A man covered a certain distance at some speed Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower; he would have taken 40 minutes more. The distance (in km) is `35` b. `40` c. `36 2/3` d. `37 1/2`

Options

A. 35
B. `36(2)/(3)`
C. `37(1)/(2)`
D. 40

Answer

Let's solve the problem step by step.

Given:
1. If the man had moved 3 km/h faster, he would have taken 40 minutes (or 2/3 hour) less.
2. If he had moved 2 km/h slower, he would have taken 40 minutes (or 2/3 hour) more.

Let:
- \( d \) be the distance (in km).
- \( s \) be the original speed (in km/h).

### Step 1: Formulate the equations

#### Faster Speed Scenario
If the man had moved 3 km/h faster, his speed would be \( s + 3 \) km/h and he would have taken 40 minutes less.

Original time taken = \( \frac{d}{s} \)

New time taken = \( \frac{d}{s+3} \)

Given that the new time is 40 minutes (or \( \frac{2}{3} \) hour) less than the original time:
\[ \frac{d}{s} - \frac{d}{s+3} = \frac{2}{3} \]

#### Slower Speed Scenario
If the man had moved 2 km/h slower, his speed would be \( s - 2 \) km/h and he would have taken 40 minutes more.

New time taken = \( \frac{d}{s-2} \)

Given that the new time is 40 minutes (or \( \frac{2}{3} \) hour) more than the original time:
\[ \frac{d}{s-2} - \frac{d}{s} = \frac{2}{3} \]

### Step 2: Solve the equations

#### Equation 1:
\[ \frac{d}{s} - \frac{d}{s+3} = \frac{2}{3} \]

Multiply through by \( s(s+3) \):
\[ d(s+3) - ds = \frac{2}{3} s(s+3) \]
\[ ds + 3d - ds = \frac{2}{3} s^2 + \frac{2}{3} 3s \]
\[ 3d = \frac{2}{3} s^2 + 2s \]
Multiply through by 3:
\[ 9d = 2s^2 + 6s \]
\[ 9d = 2s(s + 3) \] ... (1)

#### Equation 2:
\[ \frac{d}{s-2} - \frac{d}{s} = \frac{2}{3} \]

Multiply through by \( s(s-2) \):
\[ d(s) - d(s-2) = \frac{2}{3} s(s-2) \]
\[ ds - ds + 2d = \frac{2}{3} s^2 - \frac{2}{3} 2s \]
\[ 2d = \frac{2}{3} s^2 - \frac{4}{3} s \]
Multiply through by 3:
\[ 6d = 2s^2 - 4s \]
\[ 6d = 2s(s - 2) \] ... (2)

### Step 3: Solve for \( s \) and \( d \)

From equations (1) and (2):
\[ 9d = 2s(s + 3) \]
\[ 6d = 2s(s - 2) \]

Divide the first equation by the second:
\[ \frac{9d}{6d} = \frac{2s(s + 3)}{2s(s - 2)} \]
\[ \frac{3}{2} = \frac{s + 3}{s - 2} \]

Cross-multiply:
\[ 3(s - 2) = 2(s + 3) \]
\[ 3s - 6 = 2s + 6 \]
\[ s = 12 \]

Substitute \( s = 12 \) into either equation (1) or (2):
\[ 9d = 2 \cdot 12 \cdot (12 + 3) \]
\[ 9d = 2 \cdot 12 \cdot 15 \]
\[ 9d = 360 \]
\[ d = 40 \]

So, the distance is \( \boxed{40} \) km.