Question

Ankita reaches her destination 20 minutes late if her speed is 3 km per hour and reaches 10 minutes before time if her speed is 4 km per hour. Find the distance of her destination from her starting point

• A. 5 Km
• B. 2 Km
• C. 4 Km
• D. 6 Km

Hint:

A faster shortcut formula for such problems is: \[ \text{Distance} = \frac{s_1 \cdot s_2}{|s_1 - s_2|} \times (\text{Difference in arrival times in hours}) \] Here, difference in time $= 20 - (-10) = 30 \text{ minutes} = 0.5 \text{ hours}$.
Distance $= \frac{3 \times 4}{|3 - 4|} \times 0.5 = 12 \times 0.5 = 6 \text{ km}$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This is a standard time, speed, and distance problem where the distance remains constant, but changes in speed result in different travel times.
A change in speed alters the time taken, and the relationship can be modeled algebraically.
Key Formula or Approach:
The fundamental relationship is: \[ \text{Distance} = \text{Speed} \times \text{Time} \] Let $d$ be the distance to the destination, and $t$ be the exact scheduled time in hours.
Ensure all time units are consistent (converting minutes to hours).

Step 2: Detailed Explanation:
Let's convert the given time variances into hours:
- 20 minutes late = $\frac{20}{60}$ hours = $\frac{1}{3}$ hours
- 10 minutes early = $\frac{10}{60}$ hours = $\frac{1}{6}$ hours
Let's construct equations for both cases:
Case 1: At speed $s_1 = 3 \text{ km/h}$, she is 20 minutes late: \[ \text{Time taken } t_1 = \frac{d}{3} \implies t_1 = t + \frac{1}{3} \] \[ t = \frac{d}{3} - \frac{1}{3} \quad \text{--- (Equation 1)} \] Case 2: At speed $s_2 = 4 \text{ km/h}$, she is 10 minutes early: \[ \text{Time taken } t_2 = \frac{d}{4} \implies t_2 = t - \frac{1}{6} \] \[ t = \frac{d}{4} + \frac{1}{6} \quad \text{--- (Equation 2)} \] Since $t$ is the same in both cases, equate Equation 1 and Equation 2: \[ \frac{d}{3} - \frac{1}{3} = \frac{d}{4} + \frac{1}{6} \] Rearrange terms to isolate the variable $d$: \[ \frac{d}{3} - \frac{d}{4} = \frac{1}{6} + \frac{1}{3} \] Find common denominators for both sides: \[ \frac{4d - 3d}{12} = \frac{1 + 2}{6} \] \[ \frac{d}{12} = \frac{3}{6} \] \[ \frac{d}{12} = \frac{1}{2} \] Solve for $d$: \[ d = 12 \times \frac{1}{2} = 6 \text{ km} \]

Step 3: Final Answer:
The distance of her destination from her starting point is 6 Km.