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Last year Harolds average time to finish the qualifying even [#permalink]

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20 Dec 2010, 15:36

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Last year Harolds average time to finish the qualifying event was three hours. If he knows that he can increase his speed this year by 20%, how many minutes should it take him to complete the event?

Last year Harolds average time to finish the qualifying event was three hours. If he knows that he can increase his speed this year by 20%, how many minutes should it take him to complete the event?

(A) 150 (B) 144 (C) 120 (D) 90 (E) 36

So we know that Rate \(= \frac{A}{T}\) where A represents either a distance or a job or whatever. Let's assume that the qualifying event = A, here.

In the first year, we have Rate \(= \frac{A}{3*60} = \frac{A}{180}\) per minute. He increases this speed by 20%, which means that new rate \(= \frac{120}{100}*\frac{A}{180} = \frac{A}{150}\)

From this you can see that the new time is 150 minutes.

Last year Harolds average time to finish the qualifying event was three hours. If he knows that he can increase his speed this year by 20%, how many minutes should it take him to complete the event?

(A) 150 (B) 144 (C) 120 (D) 90 (E) 36

So we know that Rate \(= \frac{A}{T}\) where A represents either a distance or a job or whatever. Let's assume that the qualifying event = A, here.

In the first year, we have Rate \(= \frac{A}{3*60} = \frac{A}{180}\) per minute. He increases this speed by 20%, which means that new rate \(= \frac{120}{100}*\frac{A}{180} = \frac{A}{150}\)

From this you can see that the new time is 150 minutes.

Time taken to complete the work = 3hrs Let the total work be 300 (assumed value. Its always better if u assume good values if the question consists of percentage or percentage incease) so the speed = 300/3 = 100

Now the speed is increased by 20%. so the new speed = 120 and the total work is still the same = 300. There for the new time taken to complete the event = 300/120 = 5/2 hrs. = 2hrs 30 mins = 150 mins Answer A
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Last year Harolds average time to finish the qualifying event was three hours. If he knows that he can increase his speed this year by 20%, how many minutes should it take him to complete the event?

(A) 150 (B) 144 (C) 120 (D) 90 (E) 36

Time, rate and job in work problems are in the same relationship as time, speed (rate) and distance in rate problems.

So if the rate is 1.2 times more then the time needed to complete the same job will be 1.2 times less: he needs 180 minutes with original rate to do the job --> he'll need 180/1.2=150 minutes to complete the same jon with the rate increased 1.2 times.

Last year Harolds average time to finish the qualifying event was three hours. If he knows that he can increase his speed this year by 20%, how many minutes should it take him to complete the event?

(A) 150 (B) 144 (C) 120 (D) 90 (E) 36

Think in terms of ratios, if you will...

For same distance/work, Old Speed : New Speed = 5 : 6 (An increase of 20%) So Old Time : New Time = 6 : 5 = 3 hrs: 2.5 hrs (The ratio will be inverse if work is same because Work = Rate * Time) New Time = 2.5*60 = 150 minutes
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Re: Last year Harolds average time to finish the qualifying even [#permalink]

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21 Oct 2013, 12:23

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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Last year Harolds average time to finish the qualifying even [#permalink]

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29 Mar 2014, 03:03

Simple And sober approach:

We know that , Rate * Time = Work, now work being constant we can write.

Time = Work/Rate

Instance 1: 3=w/r

Instance 2: t=w/1.2r

Dividing 1/2 => t=2.5 hours = 150 minutes

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Re: Last year Harolds average time to finish the qualifying even [#permalink]

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18 Apr 2015, 04:21

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Last year Harolds average time to finish the qualifying even [#permalink]

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14 May 2015, 05:23

Bunuel wrote:

rxs0005 wrote:

Last year Harolds average time to finish the qualifying event was three hours. If he knows that he can increase his speed this year by 20%, how many minutes should it take him to complete the event?

(A) 150 (B) 144 (C) 120 (D) 90 (E) 36

Time, rate and job in work problems are in the same relationship as time, speed (rate) and distance in rate problems.

So if the rate is 1.2 times more then the time needed to complete the same job will be 1.2 times less: he needs 180 minutes with original rate to do the job --> he'll need 180/1.2=150 minutes to complete the same jon with the rate increased 1.2 times.

Why is my approach not working, could you check where it is flawed?

Rate * Time = Job Done 1/180 * 180 = 1 1/144 * 144 = 1 (New Rate is 20% faster, hence 180 - 0.2*180 = 144 (NOT 150)). What's wrong with that? I see that the difference here is, you divide 180 by 1.2 - but i don't get it why "20% faster" is 180/1.2 instead of 180 - 0.2*180 ...

Please help

Thank you very much

Answer: A.
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Re: Last year Harolds average time to finish the qualifying even [#permalink]

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29 Jun 2016, 10:50

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Last year Harolds average time to finish the qualifying even [#permalink]

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24 Jan 2017, 11:49

rxs0005 wrote:

Last year Harolds average time to finish the qualifying event was three hours. If he knows that he can increase his speed this year by 20%, how many minutes should it take him to complete the event?

(A) 150 (B) 144 (C) 120 (D) 90 (E) 36

work completed in 3hr it means rate or speed =1/3 rate after increase of 20% =1/3*6/5 = 2/5 thus time needed =5/2 =2.5 hrs =150mnts