Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2512:30 AM EDT
-01:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
Apr 2601:30 AM EDT
-02:30 AM EDT
Learn how Keshav, a Chartered Accountant, scored an impressive 705 on GMAT in just 30 days with GMATWhiz's expert guidance. In this video, he shares preparation tips and strategies that worked for him, including the mock, time management, and more.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
May 0111:00 AM PDT
-12:00 PM PDT
Free- full-length test + 15 concept videos + 150+ short videos + study plan!
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 456
sessions
History
Date
Time
Result
Not Attempted Yet
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
springfield rules.jpg [ 32.07 KiB | Viewed 6522 times ]
