Author 
Message 
TAGS:

Hide Tags

Director
Joined: 07 Jun 2004
Posts: 603
Location: PA

Last year Harolds average time to finish the qualifying even
[#permalink]
Show Tags
20 Dec 2010, 16:36
Question Stats:
67% (01:37) correct 33% (01:37) wrong based on 519 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
If the Q jogged your mind do Kudos me : )



SVP
Status: Three Down.
Joined: 09 Jun 2010
Posts: 1856
Concentration: General Management, Nonprofit

Re: Rate intersting
[#permalink]
Show Tags
20 Dec 2010, 17:13
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: 104
Location: Calicut, India

Re: Rate intersting
[#permalink]
Show Tags
20 Dec 2010, 19:32
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 Answer A
_________________
If u think this post is useful plz feed me with a kudo



Math Expert
Joined: 02 Sep 2009
Posts: 50042

Re: Rate intersting
[#permalink]
Show Tags
21 Dec 2010, 01:53



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8406
Location: Pune, India

Re: Rate intersting
[#permalink]
Show Tags
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
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Director
Joined: 03 Aug 2012
Posts: 753
Concentration: General Management, General Management
GMAT 1: 630 Q47 V29 GMAT 2: 680 Q50 V32
GPA: 3.7
WE: Information Technology (Investment Banking)

Re: Last year Harolds average time to finish the qualifying even
[#permalink]
Show Tags
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
Rgds, TGC!



Retired Moderator
Joined: 29 Apr 2015
Posts: 848
Location: Switzerland
Concentration: Economics, Finance
WE: Asset Management (Investment Banking)

Re: Last year Harolds average time to finish the qualifying even
[#permalink]
Show Tags
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: wordtranslationsrateswork104208.html?hilit=relationship%20rate#p812628Hope 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 ... Please help Thank you very much Answer: A.
_________________
Saving was yesterday, heat up the gmatclub.forum's sentiment by spending KUDOS!
PS Please send me PM if I do not respond to your question within 24 hours.



VP
Joined: 07 Dec 2014
Posts: 1104

Last year Harolds average time to finish the qualifying even
[#permalink]
Show Tags
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: 317

Re: Last year Harolds average time to finish the qualifying even
[#permalink]
Show Tags
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: 142
Location: India
Concentration: General Management, Strategy
WE: Information Technology (Computer Software)

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



VP
Joined: 05 Mar 2015
Posts: 1000

Re: Last year Harolds average time to finish the qualifying even
[#permalink]
Show Tags
24 Jan 2017, 12:49
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 Ans A



NonHuman User
Joined: 09 Sep 2013
Posts: 8535

Re: Last year Harolds average time to finish the qualifying even
[#permalink]
Show Tags
27 Jul 2018, 11:23
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). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: Last year Harolds average time to finish the qualifying even &nbs
[#permalink]
27 Jul 2018, 11:23






