Q: Yesterday it took a certain plane 3 hours to fly from City A to City B at an average speed of 400 miles per hour. Today the same plane flew from City A to City B along the same route at an average speed of 450 miles per hour. How many more yesterday than it took today?
A. 10
B. 15
C. 20
D. 25
E. 30
I reached till this point:
1200/450 = 8/3
OA C
Can someone please let me know how to proceed ahead. I am sure I am missing onto something here.
Thanks.
Please take a look at this PS Speed-Time problem.
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- eaakbari
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I prefer making a quick table in DST questions however easy the question. It helps avoid lousy mistakes.
D S T
d 400 3
d 450 t
Distance is the same and & since D = S* T
400 * 3 = 450 * t
Solving for t
t = 3 *400 / 450 = 8/3
Time difference = 3-t = 3 - 8/3 = 1/3 (hours)
1/3 * 60 = 20 minutes
Hence C
D S T
d 400 3
d 450 t
Distance is the same and & since D = S* T
400 * 3 = 450 * t
Solving for t
t = 3 *400 / 450 = 8/3
Time difference = 3-t = 3 - 8/3 = 1/3 (hours)
1/3 * 60 = 20 minutes
Hence C
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You're missing part of the question. Presumably, we want to find out how many minutes more the first trip was.aman88 wrote:Q: Yesterday it took a certain plane 3 hours to fly from City A to City B at an average speed of 400 miles per hour. Today the same plane flew from City A to City B along the same route at an average speed of 450 miles per hour. How many more yesterday than it took today?
A. 10
B. 15
C. 20
D. 25
E. 30
First find the distance from A to B
Distance = (rate)(time)
= (400)(3)
= 1200 miles
Now find the time for the second trip.
Time = (distance)/(rate)
= 1200/450
= 8/3 hours
= 2 2/3 hours
So, the first trip took 3 hours, and the second trip took 2 2/3 hours.
So, the first trip took 1/3 hours longer.
1/3 hours = [spoiler]20 minutes = C[/spoiler]
Cheers,
Brent
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Distance between City A and City B will be same in both the cases, so
Distance in Case 1 = Distance in Case 2
400*3 = 450*t, solving for t we get t = 8/3
Difference in times = 3-(8/3) = 1/3 of an hour i.e. 20 minutes.
Distance in Case 1 = Distance in Case 2
400*3 = 450*t, solving for t we get t = 8/3
Difference in times = 3-(8/3) = 1/3 of an hour i.e. 20 minutes.
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Faster rate : slower rate = 450/400 = 9/8.aman88 wrote:Q: Yesterday it took a certain plane 3 hours to fly from City A to City B at an average speed of 400 miles per hour. Today the same plane flew from City A to City B along the same route at an average speed of 450 miles per hour. How many more yesterday than it took today?
A. 10
B. 15
C. 20
D. 25
E. 30
Since rate and time are RECIPROCALS, faster time : slower time = 8/9.
Since the faster time is 8/9 of the slower time, the faster time requires 1/9 fewer minutes than the slower time (180 minutes):
(1/9)180 = 20.
The correct answer is C.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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