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04 Nov 2014, 00:03
Kudos
Bookmarks
C
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:
72% (02:39) correct
28%
(02:35)
wrong
based on 669
sessions
History
Date
Time
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Not Attempted Yet
A train travels from city A to city B. The average speed of the train is 60 miles/hr and it travels the first quarter of the trip at a speed of 90 mi/hr. What is the speed of the train in the remaining trip?
A. 30
B. 45
C. 54
D. 72
E. 90
A. 30
B. 45
C. 54
D. 72
E. 90
10 Nov 2014, 20:00
Kudos
Bookmarks
Refer diagram below:
speed.png [ 3.39 KiB | Viewed 19850 times ]
Setting up the time equation:
Total time = Time required in first quarter + Time required in the remaining journey
\(\frac{d}{60} = \frac{d}{4*90} + \frac{3d}{4*s}\)
\(s = \frac{90*3}{5} = 54\)
Answer = C
Attachment:
speed.png [ 3.39 KiB | Viewed 19850 times ]
Setting up the time equation:
Total time = Time required in first quarter + Time required in the remaining journey
\(\frac{d}{60} = \frac{d}{4*90} + \frac{3d}{4*s}\)
\(s = \frac{90*3}{5} = 54\)
Answer = C
General Discussion
30 Jan 2015, 06:23
Kudos
Bookmarks
Using the RTD chart, with A being the first quarter, B the rest of the trip and All the combined "trip".
_______R_____T______D
A.........90........3..........270
B.........54.......15.........810
All........60.......18........1080
Let me explain how we fill in the chart:
1) Starting with what we know, we add 90 and 60
2) Picking an easy number for the total distance (6*9=54), so I chose 540. Multiply by 2 to get the whole trip, and we get 1080. Add 1080 for all-distance.
3) 1/4 of the whole distance happened at a Rate of 90. So, 1080/4=270, add 270 under A-D.
4) 2080 - 270 = 810, for the rest of the trip. Add 810 under B-D.
5) Claculate the individual Times for A and All, using T=D/R. Add the results, 3 and 18, under A-T and All-T.
6) 18-3=15, this is the remaining Time for B. Add 15 under B-T.
7) Finally, 15R=810 -->R=810/15 --> R= 54 ANS C
_______R_____T______D
A.........90........3..........270
B.........54.......15.........810
All........60.......18........1080
Let me explain how we fill in the chart:
1) Starting with what we know, we add 90 and 60
2) Picking an easy number for the total distance (6*9=54), so I chose 540. Multiply by 2 to get the whole trip, and we get 1080. Add 1080 for all-distance.
3) 1/4 of the whole distance happened at a Rate of 90. So, 1080/4=270, add 270 under A-D.
4) 2080 - 270 = 810, for the rest of the trip. Add 810 under B-D.
5) Claculate the individual Times for A and All, using T=D/R. Add the results, 3 and 18, under A-T and All-T.
6) 18-3=15, this is the remaining Time for B. Add 15 under B-T.
7) Finally, 15R=810 -->R=810/15 --> R= 54 ANS C
