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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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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]
1
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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]
1
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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]
1
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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]
1
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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]
1
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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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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]
1
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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]
1
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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]
Expert Reply
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. \(\frac{60^2}{3590}\)


Are You Up For the Challenge: 700 Level Questions


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]
Can somebody explain why "increased his pace" actually translated into decreasing the pace?!?!?!
His original pace is 2400s/m, why do we have to subtract 10 if his pace increased? Doesn't he get slower if the pace increases?
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Re: In March, Kurt ran an average of 1.5 miles an hour. If by June he had [#permalink]
Can use the Unitary Method/Ratios

3,600 seconds = 1 hour = (60)^2 seconds

In June:

Run 1.5 miles —————> 1 hour or 3,600 seconds

* 2/3 —————————> * 2/3
___________________________
run (2/3)(1.5) = 1 mile—————> (2/3)(3,600) = 2,400 seconds

Rate in June = 1 mile/2,400 sec


Then, he improves his time by 10 seconds

New rate = 1 mile/2,390 sec

How long to complete 1 mile (in hours)?

T = D/R

T in seconds = 1 mile / (1/2,390)

T = 2,390 seconds

To convert from seconds to hour:

(2,390 sec) * (1 hour / 3,600 sec) =

2,390 / (60)^2

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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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