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08 Jul 2021, 03:45
Kudos
Bookmarks
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
79% (02:23) correct
21%
(02:45)
wrong
based on 457
sessions
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Joe drove from Springfield to Shelbyville at x miles per hour. He then drove from Shelbyville to Bakersfield at (1.5)x miles per hour. If the distance between Springfield and Shelbyville is twice the distance between Shelbyville and Bakersfield, what was Joe’s average speed for the entire trip?
A. 9x/8
B. 6x/5
C. 5x/4
D. 7x/4
E. 9x/4
A. 9x/8
B. 6x/5
C. 5x/4
D. 7x/4
E. 9x/4
Let the distance between Shelbyville and Bakersfield = D
Therefore, the distance between Springfield and Shelbyville = 2 * (the distance between Shelbyville and Bakersfield) = 2D
Speed between Springfield & Shelbyville = x miles per hour
Speed between Shelbyville to Bakersfield = 1.5x miles per hour
Let the Time to cover the distance between Shelbyville and Bakersfield = T'
Let the Time to cover the distance between Springfield and Shelbyville = T"
speed * time = distance
T' = 2D/x, and
T" = D/1.5x
Total Distance covered = D+2D = 3D
Total Time = T'+T"= 2D/x + D/1.5x = 4D/1.5x
Avg. Speed = Total Distance/Total Time = 3D*1.5x/4D = 4.5x/4 = 45x/40 = 9x/8 (A)
Therefore, the distance between Springfield and Shelbyville = 2 * (the distance between Shelbyville and Bakersfield) = 2D
Speed between Springfield & Shelbyville = x miles per hour
Speed between Shelbyville to Bakersfield = 1.5x miles per hour
Let the Time to cover the distance between Shelbyville and Bakersfield = T'
Let the Time to cover the distance between Springfield and Shelbyville = T"
speed * time = distance
T' = 2D/x, and
T" = D/1.5x
Total Distance covered = D+2D = 3D
Total Time = T'+T"= 2D/x + D/1.5x = 4D/1.5x
Avg. Speed = Total Distance/Total Time = 3D*1.5x/4D = 4.5x/4 = 45x/40 = 9x/8 (A)
15 Jul 2021, 14:07
Kudos
Bookmarks
My kind of question!
Let's start with the basic formula for distance
Distance = Rate x Time
This creates
Time = \(\frac{Distance}{Rate}\)
and
Rate = \(\frac{Distance}{Time}\)
These are the equations we will use to answer this question
Joe's trip has two legs, Springfield to Shelbyville, and Shelbyville to Bakersfield
We are told that the distance from Springfield to Shelbyville is twice the distance from Shelbyville to Bakersfield
Use variables for unknowns, in this case y will do
D1 = 2y
D2 = y
For the first leg of the trip, Joe was traveling x mph
R1 = x
For the second part of the trip, he was traveling 1.5x mph or \((\frac{3}{2})\)x mph
R2 = \((\frac{3}{2})\)x
Time = \(\frac{Distance}{Rate}\)
\(\\
T1 = \frac{D1}{R1} =\frac{ 2y}{x}\\
\)
\(\\
T2 = \frac{D2}{R2} =\frac{ y}{1.5x}\\
\)
Rate = \(\frac{Distance}{Time}\)
Total Rate = \(\frac{Total Distance}{Total Time}\)
Total Distance = D1 + D2 = 2y + y = 3y
Total Time = T1 + T2 = \(\frac{2y}{x }\) + \(\frac{y}{(3/2)x}\) = \(\frac{2y}{x }\) + \(\frac{2y}{3x }\) = \(\frac{6y}{3x }\) + \(\frac{2y}{3x }\) = \(\frac{8y}{3x }\)
Total Rate = \(\frac{3y}{(8y/3x)}\) = \(\frac{9xy}{8y }\)
Cancel the y from top and bottom to get \(\frac{9x}{8 }\)
The answer is (A)
The reason Joe drove to Shelbyville:
Let's start with the basic formula for distance
Distance = Rate x Time
This creates
Time = \(\frac{Distance}{Rate}\)
and
Rate = \(\frac{Distance}{Time}\)
These are the equations we will use to answer this question
Joe's trip has two legs, Springfield to Shelbyville, and Shelbyville to Bakersfield
We are told that the distance from Springfield to Shelbyville is twice the distance from Shelbyville to Bakersfield
Use variables for unknowns, in this case y will do
D1 = 2y
D2 = y
For the first leg of the trip, Joe was traveling x mph
R1 = x
For the second part of the trip, he was traveling 1.5x mph or \((\frac{3}{2})\)x mph
R2 = \((\frac{3}{2})\)x
Time = \(\frac{Distance}{Rate}\)
\(\\
T1 = \frac{D1}{R1} =\frac{ 2y}{x}\\
\)
\(\\
T2 = \frac{D2}{R2} =\frac{ y}{1.5x}\\
\)
Rate = \(\frac{Distance}{Time}\)
Total Rate = \(\frac{Total Distance}{Total Time}\)
Total Distance = D1 + D2 = 2y + y = 3y
Total Time = T1 + T2 = \(\frac{2y}{x }\) + \(\frac{y}{(3/2)x}\) = \(\frac{2y}{x }\) + \(\frac{2y}{3x }\) = \(\frac{6y}{3x }\) + \(\frac{2y}{3x }\) = \(\frac{8y}{3x }\)
Total Rate = \(\frac{3y}{(8y/3x)}\) = \(\frac{9xy}{8y }\)
Cancel the y from top and bottom to get \(\frac{9x}{8 }\)
The answer is (A)
The reason Joe drove to Shelbyville:
Attachments
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