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Car A, travelling at 60kmph leaves City X for City Y at 5pm. Car B lea

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Car A, travelling at 60kmph leaves City X for City Y at 5pm. Car B lea  [#permalink]

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New post 17 Mar 2017, 04:41
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Question Stats:

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Car A, travelling at 60kmph leaves City X for City Y at 5pm. Car B leaves city X for City Y at 7pm and travels at 100 kmph. When will Car B overtake Car A?

A) 9:45 p.m

B) 9:55 p.m

C) 10:00 p.m

D) 10:05 p.m

E) 7:55 p.m

Source: www.f1gmat.com
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Re: Car A, travelling at 60kmph leaves City X for City Y at 5pm. Car B lea  [#permalink]

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New post 17 Mar 2017, 06:00
1
vikasp99 wrote:
Car A, travelling at 60kmph leaves City X for City Y at 5pm. Car B leaves city X for City Y at 7pm and travels at 100 kmph. When will Car B overtake Car A?

A) 9:45 p.m

B) 9:55 p.m

C) 10:00 p.m

D) 10:05 p.m

E) 7:55 p.m

Source: http://www.f1gmat.com

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Sent from my ONE E1003 using GMAT Club Forum mobile app
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Re: Car A, travelling at 60kmph leaves City X for City Y at 5pm. Car B lea  [#permalink]

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New post 17 Mar 2017, 06:00
appu2693 wrote:
vikasp99 wrote:
Car A, travelling at 60kmph leaves City X for City Y at 5pm. Car B leaves city X for City Y at 7pm and travels at 100 kmph. When will Car B overtake Car A?

A) 9:45 p.m

B) 9:55 p.m

C) 10:00 p.m

D) 10:05 p.m

E) 7:55 p.m

Source: http://www.f1gmat.com

Image

Sent from my ONE E1003 using GMAT Club Forum mobile app

Option C 10pm

Sent from my ONE E1003 using GMAT Club Forum mobile app
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Re: Car A, travelling at 60kmph leaves City X for City Y at 5pm. Car B lea  [#permalink]

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New post 19 Mar 2017, 06:36
2
Car A takes a lead of 2 hrs.
So the distance traveled by car A in 2 hours=60*2=120 km
This is the gap between the two cars which needs to be covered by car B in order to overtake car A
Rate with which this gap will be closed = 100-60=40 k/hr
using the formula
D=RT
120=40*T=3 hrs
7+3=10 pm
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Re: Car A, travelling at 60kmph leaves City X for City Y at 5pm. Car B lea  [#permalink]

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New post 20 Mar 2017, 02:31
vikasp99 wrote:
Car A, travelling at 60kmph leaves City X for City Y at 5pm. Car B leaves city X for City Y at 7pm and travels at 100 kmph. When will Car B overtake Car A?

A) 9:45 p.m

B) 9:55 p.m

C) 10:00 p.m

D) 10:05 p.m

E) 7:55 p.m

Source: http://www.f1gmat.com


At 7pm car A will be 50*2 = 100km away
relative speed = 100-60 = 40kmph

time taken to cover this distance with this relative speed = 100/40 = 2.5 hours or 2 hrs 30 minutes

therefore at 7 +2h 30 minutes = 9:30 it will overtake
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Car A, travelling at 60kmph leaves City X for City Y at 5pm. Car B lea  [#permalink]

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New post 07 Apr 2017, 17:22
vikasp99 wrote:
Car A, travelling at 60kmph leaves City X for City Y at 5pm. Car B leaves city X for City Y at 7pm and travels at 100 kmph. When will Car B overtake Car A?

A) 9:45 p.m

B) 9:55 p.m

C) 10:00 p.m

D) 10:05 p.m

E) 7:55 p.m



Common sense approach:

Car A -

Speed: 60km/h
Starts at 5pm;

@6pm: 60km travelled
@7pm: 120km travelled
@8pm: 180km travelled
@9pm: 240km travelled
@10pm: 300km travelled


Car B

Speed: 100km/h
Starts at 7pm;

@8pm: 100km travelled
@9pm: 200km travelled
@10pm: 300km travelled //This is where the overtake begins
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Car A, travelling at 60kmph leaves City X for City Y at 5pm. Car B lea  [#permalink]

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New post 07 Apr 2017, 19:34
vikasp99 wrote:
Car A, travelling at 60kmph leaves City X for City Y at 5pm. Car B leaves city X for City Y at 7pm and travels at 100 kmph. When will Car B overtake Car A?

A) 9:45 p.m

B) 9:55 p.m

C) 10:00 p.m

D) 10:05 p.m

E) 7:55 p.m

Source: http://www.f1gmat.com


60t=100(t-2)
t=5 hours
5:00pm+5 hours=10:00pm
C
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Car A, travelling at 60kmph leaves City X for City Y at 5pm. Car B lea  [#permalink]

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New post 08 Apr 2017, 14:01
I think the easier method to solve this problem is process elimination

The answer is C by process elimination

PROOF:
By 7pm, Car A has driven 60 x 2 = 120

By 10 pm,
Car A has driven
120 + (3 x 60)
120 + 180
300

Car B has driven
100 x 3
300

Therefore by 10pm, Car B will over Car A

The answer is C
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Re: Car A, travelling at 60kmph leaves City X for City Y at 5pm. Car B lea  [#permalink]

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New post 09 Apr 2017, 21:27
ByjusGMATapp wrote:
vikasp99 wrote:
Car A, travelling at 60kmph leaves City X for City Y at 5pm. Car B leaves city X for City Y at 7pm and travels at 100 kmph. When will Car B overtake Car A?

A) 9:45 p.m

B) 9:55 p.m

C) 10:00 p.m

D) 10:05 p.m

E) 7:55 p.m

Source: http://www.f1gmat.com


At 7pm car A will be 50*2 = 100km away
relative speed = 100-60 = 40kmph

time taken to cover this distance with this relative speed = 100/40 = 2.5 hours or 2 hrs 30 minutes

therefore at 7 +2h 30 minutes = 9:30 it will overtake



You made a typo in the first sentence - since speed is 60 kmph; in 2 hours it will cover 120 kms and not 100 kms.
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Re: Car A, travelling at 60kmph leaves City X for City Y at 5pm. Car B lea  [#permalink]

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New post 18 Apr 2017, 15:18
vikasp99 wrote:
Car A, travelling at 60kmph leaves City X for City Y at 5pm. Car B leaves city X for City Y at 7pm and travels at 100 kmph. When will Car B overtake Car A?

A) 9:45 p.m

B) 9:55 p.m

C) 10:00 p.m

D) 10:05 p.m

E) 7:55 p.m


We have a catch-up rate problem in which we can use the following formula:

distance of car A = distance of car B

Since car A leaves at 5 p.m. and car B leaves at 7 p.m., we can let the time of car A = t + 2 and the time of car B = t.

Since car A was traveling at 60 kmph and car B was traveling at 100 kmph, the distance of car A is 60(t + 2) = 60t + 120 and the distance of car B is 100t. Thus:

distance of car A = distance of car B

60t + 120 = 100t

120 = 40t

3 = t

Thus, car B overtook car A at 7 p.m. + 3 hours = 10 p.m.

Answer: C
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Car A, travelling at 60kmph leaves City X for City Y at 5pm. Car B lea  [#permalink]

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New post 22 Apr 2017, 06:01
vikasp99 wrote:
Car A, travelling at 60kmph leaves City X for City Y at 5pm. Car B leaves city X for City Y at 7pm and travels at 100 kmph. When will Car B overtake Car A?

A) 9:45 p.m

B) 9:55 p.m

C) 10:00 p.m

D) 10:05 p.m

E) 7:55 p.m

Source: http://www.f1gmat.com


Overtaking means outdistancing/leaving behind the other vehicle in question. At 10:00 PM the car A and car B are at the same point and no one has over taken the other. It only happens post 10:00 PM. So in absence of a better answer I marked D. Where did I go wrong?
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Re: Car A, travelling at 60kmph leaves City X for City Y at 5pm. Car B lea  [#permalink]

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New post 22 Apr 2017, 06:20
RK84 wrote:
vikasp99 wrote:
Car A, travelling at 60kmph leaves City X for City Y at 5pm. Car B leaves city X for City Y at 7pm and travels at 100 kmph. When will Car B overtake Car A?

A) 9:45 p.m

B) 9:55 p.m

C) 10:00 p.m

D) 10:05 p.m

E) 7:55 p.m

Source: http://www.f1gmat.com


Overtaking means outdistancing/leaving behind the other vehicle in question. At 10:00 PM the car A and car B are at the same point and no one has over taken the other. It only happens post 10:00 PM. So in absence of a better answer I marked D. Where did I go wrong?


By 10 pm on the dot car A and car B will be at the same distance. However, 1 or so second later, car B will be ahead of car A because car B is driving at a faster rate than car A. hope this helps.
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Re: Car A, travelling at 60kmph leaves City X for City Y at 5pm. Car B lea  [#permalink]

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New post 25 Apr 2017, 11:41
vikasp99 wrote:
Car A, travelling at 60kmph leaves City X for City Y at 5pm. Car B leaves city X for City Y at 7pm and travels at 100 kmph. When will Car B overtake Car A?

A) 9:45 p.m

B) 9:55 p.m

C) 10:00 p.m

D) 10:05 p.m

E) 7:55 p.m

Source: http://www.f1gmat.com


Two methods

1st:

For car A, SxT=60T
For car B, SxT=100(T-2)
60T=100T-200
40T=200
T=5h
5h+5h=10h

2nd:

When car B starts, it's 120km behind. Relative speed is 40kmph.
120/40=3h
7pm+3h=10pm.

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Re: Car A, travelling at 60kmph leaves City X for City Y at 5pm. Car B lea  [#permalink]

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Re: Car A, travelling at 60kmph leaves City X for City Y at 5pm. Car B lea &nbs [#permalink] 26 Jul 2018, 15:13
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