Bunuel
If Carl must travel the 97 miles to Santa Monica for a business meeting at 10:00 a.m., what time must he leave his apartment to ensure that he will be on time for his meeting?
(1) Carl traveled at an average speed of 53 miles per hour for the first third of the trip.
(2) Carl's average speed for the entire trip was 53 miles per hour.
Let's understand the question stem this way;
We are given that Carl has a business meeting @ 10:00 a.m. which is 97 miles away, and are asked to find the time he leaves his apartment to be on time to this meeting.So we are required to find in what time after he leaves he will be able to cover 97 miles.
Thus, keeping in mind
Speed x Time = Distance, we need to calculate \(Time(hours) = \frac{97 Miles}{Speed(Miles per hour)}\).
For this we require average speed for the entire trip.
Taking statement 2 first,
It is
sufficient as it directly gives us the average speed for the entire trip. Hence, the time can be easily calculated. (Therefore it is not necessary to solve further.)
But for reference (if required), \(Time = \frac{97 miles}{53 miles per hour}\), which gives Time = 1.83 Hours Approx or 1hours 50 mins approx.
Therefore, Carl must have left his appartment not later than 8:10 a.m.
Hence eliminate, Option C & E.
Taking statement 1,
It is
Insufficient, as it does not fulfill our requirement and provides the average speed for only First one third of the entire trip, what about the average speed of the other two-third of the entire trip. The speed for the rest of the trip may variate and is uncertain. Since, we cannot define the average speed for the entire speed, we cannot calculate the time required.
Hence eliminate, Option A & D also.
OA is B.
Please Hit Kudos, if it helps