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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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Senior Manager
Joined: 03 Mar 2010
Posts: 374
Schools: Simon '16 (M$) In March, Kurt ran an average of 1.5 miles an hour. If by June he had [#permalink] ### Show Tags Updated on: 06 Oct 2014, 01:42 5 00:00 Difficulty: 65% (hard) Question Stats: 61% (02:13) correct 39% (02:03) wrong based on 87 sessions ### HideShow timer Statistics 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. 3590/60^2 B. 2410/60^2 C. 2390/60^2 D. 3586/60 E. 60^2/3590 Spoiler: :: OE from Princeton Review 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. _________________ My dad once said to me: Son, nothing succeeds like success. Originally posted by jamifahad on 01 Sep 2011, 04:32. Last edited by Bunuel on 06 Oct 2014, 01:42, edited 3 times in total. Edited the topic ##### Most Helpful Community Reply Manager Joined: 28 Jul 2011 Posts: 181 Re: In March, Kurt ran an average of 1.5 miles an hour. If by June he had [#permalink] ### Show Tags 01 Sep 2011, 14:05 3 2 1500 in 3600 sec 1000 in x sec x = 2400 In june = 2400-10 = 2390 so 2390 / 60 * 60 I also got 2390/60^2 Fluke, whats the mistake u r pointing out? ##### General Discussion Retired Moderator Joined: 20 Dec 2010 Posts: 1810 Re: In March, Kurt ran an average of 1.5 miles an hour. If by June he had [#permalink] ### Show Tags 01 Sep 2011, 05:12 jamifahad wrote: Post timing if possible. 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.3590/60^2 B.2410/60^2 C.3890/60^2 D.3586/60 E.60^2/3590 I got $$\frac{2390}{(60)^2}$$ as my answer. I can't figure out my mistake. _________________ Senior Manager Joined: 03 Mar 2010 Posts: 374 Schools: Simon '16 (M$)
Re: In March, Kurt ran an average of 1.5 miles an hour. If by June he had  [#permalink]

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01 Sep 2011, 05:22
Answer choices for the question is given wrong in Princeton Review. But OE explains the correct answer choice which is not given in answer choices. Anyway, I have rectified answer choices.
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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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01 Sep 2011, 17:22
Post timing if possible.

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

I like this question, which needs a lot of carefulness though its not a difficult one.

Agree with the answer =====> C. 2390/60^2
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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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01 Sep 2011, 21:41
kuttingchai wrote:
1500 in 3600 sec
1000 in x sec

x = 2400

In june = 2400-10 = 2390

so 2390 / 60 * 60

I also got 2390/60^2

Fluke, whats the mistake u r pointing out?

Everything is good now as jamifahad corrected the option C. The answer was not there before, so I was little perplexed.
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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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06 Sep 2011, 13:16
yea! i got a similar ans too 2392/(60^2) and was a little worried after reading your 1st post
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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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06 Jan 2019, 04:03
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Re: In March, Kurt ran an average of 1.5 miles an hour. If by June he had &nbs [#permalink] 06 Jan 2019, 04:03
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