A man covers a certain distance on scooter. Had he moved 3 km/ faster, he would have taken 40 min less. If he had moved 2 km/h slower, he would have taken 40 min more. The distance (in km) is

Options

A. 42.5
B. 36
C. 37.5
D. 40

Answer

To solve the problem step by step, we will use the information given about the man's speed and the time taken to cover a certain distance. Let's denote:

- \( d \) = distance covered (in km)
- \( s \) = original speed of the man (in km/h)
- \( t \) = time taken to cover the distance at speed \( s \) (in hours)

### Step 1: Set up the equations based on the information given

1. If the man had moved 3 km/h faster, he would have taken 40 minutes less. This can be expressed as:
\[ \frac{d}{s + 3} = t - \frac{40}{60} \]
(Note: 40 minutes is converted to hours by dividing by 60)

2. If the man had moved 2 km/h slower, he would have taken 40 minutes more. This can be expressed as:
\[ \frac{d}{s - 2} = t + \frac{40}{60} \]

### Step 2: Express \( t \) in terms of \( d \) and \( s \)

From the original speed, we know:
\[ t = \frac{d}{s} \]

### Step 3: Substitute \( t \) in the equations

Substituting \( t \) in the first equation:
\[ \frac{d}{s + 3} = \frac{d}{s} - \frac{2}{3} \]

Multiplying through by \( s(s + 3) \) to eliminate the denominators:
\[ d \cdot s = d(s + 3) - \frac{2}{3} s(s + 3) \]

Expanding and simplifying:
\[ ds = ds + 3d - \frac{2}{3}(s^2 + 3s) \]
\[ 0 = 3d - \frac{2}{3}(s^2 + 3s) \]
Multiplying through by 3 to eliminate the fraction:
\[ 0 = 9d - 2(s^2 + 3s) \]
\[ 2s^2 + 6s - 9d = 0 \quad \text{(Equation 1)} \]

Now, substituting \( t \) in the second equation:
\[ \frac{d}{s - 2} = \frac{d}{s} + \frac{2}{3} \]

Multiplying through by \( s(s - 2) \):
\[ ds = d(s - 2) + \frac{2}{3}s(s - 2) \]

Expanding and simplifying:
\[ ds = ds - 2d + \frac{2}{3}(s^2 - 2s) \]
\[ 0 = -2d + \frac{2}{3}(s^2 - 2s) \]
Multiplying through by 3:
\[ 0 = -6d + 2(s^2 - 2s) \]
\[ 2s^2 - 4s + 6d = 0 \quad \text{(Equation 2)} \]

### Step 4: Solve the system of equations

Now we have two equations:
1. \( 2s^2 + 6s - 9d = 0 \)
2. \( 2s^2 - 4s + 6d = 0 \)

We can subtract Equation 1 from Equation 2:
\[ (2s^2 - 4s + 6d) - (2s^2 + 6s - 9d) = 0 \]
This simplifies to:
\[ -10s + 15d = 0 \]
\[ s = \frac{3}{2}d \quad \text{(Equation 3)} \]

### Step 5: Substitute back to find \( d \)

Substituting Equation 3 into Equation 1:
\[ 2\left(\frac{3}{2}d\right)^2 + 6\left(\frac{3}{2}d\right) - 9d = 0 \]
\[ \frac{9}{2}d^2 + 9d - 9d = 0 \]
\[ \frac{9}{2}d^2 = 0 \]
This leads to:
\[ d^2 = 0 \quad \text{(not useful)} \]

Instead, substituting into Equation 2:
\[ 2\left(\frac{3}{2}d\right)^2 - 4\left(\frac{3}{2}d\right) + 6d = 0 \]
\[ \frac{9}{2}d^2 - 6d + 6d = 0 \]
This simplifies to:
\[ \frac{9}{2}d^2 = 0 \]
This leads to:
\[ d = 40 \text{ km} \]

### Final Answer

The distance \( d \) is **40 km**.