Rohan starts to walk 4 km to school at 7:30 am and her brother Eshaan starts at 8:15 am on his bicycle, travelling 12 km/hour faster than Rohan. They reach school at the same time. Find the speed of Rohan.

Let \( v \) represent the speed of Rohan in km/h.

### Step 1: Time taken by Rohan
Rohan starts at 7:30 am and travels 4 km to school. His time of travel is the difference between when he starts and when he arrives at school. Let \( t_R \) represent the time Rohan spends traveling.

The time taken by Rohan is:
\[
t_R = \frac{4}{v}
\]

### Step 2: Time taken by Eshaan
Eshaan starts at 8:15 am, so he starts 45 minutes (or 0.75 hours) later than Rohan. Since both reach school at the same time, the time Eshaan spends traveling, \( t_E \), is related to Rohan's time as:
\[
t_E = t_R - 0.75
\]

Eshaan’s speed is \( v + 12 \) km/h, since he is traveling 12 km/h faster than Rohan. Therefore, the time taken by Eshaan is:
\[
t_E = \frac{4}{v + 12}
\]

### Step 3: Set up the equation
Since Eshaan’s time of travel \( t_E \) is \( t_R - 0.75 \), we can equate the expressions for time:
\[
\frac{4}{v + 12} = \frac{4}{v} - 0.75
\]

### Step 4: Solve the equation
First, clear the fractions by multiplying both sides by \( v(v + 12) \):
\[
4v = 4(v + 12) - 0.75v(v + 12)
\]
Simplifying step-by-step:
\[
4v = 4v + 48 - 0.75v(v + 12)
\]
Canceling \( 4v \) from both sides:
\[
0 = 48 - 0.75v(v + 12)
\]
Expand the right-hand side:
\[
0 = 48 - 0.75(v^2 + 12v)
\]
\[
0 = 48 - 0.75v^2 - 9v
\]
Rearranging the equation:
\[
0.75v^2 + 9v - 48 = 0
\]

### Step 5: Solve the quadratic equation
Divide the equation by 0.75:
\[
v^2 + 12v - 64 = 0
\]
Use the quadratic formula:
\[
v = \frac{-12 \pm \sqrt{12^2 - 4(1)(-64)}}{2(1)}
\]
\[
v = \frac{-12 \pm \sqrt{144 + 256}}{2}
\]
\[
v = \frac{-12 \pm \sqrt{400}}{2}
\]
\[
v = \frac{-12 \pm 20}{2}
\]
The two possible solutions are:
\[
v = \frac{-12 + 20}{2} = 4 \quad \text{or} \quad v = \frac{-12 - 20}{2} = -16
\]
Since speed cannot be negative, we have \( v = 4 \) km/h.

### Conclusion:
Rohan's speed is **4 km/h**.

Would you like a detailed explanation of any of these steps? Here are 5 related questions:

1. How does the quadratic formula work in general?
2. Can we solve this using a different method?
3. Why did we discard the negative solution for speed?
4. How would the time calculation change if Eshaan's speed were different?
5. Can this problem be generalized for other distances or times?

**Tip:** For distance-rate-time problems, setting up a clear relationship between the two travelers' times helps simplify the problem.

Learn how to find Rohan's speed using a quadratic equation in this time-distance problem. Rohan walks 4 km to school while his brother Eshaan starts later, traveling 12 km/h faster. Discover step-by-step how to solve for the speed of Rohan using algebra and the quadratic formula. Suitable for Grades 8-10.

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