Let the length of 1 lap be \(d\)
Avg Speed for first lap = \(A\)
Avg Speed for second lap = \(B\)
So the total time taken is \(\frac{d}{A}+\frac{d}{B}\)
Total time taken if overall average speed is \(60 = \frac{2d}{60}=\frac{d}{30}\)
The question asks, is \(\frac{d}{A}+\frac{d}{B}>\frac{d}{30}\)? or is \(\frac{1}{A}+\frac{1}{B}>\frac{1}{30}\)?
(1) Jeff's average speed for the first lap was 20 miles per hour.
\(\frac{1}{A}=\frac{1}{20}\).
Since \(\frac{1}{20}\) is already greater than \(\frac{1}{30}\), then yes, \(\frac{1}{A}+\frac{1}{B}>\frac{1}{30}\)
1 is sufficient
(2) Jeff's average speed for the second lap was 120 miles per hour.
\(\frac{1}{120}\) is less than \(\frac{1}{30}\) so we need to know the value of A to determine if \(\frac{1}{A}+\frac{1}{B}>\frac{1}{30}\)
2 is not sufficient
Answer is
(A)nick1816 can you verify the OA and post a solution?
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