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09 Apr 2024, 01:30
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
61% (01:50) correct
39%
(01:40)
wrong
based on 171
sessions
History
Date
Time
Result
Not Attempted Yet
Jim traveled from X to Y, and arrived at Z. If the average speed for the whole trip was 60 miles per hour, what is the average speed for the trip from Y to Z?
(1) The average speed for the trip from X to Y is 55 miles per hour.
(2) It took Jim 0.5 hour from Y to Z.
(1) The average speed for the trip from X to Y is 55 miles per hour.
(2) It took Jim 0.5 hour from Y to Z.
poojaarora1818
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Updated on: 09 Apr 2024, 09:15
Originally posted by poojaarora1818 on 09 Apr 2024, 04:29.
Last edited by poojaarora1818 on 09 Apr 2024, 09:15, edited 1 time in total.
Last edited by poojaarora1818 on 09 Apr 2024, 09:15, edited 1 time in total.
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Jim traveled from X to Y, and arrived at Z. If the average speed for the whole trip was 60 miles per hour, what is the average speed for the journey from Y to Z?
(1) The average speed for the trip from X to Y is 55 miles per hour.
(2) It took Jim 0.5 hours from Y to Z.
Statement 1
(1) The average speed for the trip from X to Y is 55 miles per hour. Insufficient
Statement 2
(2) It took Jim 0.5 hours from Y to Z. Insufficient
Even if we combine both the statement we won't get a conclusive answer.
Hence, Option E
(1) The average speed for the trip from X to Y is 55 miles per hour.
(2) It took Jim 0.5 hours from Y to Z.
Statement 1
(1) The average speed for the trip from X to Y is 55 miles per hour. Insufficient
Statement 2
(2) It took Jim 0.5 hours from Y to Z. Insufficient
Even if we combine both the statement we won't get a conclusive answer.
Hence, Option E
poojaarora1818
poojaarora1818 be careful here. Without knowing anything about the distances one won't be able to solve this question. Just to demonstrate it quickly, I will focus on just statement (1).
From the question stem we know that the average speed for the entire distance is 60mph (X to Y and Y to Z). Statement one tells us that he travelled at an average speed of 55mph between X and Y.
Let's say that the entire distance is 600 miles, and therefore the entire trip took him 10 hours. Let's say that X to Y is 300 miles and Y to Z is 300 miles. As he went an average of 55mph in the first leg of the trip, he would have to go 65mph in the second leg for the average speed to be 60mph over the entire trip. This is easily deduced because the two legs are of equal distance.
What if the total distance remained the same, but this time X to Y is 110 miles long and Y to Z is 490 miles long. Well the total time it took to travel X to Y to Z is still 10 hours as above, this time, the first leg took him only 2 hours, which means to have an average of 60mph over the whole trip he needs to travel 490 miles in 8 hours. \(\frac{490}{8} = 61.25\) which is a slower average speed over the second leg than in the first example. Two different examples, both of which use the info provided in the question stem and statement (1), however, with different lengths produce different answers.
Solving the question:
(1) The average speed for the trip from X to Y is 55 miles per hour.
Without knowing more about the distance of X to Y in relation to X and Z, this statement does not offer much.
INSUFFICIENT
(2) It took Jim 0.5 hour from Y to Z.
This statement also does not offer anything that'll help solve for the average speed between Y to Z.
INSUFFICIENT
(1+2)
Putting the two statements together still does not provide sufficient info with which to solve this question.
Let the distance X to Y = a, and Y to Z = b.
Then we know that the trip in its entirety had a distance of: \(a + b\), avg. speed of: \(60\) and took: \(\frac{a+b}{60}\)
Then we know that the trip for X to Y had a distance of: \(a\), an average speed of: \(55\) and took: \(\frac{a}{55}\)
While the trip for Y to Z had a distance of: \(b\), took: \(\frac{1}{2}\) and therefore had average speed of: \(2b\)
If we use the time (as we can add it nicely) to create an equation: (time taken to cover a)+(time taken to cover b) = (time taken to cover a + b)
\(0.5 + \frac{a}{55} =\frac{a+b}{60}\) now we have two unknowns and neither can isolated to solve for an actual value.
INSUFFICIENT
ANSWER E
