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In March, Kurt ran an average of 1.5 miles an hour. If by June he had

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In March, Kurt ran an average of 1.5 miles an hour. If by June he had  [#permalink]

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New post 19 Jan 2020, 22:09
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In March, Kurt ran an average of 1.5 miles an hour. If by June he had increased his pace by 10 seconds per mile, then which of the following expresses the number of hours it would take Kurt to complete one mile in June?


A. \(\frac{2390}{60^2}\)

B. \(\frac{2410}{60^2}\)

C. \(\frac{3586}{60^2}\)

D. \(\frac{3590}{60^2}\)

E. \(\frac{60^2}{3590}\)


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Re: In March, Kurt ran an average of 1.5 miles an hour. If by June he had  [#permalink]

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New post 20 Jan 2020, 01:45
1
1
There are information provided
- K used to run speed @ 1.5 miles/ 1 hour
- Now he runs faster
- He runs 10 secs faster per 1 mile
- And question ask about time in "hours" it uses to run in "1 mile"

So we should focus on question which is "per hours" and "per 1 mile"

Let's try to calculate
1. We convert the first speed into "1 mile" unit
1.5 mile per 1 hour
1 mile per 1*1/1.5 hour = 2/3 hour = 40 mins

2. We convert mins into secs so that we can calculate it with the pace which is "secs per 1 mile"
40mins = 2,400 secs

3. We reduce the the secs by 10 secs as the question provided information "he had increased his pace by 10 seconds per mile"
So the time spent after the increasing speed is
2,400 - 10 = 2,390 secs per 1 mile
(**BE CAREFUL**; 10 secs FASTER means that he runs faster and the time spent should be reduced NOT increased.)

4. Now we convert our calculation into what question want. The question want "hours" spent per "1 mile"
what we have is 2,390 secs per 1 mile.
we convert secs per mile into hour per mile by divide it with 60^2
(**TRICK**; sometime we are not sure whether we should divide or multiply when we have to convert some units. The easy way that I do is just think like should it have to go larger or smaller? eg. convert 100,000 secs into hrs. Just think like should hrs be a lot larger than 100k secs or a lot smaller than 100k secs? larger NO! we go smaller, then we just divide it with 60*60. For me think like this can easily eliminate the confusion during rush time)

so the answer is = 2,390/60^2
I believe that answer is "A" (Please give me suggestion if I calculate something wrong :lol: )


My Learning
- I believe that there is another approach where we start to convert from 10secs/1mile into n secs / 1.5mile and add it with the 1hr/1.5mile. This approach can lead to answer too. But it requires a little more calculation. That's why I think we should focus on question. The question ask about time per 1 mile. So convert per 1mile to per 1.5mile may need another convert back to per 1mile in the end.. This little trick could reduce time using by 15-30secs.
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Re: In March, Kurt ran an average of 1.5 miles an hour. If by June he had  [#permalink]

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New post Updated on: 20 Jan 2020, 22:36
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In March, time taken to run 1 mile = \(\frac{Distance}{speed}\) = \(\frac{1}{1.5}\)
= \(\frac{2}{3}\) hour = \(\frac{2}{3}*60*60\) = 2400 seconds per mile

If by June he had increased his pace by 10 seconds per mile
--> Time taken to travel 1 mile in June = 2400 - 10 = 2390 seconds
--> Number of hours = \(\frac{2390}{60^2}\)

Option A

Originally posted by Dillesh4096 on 20 Jan 2020, 02:12.
Last edited by Dillesh4096 on 20 Jan 2020, 22:36, edited 1 time in total.
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Re: In March, Kurt ran an average of 1.5 miles an hour. If by June he had  [#permalink]

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New post 20 Jan 2020, 05:47
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Quote:
In March, Kurt ran an average of 1.5 miles an hour. If by June he had increased his pace by 10 seconds per mile, then which of the following expresses the number of hours it would take Kurt to complete one mile in June?

A. 2390/60^2
B. 2410/60^2
C. 3586/60^2
D. 3590/60^2
E. 60^2/3590


r=d/t=1.5/1=1.5
d=1:t=d/r=1/1.5=2/3hours*60*60=2400secs
r(secs)=2400-10=2390
r(hrs)=2390secs/60^2

d=1:t=d/r=1/2390/60^2=1/1*2390/60^2=2390/60^2

Ans (A)
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Re: In March, Kurt ran an average of 1.5 miles an hour. If by June he had  [#permalink]

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New post 20 Jan 2020, 07:37
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In March, Kurt ran an average of 1.5 miles an hour. If by June he had increased his pace by 10 seconds per mile, then which of the following expresses the number of hours it would take Kurt to complete one mile in June?

In march, he ran 1.5 miles per 3600 seconds meaning he can run 1 miles in 2400 seconds
In april he increase his pace by 10 seconds per mile meaning he ran faster 10 seconds per mile meaning
2400-10 =2390 seconds per mile
2390 sec (1min/60sec) (1 hr/60 min)
therefore 2390/60^2
A
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Re: In March, Kurt ran an average of 1.5 miles an hour. If by June he had  [#permalink]

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New post 20 Jan 2020, 07:44
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Initially, Kurt ran an average of 1.5 miles an hour.
--> In one mile, he spent 2/3 of an hour (=2,400 secs)

By June he had increased his pace by 10 seconds per mile.
--> in June, he spent (2400-10) = 2,390 sec per mile = 2, 390/60^2 hour per mile

FINAL ANSWER IS (A)

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Re: In March, Kurt ran an average of 1.5 miles an hour. If by June he had  [#permalink]

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New post 20 Jan 2020, 08:12
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In March
1.5 miles run in 1 hour = 3600 seconds
1 mile run in 1/1.5 hour = 2400 seconds

In June
1 mile run in 2400-10 = 2390 seconds = 2390/60^2 hours

IMO A

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Re: In March, Kurt ran an average of 1.5 miles an hour. If by June he had  [#permalink]

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New post 20 Jan 2020, 09:50
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In March, Kurt ran an average of 1.5 miles an hour. If by June he had increased his pace by 10 seconds per mile, then which of the following expresses the number of hours it would take Kurt to complete one mile in June?

A. \(\frac{2390}{60^2}\)

B. \(\frac{2410}{60^2}\)

C. \(\frac{3586}{60^2}\)

D. \(\frac{3590}{60^2}\)

E. \(\frac{60^2}{3590}\)

Average speed of Kurt, Sk = 1.5mph
Time taken to cover 1 mile = 1/1.5 hour = 40 minutes = 2400 sec

Since he improves by 10 sec, new time taken to cver 1 mile = 2400-10 = 2390 sec = \(\frac{2390}{60^2}\)

Answer A.
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Re: In March, Kurt ran an average of 1.5 miles an hour. If by June he had  [#permalink]

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New post 20 Jan 2020, 17:25
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1.5 miles in an hr....i.e for 1mile he took 2/3hrs...i.e 2400sec
But he had increased his pace by 10 seconds per mile....so he can complete a mile in 2390sec.....i.e 2390/3600 hrs

OA:A
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Re: In March, Kurt ran an average of 1.5 miles an hour. If by June he had  [#permalink]

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New post 20 Jan 2020, 18:53
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Ans: A

1.5 miles time 60*60 sec
1 miles time 2400 sec

now, in June he takes=(2400-10)=2390sec=2390/(60^2) hr
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Re: In March, Kurt ran an average of 1.5 miles an hour. If by June he had  [#permalink]

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New post 20 Jan 2020, 20:18
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In March, Kurt ran at a speed of 1.5mph. So if Kurt were to run for a mile, he would take the time t hours = 1/1.5 = 2/3 hours
So we now know that Kurt has improved his pace of running in June by running a mile 10 seconds faster, implying that his speed has increased and that he takes lesser time to cover 1mile in his runs.
So his new time to cover a mile = 2/3hrs - 10/3600 = (2400-10)/3600 = 2390/3600 hours

A is the answer.
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Re: In March, Kurt ran an average of 1.5 miles an hour. If by June he had  [#permalink]

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New post 20 Jan 2020, 21:32
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1.5 miles — 1 hour (3600 seconds )
1 mile — x hour
—> \(x =\frac{( 1 hour* mile)}{( 1.5 miles) }=
40 minutes = 2400 second\)

Increased his pace by 10 seconds per mile —> 2400–10 = 2390
—> it would take 2390 seconds to compete 1 mile in June or \(\frac{2390}{3600}\) hours

The answer is A .

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Re: In March, Kurt ran an average of 1.5 miles an hour. If by June he had  [#permalink]

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New post 21 Jan 2020, 23:00
Bunuel wrote:

Competition Mode Question



In March, Kurt ran an average of 1.5 miles an hour. If by June he had increased his pace by 10 seconds per mile, then which of the following expresses the number of hours it would take Kurt to complete one mile in June?


A. \(\frac{2390}{60^2}\)

B. \(\frac{2410}{60^2}\)

C. \(\frac{3586}{60^2}\)

D. \(\frac{3590}{60^2}\)

E. \(\frac{60^2}{3590}\)


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OFFICIAL EXPLANATION



The problem has two conversions to watch out for; first, it gives 1.5 miles in March but 1 mile in June second, it adds 10 seconds to his mile per hour rate. The order in which you deal with these are up to you, but they must be dealt with. First let’s deal with the 1.5 mile to 1 mile problem. Initially, he runs 1.5 miles per hour, which is the same as saying that he does 3 halves of a mile in 60 minutes, thus each half must take 20 minutes. Now we know that in March it took him 40 minutes to run a mile. Let’s now convert those minutes to seconds, 40 minutes = 2400 seconds. If by June he increased his pace by 10 seconds, that means it would take him less time to complete the mile, so in June a mile would take him 2390 seconds. Now we have the time it would take him to do a mile in June, so the last step is to convert 2390 seconds to hours. To do so we must divide 2390 by 60 to get minutes and then divide it again by 60 to convert minutes into hours.
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Re: In March, Kurt ran an average of 1.5 miles an hour. If by June he had   [#permalink] 21 Jan 2020, 23:00
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