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# Car X leaves Town A at 2 p.m. and drives toward Town B at a

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Manager
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Car X leaves Town A at 2 p.m. and drives toward Town B at a  [#permalink]

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30 May 2010, 09:49
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75% (hard)

Question Stats:

57% (02:07) correct 43% (02:46) wrong based on 492 sessions

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Car X leaves Town A at 2 p.m. and drives toward Town B at a constant rate of m miles per hour. Fifteen minutes later, Car Y begins driving from Town B to Town A at a constant rate of n miles an hour. If both Car X and Car Y drive along the same route, will Car X be closer to Town A or Town B when it passes Car Y ?

(1) Car X arrives in Town B 90 minutes after leaving city A.

(2) Car Y arrives in Town A at the same time Car X arrived in Town B.
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30 May 2010, 11:26
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dixitraghav wrote:
Car X leaves Town A at 2 p.m. and drives toward Town B at a constant rate of m miles per hour. Fifteen minutes later, Car Y begins driving from Town B to Town A at a constant rate of n miles an hour. If both Car X and Car Y drive along the same route, will Car X be closer to Town A or Town B when it passes Car Y ?

(1) Car X arrives in Town B 90 minutes after leaving city A.

(2) Car Y arrives in Town A at the same time Car X arrived in Town B.

$$A----------P----B$$ (P meeting point). Question: is $$PA>PB$$?

(1) Car X arrives in Town B 90 minutes after leaving city A. No info about car Y. Not sufficient.

(2) Car Y arrives in Town A at the same time Car X arrived in Town B --> X needs 15 min longer than Y to cover the same distance (as Y starts 15 min after X) --> {rate of X} < {rate of Y}, (m<n). Also, after they meet X and Y will need the same time to get to their respective destinations (as they arrived at the same time) --> the distance covered by Y after they meet (PA=nt) at higher speed, will obviously be more than the distance covered by X after they meet (PB=mt) at lower speed (or algebraically as m<n, mt<nt --> $$PA>PB$$). So $$PA>PB$$. Sufficient.

Hope it's clear.
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Joined: 27 Jul 2010
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07 Oct 2010, 02:55
Hi Bunuel,

I am not able to understand this statment in your solution:
Also, after they meet X and Y will need the same time to get to their respective destinations (as they arrived at the same time) --> the distance covered by Y after they meet (PA=nt) at higher speed

how is it that X and Y need the same time to get to respective destinations after meeting at one point at the same time?

thank you,
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07 Oct 2010, 03:19
wininblue wrote:
Hi Bunuel,

I am not able to understand this statment in your solution:
Also, after they meet X and Y will need the same time to get to their respective destinations (as they arrived at the same time) --> the distance covered by Y after they meet (PA=nt) at higher speed

how is it that X and Y need the same time to get to respective destinations after meeting at one point at the same time?

thank you,

Because "Car Y arrives in Town A at the same time Car X arrived in Town B". Is it possible that Car Y will need more (or less) time to get to A after the meeting and still arrive at the same time?
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08 Oct 2010, 03:51
Straight B.

Undoubtedly Bunnel's explanation is Amazing. However I stopped way early when I realized that Option B provides a solution as it gives the rate of both x and Car y. Option A is insufficient as no info is provided about the rate of car Y
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Re: Car X leaves Town A at 2 p.m. and drives toward Town B at a  [#permalink]

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07 Aug 2013, 05:34
Gian wrote:
Car X leaves Town A at 2 p.m. and drives toward Town B at a constant rate of m miles per hour. Fifteen minutes later, Car Y begins driving from Town B to Town A at a constant rate of n miles an hour. If both Car X and Car Y drive along the same route, will Car X be closer to Town A or Town B when it passes Car Y ?

(1) Car X arrives in Town B 90 minutes after leaving city A.

(2) Car Y arrives in Town A at the same time Car X arrived in Town B.

solution:

3pm...... supposed Arriving time.... 3pm
A........................................M.................B

x---> <------Y

st(2) both reached at 3pm but Y traveled 15min less.so Y has more velocity.
after meeting at any point X and Y has to cover two different distances within the same time.
distance that must be covered by Y has to be greater than that of X, because of the same time duration and the more velocity that Y has.
M = meeting point , so MA > MB . Thus X was closer to town B when it passed Y.
st(2) sufficient alone.
st(1) is nothing.
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Re: Car X leaves Town A at 2 p.m. and drives toward Town B at a  [#permalink]

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31 Dec 2013, 07:12
1
dixitraghav wrote:
Car X leaves Town A at 2 p.m. and drives toward Town B at a constant rate of m miles per hour. Fifteen minutes later, Car Y begins driving from Town B to Town A at a constant rate of n miles an hour. If both Car X and Car Y drive along the same route, will Car X be closer to Town A or Town B when it passes Car Y ?

(1) Car X arrives in Town B 90 minutes after leaving city A.

(2) Car Y arrives in Town A at the same time Car X arrived in Town B.

One needs to picture this problem

We are basically being told the following

We have 2 cars: X and Y , with different unknown speeds 'm' and 'n'

Finally, X has a head start of 15 minutes

Question is, what car will cover the greater distance in the same time

So basically, we need the relative rates to figure out whether B can make up for that lost time

Statement 1

Insuff. We know nothing about 'n' = rate of Y

Statement 2

Suff

If they arrived at the same time at their destinations then Y is faster than X for sure.
Now, do we need to find the exact answer? No, this is a DS question. I'm confident that there can only be an absolute difference between both rates so that will answer my question for sure

But just to elaborate, if they meet somewhere then since Y is faster then it will need less time to reach its destination. So for sure X will be closer to town B

Hope it helps
Cheers!
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Re: Car X leaves Town A at 2 p.m. and drives toward Town B at a  [#permalink]

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31 May 2014, 01:06
dixitraghav wrote:
Car X leaves Town A at 2 p.m. and drives toward Town B at a constant rate of m miles per hour. Fifteen minutes later, Car Y begins driving from Town B to Town A at a constant rate of n miles an hour. If both Car X and Car Y drive along the same route, will Car X be closer to Town A or Town B when it passes Car Y ?

(1) Car X arrives in Town B 90 minutes after leaving city A.

(2) Car Y arrives in Town A at the same time Car X arrived in Town B.

Statement I is insufficient as it does not compare speed of A and B.

Statement II is sufficient:

Now Car B leaves late but reaches at the same time hence Speed of B is greater than Speed of A.

Imagine both the cars meeting exactly in the middle of the road and if they reach at same time speeds must be equal which is not the case. Hence they must meet more towards town B so that car B gets more time to travel in the same time as its speed is more too.

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Re: Car X leaves Town A at 2 p.m. and drives toward Town B at a  [#permalink]

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31 May 2014, 02:45
Brunel,

I have trouble understanding the solution.
I am confused that although it is indeed given that n > m, we can't be sure that n*(t-1/4) > m*t, if t is elapsed time when both X and Y meet.
In other words, although the speed of Car Y is greater than that of X, we don't know how would the distance traversed compare between these 2 cars, since Y start 1/4 hrs late.

Thanks
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Re: Car X leaves Town A at 2 p.m. and drives toward Town B at a  [#permalink]

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31 May 2014, 04:33
kinghyts wrote:
Brunel,

I have trouble understanding the solution.
I am confused that although it is indeed given that n > m, we can't be sure that n*(t-1/4) > m*t, if t is elapsed time when both X and Y meet.
In other words, although the speed of Car Y is greater than that of X, we don't know how would the distance traversed compare between these 2 cars, since Y start 1/4 hrs late.

Thanks

You didn't read the solution carefully: t there is the time, X and Y need to reach to get to their respective destinations. The distance covered by Y after they meet (PA=nt) at higher speed, will obviously be more than the distance covered by X after they meet (PB=mt) at lower speed (or algebraically as m<n, mt<nt --> PA>PB).

Hope it's clear.
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Re: Car X leaves Town A at 2 p.m. and drives toward Town B at a  [#permalink]

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20 Oct 2016, 07:17
dixitraghav wrote:
Car X leaves Town A at 2 p.m. and drives toward Town B at a constant rate of m miles per hour. Fifteen minutes later, Car Y begins driving from Town B to Town A at a constant rate of n miles an hour. If both Car X and Car Y drive along the same route, will Car X be closer to Town A or Town B when it passes Car Y ?

(1) Car X arrives in Town B 90 minutes after leaving city A.

(2) Car Y arrives in Town A at the same time Car X arrived in Town B.

if I had looked first at both statements, I would have got the right answer in seconds...
1. doesn't tell us much
2. since Y started 15 mins later but arrived at the same time to X, it must be true that Y had a greater speed
eventually, we can solve our way to get the answer..but since it's a DS question..we don't need to!
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Re: Car X leaves Town A at 2 p.m. and drives toward Town B at a  [#permalink]

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