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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?

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
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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

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.

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.

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

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.
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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