrange
Expanding both sides gives:
\[ 100x + 1500 + 280x = 11x^2 + 165x \]
Combining like terms results in:
\[ 380x + 1500 = 11x^2 + 165x \]
Rearranging this equation leads to:
\[ 11x^2 - 215x - 1500 = 0 \]

### Step 5: Apply the Quadratic Formula
Now we will use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 11 \), \( b = -215 \), and \( c = -1500 \):
\[ x = \frac{-(-215) \pm \sqrt{(-215)^2 - 4 \cdot 11 \cdot (-1500)}}{2 \cdot 11} \]
Calculating the discriminant:
\[ (-215)^2 = 46225 \]
\[ 4 \cdot 11 \cdot 1500 = 66000 \]
Thus,
\[ b^2 - 4ac = 46225 + 66000 = 112225 \]

### Step 6: Solve for \( x \)
Now substituting back into the formula:
\[ x = \frac{215 \pm \sqrt{112225}}{22} \]
Calculating \( \sqrt{112225} = 335 \):
\[ x = \frac{215 \pm 335}{22} \]
This gives two possible solutions:
1. \( x = \frac{550}{22} = 25 \)
2. \( x = \frac{-120}{22} \) (not valid as speed cannot be negative)

Thus, the initial speed \( x = 25 \) km/h.

### Step 7: Calculate the Time for the First 100 km
Using the formula for time:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{100}{25} = 4 \text{ hours} \]

### Final Answer
The time taken to travel the first 100 km is **4 hours**.

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