A person leaves city J at 7 am and reaches city K at 5 pm. Another person leaves from city K at 4 pm and reaches city J at 10 pm. At what time (approximately) they both will meet?

Correct option is D

Given:

• Person A: Leaves city J at 7:00 am, reaches city K at 5:00 pm
• Person B: Leaves city K at 4:00 pm, reaches city J at 10:00 pm
• We are to find at what time they meet.

Solution:

Let’s assume the distance between city J and city K is same for both and consider constant speeds.

Step 1: Travel time

• Person A's total time = 7:00 am to 5:00 pm = 10 hours
• Person B's total time = 4:00 pm to 10:00 pm = 6 hours

Let the total distance between J and K = LCM of 10 and 6 = 60 km (assumption for easy calculation)

​$$\text{Speed of Person A} = \frac{60}{10} = 6 \text{ km/h} \\\ \\\text{Speed of Person B} = \frac{60}{6} = 10 \text{ km/h}$$​​

Step 2: At 4:00 pm (when Person B starts), how far has Person A traveled?

From 7:00 am to 4:00 pm = 9 hours
=> Distance by A = 9 × 6 = 54 km

So, A is 6 km away from K
B starts from K toward J at 10 km/h
A is moving from J toward K at 6 km/h
So, their relative speed = 6 + 10 = 16 km/h

Step 3: Time to meet after 4:00 pm

$$\text{Time to meet} = \frac{6}{16} = 0.375 \text{ hours} = 22.5 \text{ minutes} \\\text{Meeting time} = 4:00 \text{ pm } + 22.5 \text{ minutes} = \boxed{4:23 \text{ pm (approx)}}$$

​Final Answer: (D) 4:23 pm