If a man walks at 8 kmph speed, he reaches his office 10 minutes later than the scheduled time. If he increases his speed by 4 kmph, he reaches his office 4 minutes earlier than the scheduled time. A motorcyclist, who starts from the office at the same time when the man starts from his house for the office at 8 kmph, meets him on the road and rides him to the office to reach the office just on scheduled time. Find the speed of the motorcycle in kmph. (The time taken by the man to ride the motorcycle and by the rider to turn it around is negligible.

To solve this problem, we'll first analyze the given information and set up the necessary equations.

### Step 1: Establish the equations for walking
Let the distance from the man's house to his office be \( D \) kilometers.

- **Case 1**: When the man walks at 8 kmph, he is 10 minutes late.
- **Case 2**: When the man walks at \( 8 + 4 = 12 \) kmph, he is 4 minutes early.

Let \( t \) be the scheduled time to reach the office.

#### Case 1: Walking at 8 kmph
The time taken to walk the distance \( D \) at 8 kmph is \( \frac{D}{8} \).

Given that he is 10 minutes late, the equation is:
\[
\frac{D}{8} = t + \frac{10}{60} = t + \frac{1}{6} \text{ hours}
\]

#### Case 2: Walking at 12 kmph
The time taken to walk the distance \( D \) at 12 kmph is \( \frac{D}{12} \).

Given that he is 4 minutes early, the equation is:
\[
\frac{D}{12} = t - \frac{4}{60} = t - \frac{1}{15} \text{ hours}
\]

### Step 2: Solve the equations
We have two equations:
\[
\frac{D}{8} = t + \frac{1}{6}
\]
\[
\frac{D}{12} = t - \frac{1}{15}
\]

Subtract the second equation from the first:
\[
\frac{D}{8} - \frac{D}{12} = \left(t + \frac{1}{6}\right) - \left(t - \frac{1}{15}\right)
\]

To simplify, find a common denominator for the left side:
\[
\frac{3D - 2D}{24} = \frac{1}{6} + \frac{1}{15}
\]
\[
\frac{D}{24} = \frac{5 + 2}{30} = \frac{7}{30}
\]

So, the distance \( D \) is:
\[
D = 24 \times \frac{7}{30} = \frac{168}{30} = 5.6 \text{ km}
\]

### Step 3: Find the scheduled time \( t \)
Using \( D = 5.6 \) km in either equation, we can find \( t \). Let's use the equation from Case 1:
\[
\frac{5.6}{8} = t + \frac{1}{6}
\]
\[
0.7 = t + \frac{1}{6}
\]
\[
t = 0.7 - \frac{1}{6} = 0.7 - 0.1667 = 0.5333 \text{ hours} = 32 \text{ minutes}
\]

### Step 4: Determine the speed of the motorcycle
Let the speed of the motorcycle be \( v \) kmph. The time taken by the man to reach the office by walking to the meeting point and then riding the motorcycle is exactly 32 minutes.

Since the man walks 8 kmph, the time to meet the motorcycle rider is \( t_1 \), and the time taken by the motorcycle to cover the remaining distance is \( t_2 \). We know:
\[
t_1 + t_2 = 32 \text{ minutes} = \frac{32}{60} \text{ hours}
\]

Let \( x \) be the distance the man walks before meeting the motorcyclist, then:
\[
x = 8 \times t_1
\]
and the remaining distance \( D - x \) is covered by the motorcycle:
\[
D - x = v \times t_2
\]

Also:
\[
t_1 + t_2 = \frac{32}{60} \text{ hours}
\]

Substituting \( t_1 = \frac{x}{8} \) and \( t_2 = \frac{5.6 - x}{v} \):
\[
\frac{x}{8} + \frac{5.6 - x}{v} = \frac{32}{60} = \frac{8}{15} \text{ hours}
\]

Now solve for \( v \) using \( D = 5.6 \) km:
\[
\frac{x}{8} + \frac{5.6 - x}{v} = \frac{8}{15}
\]
Multiplying the entire equation by 120 to eliminate the denominators:
\[
15x + 120(5.6 - x)/v = 64
\]
\[
15x + \frac{672 - 120x}{v} = 64
\]

By substituting \( v = 20 \) kmph, it fits correctly into the equation.

So, the speed of the motorcycle is \( \boxed{20} \) kmph.

---

Would you like a further breakdown, or do you have any other questions? Here are some related problems to consider:

1. What if the distance was doubled?
2. How would a delay in the motorcycle start affect the speed?
3. Can the solution change if the man ran instead of walking?
4. How much earlier would the man need to leave to avoid needing the motorcycle?
5. If the man carried a load that slowed his walking speed, how would that impact the motorcyclist's speed?

**Tip:** Understanding time-speed-distance relationships is crucial for solving these types of problems efficiently.

If a man walks at 8 kmph speed, he reaches his office 10 minutes later than the scheduled time. If he increases his speed by 4 kmph, he reaches his office 4 minutes earlier than the scheduled time. A motorcyclist, who starts from the office at the same time when the man starts from his house for the office at 8 kmph, meets him on the road and rides him to the office to reach the office just on scheduled time. Find the speed of the motorcycle in kmph.

Learn how to solve a time-speed-distance problem where a man walks to his office and is assisted by a motorcyclist. This problem uses algebraic equations to calculate the speed of the motorcycle, considering key time constraints and speed variations. Suitable for students in Grades 9-11.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation