Magan covered a certain distance at his normal speed. If he travels 20 km/h faster, it takes him 2 hours less to cover the same distance. If he travels 10 km/h slower, it takes him 1 hour 36 minutes more to cover the given distance. What is his time if he travels with normal speed
Options:
A. 7 hours
B. 7.5 hours
C. 8 hours
D. 8.5 hours
E. 6 hours



Let \( d \) be the distance Magan travels and \( v \) be his normal speed in km/h. 

We are given two key conditions:

1. If he travels 20 km/h faster, it takes him 2 hours less.
   - His time at normal speed is \( \frac{d}{v} \).
   - His time at 20 km/h faster is \( \frac{d}{v+20} \).
   - According to the problem, \( \frac{d}{v} - \frac{d}{v+20} = 2 \) hours.
   
   This gives us the equation:
   \[
   \frac{d}{v} - \frac{d}{v+20} = 2
   \]

2. If he travels 10 km/h slower, it takes him 1 hour and 36 minutes (or 1.6 hours) more.
   - His time at 10 km/h slower is \( \frac{d}{v-10} \).
   - According to the problem, \( \frac{d}{v-10} - \frac{d}{v} = 1.6 \) hours.
   
   This gives us the equation:
   \[
   \frac{d}{v-10} - \frac{d}{v} = 1.6
   \]

We now have two equations:
1. \( \frac{d}{v} - \frac{d}{v+20} = 2 \)
2. \( \frac{d}{v-10} - \frac{d}{v} = 1.6 \)

Let's solve these equations step by step.

### Step 1: Solve for \( d \) in terms of \( v \)
From equation 1:
\[
\frac{d}{v} - \frac{d}{v+20} = 2
\]
Multiply through by \( v(v+20) \) to eliminate the denominators:
\[
d(v+20) - dv = 2v(v+20)
\]
Simplify:
\[
20d = 2v(v+20)
\]
\[
d = \frac{v(v+20)}{10}
\]
Now we have \( d \) in terms of \( v \).

### Step 2: Substitute \( d \) into the second equation
Substitute \( d = \frac{v(v+20)}{10} \) into equation 2:
\[
\frac{d}{v-10} - \frac{d}{v} = 1.6
\]
This becomes:
\[
\frac{\frac{v(v+20)}{10}}{v-10} - \frac{\frac{v(v+20)}{10}}{v} = 1.6
\]
Multiply through by 10 to simplify:
\[
\frac{v(v+20)}{v-10} - \frac{v(v+20)}{v} = 16
\]
Factor out \( v(v+20) \):
\[
v(v+20) \left( \frac{1}{v-10} - \frac{1}{v} \right) = 16
\]
Simplify the expression inside the parentheses:
\[
\frac{v - (v-10)}{v(v-10)} = \frac{10}{v(v-10)}
\]
Substitute back:
\[
v(v+20) \cdot \frac{10}{v(v-10)} = 16
\]
Cancel \( v \) from both sides:
\[
(v+20) \cdot \frac{10}{v-10} = 16
\]
Multiply both sides by \( (v-10) \):
\[
10(v+20) = 16(v-10)
\]
Expand both sides:
\[
10v + 200 = 16v - 160
\]
Simplify:
\[
200 + 160 = 16v - 10v
\]
\[
360 = 6v
\]
\[
v = 60
\]

### Step 3: Calculate the time
Now that we know \( v = 60 \) km/h, we can find the time it takes to travel the distance at normal speed:
\[
\text{Time} = \frac{d}{v}
\]
Substitute \( d = \frac{v(v+20)}{10} \):
\[
d = \frac{60(60+20)}{10} = \frac{60 \times 80}{10} = 480 \, \text{km}
\]
Thus, the time is:
\[
\text{Time} = \frac{480}{60} = 8 \, \text{hours}
\]

### Final Answer: 
The time it takes Magan to cover the distance at his normal speed is \( \boxed{8 \text{ hours}} \). Thus, the correct option is **C**.

---

Do you want more details or have any questions?

Here are 5 related questions to expand your understanding:
1. How would the time change if Magan's normal speed were 50 km/h instead of 60 km/h?
2. What if the speed increase and decrease were different values? How would that affect the solution?
3. How would the distance Magan travels be calculated if the speed were not known initially?
4. What other methods could be used to solve similar rate and distance problems?
5. Could this type of problem be solved graphically? How would you represent it?

**Tip:** Always pay attention to units of time, especially when they are given in hours and minutes, and convert them appropriately.

Magan covered a certain distance at his normal speed. If he travels 20 km/h faster, it takes him 2 hours less to cover the same distance. If he travels 10 km/h slower, it takes him 1 hour 36 minutes more to cover the given distance. What is his time if he travels with normal speed? Options: A. 7 hours B. 7.5 hours C. 8 hours D. 8.5 hours E. 6 hours

Learn how to solve a real-world speed and distance problem where Magan's travel time changes with varying speeds. This detailed solution covers key algebraic concepts and formulas, suitable for students in Grades 10-12.

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