rishianand7 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 1: Car X arrives in Town B 90 minutes after leaving city A.
No information about Car Y.
INSUFFICIENT.

Statement 2: Car Y arrives in Town A at the same time Car X arrived in Town B.
To see the situation more clearly, we can plug in values.

Let the distance between the towns = 60 miles.

Let X's rate = 60 miles per hour.
Time for X to travel from Town A to Town B = d/r = 60/60 = 1 hour.

Since Y leaves 15 minutes after X and travels the same distance, Y's total time = 3/4 of an hour.
Thus, Y's rate = d/t = 60/(3/4) = 80 miles per hour.

Rate ratio for X and Y = 60:80 = 3:4.
Implication:
When X and Y travel toward each other, of every 7 miles traveled, X will travel 3 miles and Y will travel 4 miles.
Thus, of the distance between X and Y, the fraction traveled by X = 3/7.

From 2pm to 2:15pm, the distance traveled by X = r*t = 60*(1/4) = 15 miles.
Remaining distance between X and Y = 60-15 = 45 miles.
When Y leaves -- with the result that X and Y are now traveling toward each other -- X will travel 3/7 of the distance between them:
(3/7) * 45 ≈ (1/2) * 45 ≈ 22.5 miles.
Total distance traveled by X when the two cars meet = 15+22.5 = 37.5 -- more than 1/2 the distance between Town A and Town B.

The case above illustrates that -- when X and Y meet -- X will have traveled more than 1/2 the distance between Town A and Town B and thus will be closer to Town B than to Town A.
SUFFICIENT.

The correct answer is B.

We could also use REASON to evaluate statement 2.
Since Y leaves 15 minutes later and completes the trip at the same time as X, Y is FASTER than X.
If X and Y meet at the HALFWAY point, then Y -- the FASTER car -- will complete the remaining half of the trip BEFORE X.
To increase Y's time so that the two cars complete the trip at the same time, we must increase Y's DISTANCE after the two cars meet.
To increase Y's distance after the two cars meet, Y must be LESS THAN HALFWAY between the two towns, implying that X must be MORE THAN HALFWAY between the two towns.
Thus, when the two cars meet, X will have traveled more than half the distance between the towns, with the result that X will be closer to Town B than to Town A.
SUFFICIENT.