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Bunuel
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A thief escapes from city A at 2 pm and flees towards city B at 40 kmp 03 Jan 2020, 02:11
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D
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Difficulty:

65% (hard)

Question Stats:

66% (02:35) correct 34% (02:51) wrong based on 282 sessions

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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
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D
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A thief escapes from city A at 2 pm and flees towards city B at 40 kmp 03 Jan 2020, 11:44
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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
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At 3 pm, the thief is 40 km away from the police in city A. The police from city A is 10 kmph faster than that of the thief, it will take 40 km / (10 km/h) = 4h for the police in city A to catch up. So we need both A and B to travel 4 hours, (50 + 60) km /h * 4 h = 440 km.

Ans: D
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A thief escapes from city A at 2 pm and flees towards city B at 40 kmp 08 Jan 2024, 05:09
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Thief has a head start of one hour. Within that one hour thief has travelled

d = speed x time

= 40 x 1

40km


Time it will take police team A to catch up with him


t =40/10

= 4hours


Distance police team will cover in the given period

d = 50 x 4

= 200km

Now to get to that point thief, will have to travel 160km.


Police team B will approach thief from opposite direction.

Let x be the distance they will cover to catch thief.

And time thief will cover 160km

160/40

=4hours.


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.

Answer D
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A thief escapes from city A at 2 pm and flees towards city B at 40 kmp 07 Mar 2024, 05:07
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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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A thief escapes from city A at 2 pm and flees towards city B at 40 kmp 07 Mar 2024, 09:26
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RajibDhara
­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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­
Here is a better approach to stick to: 

https://gmatclub.com/forum/a-thief-esca ... l#p2433823­
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A thief escapes from city A at 2 pm and flees towards city B at 40 kmp 01 Jan 2026, 11:20
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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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A thief escapes from city A at 2 pm and flees towards city B at 40 kmp 02 Jan 2026, 00:19
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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
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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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A thief escapes from city A at 2 pm and flees towards city B at 40 kmp 03 Jan 2026, 11:22
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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
Show more
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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