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

can anybody explain it

By pure logic, I think it should be "B".

Because, if CarX and Y both reach their respective destinations at the same time. Car X will have to be more than half the distance at the meeting point. If not, then Y will reach its destination earlier than X does, because X could not reach even half the distance with the extra 15 minutes, it will surely not cover the remaining more than half the distance in the same time Y covers less than half of the distance. Is it not?

Kaplan must have hired some math doctorates lately, eh?

What's the OA, if at all there is one?

IMO B.

1 - dont have any information about the 2nd car. INS.
2 - if they drive in constant speeds, it means they pass each other only once (if it wasnt constant they could have passing each other 10 times on the way so answer would be E) if they pass each other only when arriving to town B - it clearly gives us the answer.

Should be B only. As explained by Fluke.

Yeah right! even after starting if Y reaches at the same time then X should be close to A. Wording of the 2) could have been more clear though- Car Y arrives in town A at same time as car X arrives in town B.

(By the way, fluke you reminded me of one of my very good friend *refer to the bolded part on your reply, sorry for OT though )

Agree with B. went with c

i think B is not a correct answer bcz
if we consider total time taken to cover A kms is 20 min then before y starts x covers almost 75 % of the distance then X is nearer to B or else if we consider it in vice versa then X is nearer to A

The OA is B.

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 ?

\(A----------P----B\) (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.

Answer: B.

Hope it's clear.

OPEN DISCUSSION OF THIS QUESTION IS HERE: car-x-leaves-town-a-at-2-p-m-and-drives-toward-town-b-at-a-95033.html