Question

A boy walks to cover certain distance. If he walked 2 km/hr faster he would have taken 1 hour less. If he had moved 1 km/hr slower he would have taken 1 hour more. The distance travelled is :

• A. 12 km
• B. 10 km
• C. 14 km
• D. 8 km

Answer 1

Answer:

A. 12 km

Let the original speed be $S$ km/h and the total distance be $D$ km.

1. Formula Shortcut

For any change in speed and time where distance remains constant, use:
$\text{Distance} = \frac{S \times (S \pm \Delta S)}{\Delta S} \times \Delta T$

Where:

• $\Delta S$ = change in speed

• $\Delta T$ = change in time

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2. Set Up the Equations

• Case 1: Moving 2 km/h faster ($+2$) saves 1 hour ($-1$).
  $D = \frac{S(S + 2)}{2} \times 1 \quad \text{--- (Equation 1)}$

• Case 2: Moving 1 km/h slower ($-1$) takes 1 hour more ($+1$).
  $D = \frac{S(S - 1)}{1} \times 1 \quad \text{--- (Equation 2)}$

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3. Find the Original Speed ($S$)

Equate both expressions since both equal the same distance $D$:
$\frac{S(S + 2)}{2} = S(S - 1)$

Cancel out the common $S$ from both sides:
$\frac{S + 2}{2} = S - 1$

Cross-multiply by 2:
$S + 2 = 2(S - 1)$
$S + 2 = 2S - 2$
$S = 4\text{ km/h}$

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4. Calculate Total Distance

Substitute $S = 4$ into Equation 2:
$D = 4 \times (4 - 1) = 4 \times 3 = \mathbf{12\text{ km}}$