psm12se wrote:If it took Carlos 1/2 hour to cycle from his house to the library yesterday, was the distance that he cycled greater than 6 miles?
(Note: 1 mine = 5280 feet)
1. The average speed at which Carlos cycled from his house to the library yesterday was greater than 16 feet per second.
2. The average speed at which Carlos cycled from his house to the library yesterday was less than 18 feet per second.
Target question:
Was the distance that he cycled greater than 6 miles?
This question could use some rephrasing. Since the two statements provide speeds in feet per second, let's first see what it means for Carlos to cycle
exactly 6 miles in 1/2 an hour.
Distance = 6 miles = (6)(5280) feet
Time = 1/2 hour = 1800 seconds.
Speed = (6)(5280)/1800 = 5280/300 = 176/10
= 17.6 feet per second.
So, in order for Carlos to travel
more than 6 miles in 1800 seconds, his average speed must be
greater than 17.6 feet per second.
Rephrased target question:
Was Carlo's average speed greater than 17.6 feet per second?
Statement 1: His average speed was greater than 16 feet per second.
So, his speed may have been greater than 17.6 feet per second, or less than 17.6 feet per second.
Since we cannot answer the
rephrased target question with certainty, statement 1 is SUFFICIENT
Statement 2: His average speed was less than 18 feet per second.
So, his speed may have been greater than 17.6 feet per second, or less than 17.6 feet per second.
Since we cannot answer the
rephrased target question with certainty, statement 2 is SUFFICIENT
Statements 1 and 2 combined:
We now know that his speed is between 16 feet per second and 18 feet per second.
So, once again, his speed may have been greater than 17.6 feet per second, or less than 17.6 feet per second.
Since we still cannot answer the
rephrased target question with certainty, the combined statements are NOT SUFFICIENT
Answer =
E
Cheers,
Brent
