rtaha2412 wrote:If Bill drove 1.5 times slower than normal and was late for school today, how long does it normally take Bill to drive to school? (Assume that each day Bill takes the same route).

1. It took Bill 15 minutes longer to drive to school today than normal.
2. The distance between home and school is 15 miles.

v -> Bill's speed
t -> time taken by Bill to reach school
d -> distance b/w school
t = d/v = ?

Statement 1 :

d/v = 15 + 3d/2v -> Sufficient

Statement 2 :
d = 15 miles

Not sufficient, as no info of v.

Hence A

rtaha2412 wrote:If Bill drove 1.5 times slower than normal and was late for school today, how long does it normally take Bill to drive to school? (Assume that each day Bill takes the same route).

1. It took Bill 15 minutes longer to drive to school today than normal.
2. The distance between home and school is 15 miles.

The wording of this question is strange: what does it mean to drive "1.5 times slower"? If the rate is 4 miles per hour, (1.5)*4 = 6, which is a faster rate, not a slower rate. I'm going to assume that the question is trying to convey the following: that the time was 1.5 times longer. In other words, that it took Bill 1.5 times as long to get to school today as it normally would.

Let x = normal time to drive to school.
Time to drive to school today = 1.5x.

Statement 1:
1.5x = x+15
.5x=15
x=30.
Sufficient.

Statement 2:
The distance does not help us to determine the normal time.
Insufficient.

The correct answer is A.