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. Represent the given problem in the form of a quadratic equation.

To represent the problem in the form of a quadratic equation, let's break it down step by step.

### Variables and Information:
- **Rohan's speed:** Let Rohan's speed be \( x \) km/h.
- **Eshaan's speed:** Since Eshaan travels 12 km/h faster than Rohan, his speed is \( x + 12 \) km/h.
- **Distance to school:** 4 km.
- **Rohan's starting time:** 7:30 am.
- **Eshaan's starting time:** 8:15 am.
- They both reach school at the same time.

### Time calculations:
- **Rohan's time:** Rohan walks the 4 km at speed \( x \) km/h. The time Rohan takes to reach school is:
  \[
  t_{\text{Rohan}} = \frac{4}{x} \, \text{hours}.
  \]
  
- **Eshaan's time:** Eshaan starts 45 minutes (or \( \frac{3}{4} \) hours) later but reaches school at the same time. So, the time Eshaan spends cycling is:
  \[
  t_{\text{Eshaan}} = \frac{4}{x+12} \, \text{hours}.
  \]
  
- Since Eshaan started 45 minutes later, Rohan has a head start of \( \frac{3}{4} \) hours. Therefore, the time Eshaan takes must be \( \frac{3}{4} \) hours less than Rohan's time:
  \[
  t_{\text{Rohan}} - t_{\text{Eshaan}} = \frac{3}{4}.
  \]

### Formulating the equation:
Using the expressions for the times:

\[
\frac{4}{x} - \frac{4}{x+12} = \frac{3}{4}.
\]

### Clearing the fractions:
Multiply the entire equation by \( 4x(x + 12) \) to eliminate the denominators:

\[
4(x + 12) - 4x = 3x(x + 12).
\]

Simplifying both sides:

\[
4x + 48 - 4x = 3x^2 + 36x.
\]

This simplifies further to:

\[
48 = 3x^2 + 36x.
\]

### Final quadratic equation:
Divide both sides by 3 to simplify:

\[
16 = x^2 + 12x.
\]

Rearranging the terms:

\[
x^2 + 12x - 16 = 0.
\]

This is the quadratic equation representing the given problem.

### Do you want details or have any questions?

---

**Relative Questions:**
1. How can we solve this quadratic equation to find Rohan's speed?
2. What happens if Eshaan's speed were 15 km/h faster than Rohan's instead of 12?
3. How does changing the distance to school affect the quadratic equation?
4. What is the significance of Rohan's head start in the equation?
5. How could the situation change if both start at the same time?

**Tip:** When setting up word problems, carefully define all variables and translate each condition step by step into an equation.

Learn how to represent and solve the problem where Rohan walks 4 km to school while Eshaan cycles 12 km/h faster, starting 45 minutes later. They both reach school at the same time. This solution explains the algebraic setup, including quadratic equations, distance-speed-time relationships, and step-by-step solving. Suitable for Grades 8-10.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation