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# Last year Harolds average time to finish the qualifying even

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Director
Joined: 07 Jun 2004
Posts: 604
Location: PA
Last year Harolds average time to finish the qualifying even  [#permalink]

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20 Dec 2010, 16:36
2
11
00:00

Difficulty:

45% (medium)

Question Stats:

67% (01:16) correct 33% (01:12) wrong based on 511 sessions

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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

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Status: Three Down.
Joined: 09 Jun 2010
Posts: 1868
Concentration: General Management, Nonprofit
Re: Rate intersting  [#permalink]

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20 Dec 2010, 17:13
4
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

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.
Manager
Joined: 25 Jan 2010
Posts: 106
Location: Calicut, India
Re: Rate intersting  [#permalink]

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20 Dec 2010, 19:32
2
whiplash2411 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

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
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Joined: 02 Sep 2009
Posts: 47918
Re: Rate intersting  [#permalink]

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21 Dec 2010, 01:53
1
1
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.

$$time*speed=distance$$ <--> $$time*rate=job \ done$$.

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.

Check this for more on work problem: word-translations-rates-work-104208.html?hilit=relationship%20rate#p812628

Hope it helps.
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Re: Rate intersting  [#permalink]

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21 Dec 2010, 13:51
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

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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Director
Joined: 03 Aug 2012
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GMAT 1: 630 Q47 V29
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Re: Last year Harolds average time to finish the qualifying even  [#permalink]

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29 Mar 2014, 04:03
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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TGC!
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Re: Last year Harolds average time to finish the qualifying even  [#permalink]

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14 May 2015, 06:23
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.

$$time*speed=distance$$ <--> $$time*rate=job \ done$$.

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?

Check this for more on work problem: word-translations-rates-work-104208.html?hilit=relationship%20rate#p812628

Hope it helps.

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 ...

Thank you very much

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Joined: 07 Dec 2014
Posts: 1064
Last year Harolds average time to finish the qualifying even  [#permalink]

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01 Jul 2016, 23:00
if his speed increases by 6/5,
his time will decrease by 5/6
5/6*180=150 minutes
Senior Manager
Joined: 02 Mar 2012
Posts: 334
Schools: Schulich '16
Re: Last year Harolds average time to finish the qualifying even  [#permalink]

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02 Jul 2016, 01:38
rate increase >time will be less, or
rate decrease>more time

here the rate increased 1.2 times..so time to finish the event will be less than 180 minutes =180/1.2=150 min
Manager
Joined: 24 Oct 2013
Posts: 145
Location: India
Concentration: General Management, Strategy
WE: Information Technology (Computer Software)
Re: Last year Harolds average time to finish the qualifying even  [#permalink]

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27 Sep 2016, 01:45
x(3) = 3x ->>> equ 1
1.20x (time) = 3x

time = 3x/1.2x *(60) because question is asking about minutes

time = 3*50 = 150 minutes
Director
Joined: 05 Mar 2015
Posts: 985
Re: Last year Harolds average time to finish the qualifying even  [#permalink]

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24 Jan 2017, 12:49
1
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

Ans A
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Posts: 7708
Re: Last year Harolds average time to finish the qualifying even  [#permalink]

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27 Jul 2018, 11:23
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# Last year Harolds average time to finish the qualifying even

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