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

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?

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

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.
Answer (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

Hence answer is B

Hope it helps
Cheers!
J

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.

Hence answer is B.

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.

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!

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.

Clearly insufficient.

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

The key here is that Town A arrived in Town A at the same time Car X arrived in Town B. Car Y started 15 minutes later, meaning Car Y needs to make up lost time. This can only happen if Car Y is going at a faster speed. If both cars arrive at their respective towns at the same time, and with Car Y driving at a faster speed, this means Car Y had more to travel than Car X when the two cars met. Therefore, Car X is closer to Town B when the two cars meet. Statement 2 is sufficient.

Answer is B.

ScottTargetTestPrep

Can you please provide a simpler explanation ?

Hi Bunnel, I still don't get it Please help !!

Solution:

It should be clear why statement one alone is not sufficient for this question. It doesn’t tell us anything about car Y, so we can assume that car Y travels so fast that despite the 15 minute head start of car X, the meeting point is closer to town A. In another scenario, we can assume that car Y travels so slow that the meeting point is closer to town B. In fact, even if the two cars traveled at the same speed, the meeting point will be closer to town B because of the 15 minute head start.

For statement two alone, we should first observe that both cars travel the same distance (which is the distance between the two cities), but car Y takes 15 minutes less time to cover that distance. Thus, car Y travels faster than car X.

Since both cars arrive at their respective destinations at the same time, we can run the problem backwards and assume that cars X and Y started at the same time from cities B and A, respectively. Since they start at the same time and car Y is faster than car X, car Y will have traveled a greater distance when the two cars meet. Thus, the meeting point of the two cars will be closer to city B.

Answer: B

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