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­So for them to catch thief

4 = 160 + x/60 + 40


400 = 160 + x

x = 240km

Therefore distance between A and B

200 + 240 = 440km.
 COULD YOU EXPLAIN ME THIS PART FROM THE STARTING
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­So for them to catch thief

4 = 160 + x/60 + 40


400 = 160 + x

x = 240km

Therefore distance between A and B

200 + 240 = 440km.
 COULD YOU EXPLAIN ME THIS PART FROM THE STARTING
­
Here is a better approach to stick to: 

https://gmatclub.com/forum/a-thief-esca ... l#p2433823­
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let distance between the towns=d and x is the distance from A at which the thief is nabbed.
so x/(50-40)=(d-x)/(40+60)-----time is constant
x/10 = (d-x)/100
d=11x only choice that is the multiple of 11 is 440 kms D
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Bunuel
A thief escapes from city A at 2 pm and flees towards city B at 40 kmph. At 3 pm, the police realize the escape and start chasing the thief at 50 kmph. Simultaneously, a police team from station B also starts towards city A to apprehend the thief at a speed of 60 kmph. What should be the distance between A and B such that both the police team nab the thief at the same time?

A. 200 km
B. 240 km
C. 400 km
D. 440 km
E. 480 km
So let's assume

Total distance is T

X is the distance between Thief from police A

so, time is equal here because both will catch the thief at same time so :

X/10 = (T-X)/100 ( D/R = T, and if you are wondering what is 10 and 100, it's relative rate for both police A and B respectively w.r.t thief )

110X = 10T

11x = T

so from here we know total distance is multiple of 11, so D is the only option here, which is multiple of 11.

Ans - D
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Bunuel
A thief escapes from city A at 2 pm and flees towards city B at 40 kmph. At 3 pm, the police realize the escape and start chasing the thief at 50 kmph. Simultaneously, a police team from station B also starts towards city A to apprehend the thief at a speed of 60 kmph. What should be the distance between A and B such that both the police team nab the thief at the same time?

A. 200 km
B. 240 km
C. 400 km
D. 440 km
E. 480 km
Deconstructing the Question
Scenario:
1. Thief starts at 2 pm from A at 40 kmph.
2. Police A starts at 3 pm from A at 50 kmph (chasing).
3. Police B starts at 3 pm from B at 60 kmph (approaching).
Condition: Both police teams nab the thief at the same time.

Step 1: Analyze the Chase from City A
At 3 pm (when Police A starts), the thief has been moving for 1 hour (2 pm to 3 pm).
\(\text{Head Start Distance} = 40 \text{ kmph} \times 1 \text{ hr} = 40 \text{ km}\).

Relative Speed (Police A catching Thief):
\(V_{rel} = 50 - 40 = 10 \text{ kmph}\).

Time required for Police A to catch the thief:
\(t = \frac{\text{Distance Gap}}{\text{Relative Speed}} = \frac{40}{10} = 4 \text{ hours}\).
So, the thief is caught 4 hours after 3 pm (at 7 pm).

Step 2: Calculate Distances
In these 4 hours (from 3 pm to 7 pm):
1. Distance covered by Police A (from City A):**
\(D_A = 50 \text{ kmph} \times 4 \text{ hrs} = 200 \text{ km}\).
This is the distance of the meeting point from City A.

2. Distance covered by Police B (from City B):**
Police B travels towards the meeting point for the same time (4 hours).
\(D_B = 60 \text{ kmph} \times 4 \text{ hrs} = 240 \text{ km}\).
This is the distance of the meeting point from City B.

Step 3: Calculate Total Distance
The total distance between A and B is the sum of the distances covered from each end to the meeting point.
\(\text{Total Distance} = D_A + D_B\)
\(\text{Total Distance} = 200 + 240 = 440 \text{ km}\).

Answer: D
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