Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
A soccer coach riding his bike at 24 km/h reaches his office 5 minutes late. Had he traveled 25% faster, he would have reached the office 4 minutes earlier than the scheduled time. How far is his office from his house?
use the answer provided to back into the answer. I.e. Start with the middle value and see whether you should go higher or lower.
I started with 24km and with 24km/hr it will take 60 minutes to arrive. Since this is 5 minutes late, I subtract 5 minutes from 60 to get a desired time of 55 minutes. The desired time is the same for both trials.
Since I know my desired time is 55 minutes, and that with a 25% increase in speed, which is 30km/hour, then I need to be 4 minutes quicker than my desired time, or actual time of 51 minutes at 30 km/hour. Divide 30 into 24 and you get .8hours or 48 minutes. The difference between my desired time of 55 minutes and 48 minutes is 7 minutes which is too great.
I know that Distance= Rate x Time and rearranged, Time = Distance/Rate. I know that my Time needs to get smaller and the only way to do that is to decrease the numerator or increase the Denominator. Since our rate is given (30 km/hour) I can't increase my denominator and therefore must decrease my numerator, Distance. Therefore move onto to a smaller distance.
The answer is 18. Work out the same way as desrcibed above.
Re: M04 Q21 - soccer coach bike ride [#permalink]
21 May 2010, 07:32
4
This post received KUDOS
Hi, my method is different:
Let D=Distance in Km and t = time in hr 5 min = 1/12 hr 4 min = 1/15 hr
Even if we have changes in time or speed, the distance travelled will be same. So, 24x(t+1/12) = 24x(5/4)x(t-1/15) (12t+1)/12 = 5/4x(15t-1)/15 12t+1 = 15t-1 t=2/3 hr
Therefore, distance = speed x time = 24 x (2/3 + 1/12) = 24 x 9/12 = 18 Km _________________
Re: M04 Q21 - soccer coach bike ride [#permalink]
13 Oct 2010, 10:18
3
This post received KUDOS
Let the Correct time for reaching Office is t minutes. Now 5 min late means he is taking (t+5) min and 4 min earlier means he is taking (t-4)min Now the distance of office is same.hence (24/60 miles per minute)(t+5)=(125/100)(24/60)(t-4) Solving we get t=40 min Therefore Distnace from office =(24/60)(40+5)=18Miles .Answer is A
Re: M04 Q21 - soccer coach bike ride [#permalink]
17 Oct 2011, 05:52
my way: V1=24km/h V2=V1*1.25=30km/h Distance1=Distance2 T1=t+5 (5 minutes late = more time to cover the distance) T2=t-4 (4 minute early = less time to cover the distance) V1T1=V2T2 24(t+5)=30(t-4) 24t+120=30t-120 240=6t t=40 (in minutes) Distance = 30*[(40-4)/60] or 24*[(40+5)/60)] = 18km
Re: M04 Q21 - soccer coach bike ride [#permalink]
17 Nov 2011, 22:19
1
This post received KUDOS
I just quickly did the following:
Speed 1 - 24 Speed 2 - 30
Time delta 9 min.
Started with choice B - 24Km
It would take Speed 1 - 1hr to do this. Speed 2 would to it in 48 mins. That is a 12 min difference too much. Since the time delta will only get more pronounced as the distance gets greater you know it must be less.
Re: M04 Q21 - soccer coach bike ride [#permalink]
17 Oct 2012, 23:59
samiam7 wrote:
A soccer coach riding his bike at 24 km/h reaches his office 5 minutes late. Had he traveled 25% faster, he would have reached the office 4 minutes earlier than the scheduled time. How far is his office from his house?
I used equations to solve and it took me longer than I would have liked.
The OA says "The best choice is backsolving." Can someone please walk me though the way they would backsolve?
Thanks
Finding the answer algebraically is simple enough.. Wont take more than a minute. Since the answer here is A, backsolving might be quicker. But that might not be the case everytime.
Let "x" be the distance and "m" be the time taken when 5 minutes late
\(\frac{x}{24}=m\)
1.25 times the speed is 30 km/hr. So \(\frac{x}{30}= m - \frac{9}{60} = \frac{x}{24} - \frac{9}{60}\)
Solving the equation is basic math which gives 18 as the answer. _________________
Did you find this post helpful?... Please let me know through the Kudos button.
Re: M04 Q21 - soccer coach bike ride [#permalink]
14 Dec 2012, 11:23
There was a post by Sriharimurthy and I really got a lot benefitted by that post. It gives idea about how to approach rate problems. Here is how I solved it:
speed distance time
late 24km/h d t+5
early 30km/h d t-4
25% faster is 24*25/100=6 Hence 24+6=30
Now, d/t+5=24 and d/t-4=30 Solving this equation, we get t=40 min. Apply this value ion the above equation and you get 18 kilometers.
gmatclubot
Re: M04 Q21 - soccer coach bike ride
[#permalink]
14 Dec 2012, 11:23