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A soccer coach riding his bike reaches his office in \(x\) hours. If he travels at 24 km/h, he reaches his office 5 minutes late. If he travels 30 km/h, he reaches his office 4 minutes early. How far is his office from his house? A. 18 km B. 24 km C. 36 km D. 40 km E. 72 km
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Re M0421
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16 Sep 2014, 00:23
Official Solution:A soccer coach riding his bike reaches his office in \(x\) hours. If he travels at 24 km/h, he reaches his office 5 minutes late. If he travels 30 km/h, he reaches his office 4 minutes early. How far is his office from his house?A. 18 km B. 24 km C. 36 km D. 40 km E. 72 km Backsolve or create an equation. Use \(D\) for distance and \(x\) for time: Notice that the rates are given in kilometre per HOUR, while time is given in minutes. Convert minutes to hours: 5 minutes is \(\frac{1}{12}\) hours and 4 minutes is \(\frac{1}{15}\) hours. \(\frac{D}{24} = x + \frac{1}{12}\). Multiply by 24 to get: \(D = 24x + 2\). \(\frac{D}{30} = x  \frac{1}{15}\). Multiply by 30 to get: \(D = 30x  2\). Subtract the second equation from the first one to get: \(6x = 4\). \(x = \frac{2}{3}\) \(D = 24*\frac{2}{3} + 2 = 18\) Answer: A
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Re: M0421
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Updated on: 07 Jul 2015, 07:20
D/24 1/12 = D/30 +1/15
By solving the equation we get D =18
How do we get the equation?
Here we are making equation for usual time.
In the first case 5 minutes late from usual time. So we have to subtract the extra time, which is 5 min or 5/60 or 1/12 hr, to get the usual time
Similarly we have to add 4/60 or 1/15 hr in the 2nd case to get the usual time.
Originally posted by Raihanuddin on 28 Nov 2014, 01:41.
Last edited by Raihanuddin on 07 Jul 2015, 07:20, edited 1 time in total.



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Re: M0421
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06 Jul 2015, 13:39
Bunuel wrote: Official Solution:
A soccer coach riding his bike at 24 km/h reaches his office 5 minutes late. If he would have reached the office 4 minutes earlier than the scheduled time by traveling 25% faster, how far is his office from his house?
A. 18 km B. 24 km C. 36 km D. 40 km E. 72 km
Backsolve or create an equation. Use \(D\) for distance and \(x\) for time: \(\frac{D}{24} = x + \frac{1}{12}\) \(\frac{D}{30} = x  \frac{1}{15}\) \(D = 24x + 2\) \(D = 30x  2\) \(6x = 4\) \(x = \frac{2}{3}\) \(D = 24*\frac{2}{3} + 2 = 18\)
Answer: A Hi Bunuel, I got wrong this question because after doing all the math I realice that I took minutes instead of hours... but it was not only the problem: I actually consume almost 3 minutes. In your comments you said that backward is another way to solve this problem, could you explain this way? Also I am getting in time trouble with this kind of word problem questions that involve some equations, what would you advice to me? Thanks a lot. Best regards. Luis Navarro Looking for 700



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Re: M0421
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07 Jul 2015, 01:18
luisnavarro wrote: Bunuel wrote: Official Solution:
A soccer coach riding his bike at 24 km/h reaches his office 5 minutes late. If he would have reached the office 4 minutes earlier than the scheduled time by traveling 25% faster, how far is his office from his house?
A. 18 km B. 24 km C. 36 km D. 40 km E. 72 km
Backsolve or create an equation. Use \(D\) for distance and \(x\) for time: \(\frac{D}{24} = x + \frac{1}{12}\) \(\frac{D}{30} = x  \frac{1}{15}\) \(D = 24x + 2\) \(D = 30x  2\) \(6x = 4\) \(x = \frac{2}{3}\) \(D = 24*\frac{2}{3} + 2 = 18\)
Answer: A Hi Bunuel, I got wrong this question because after doing all the math I realice that I took minutes instead of hours... but it was not only the problem: I actually consume almost 3 minutes. In your comments you said that backward is another way to solve this problem, could you explain this way? Also I am getting in time trouble with this kind of word problem questions that involve some equations, what would you advice to me? Thanks a lot. Best regards. Luis Navarro Looking for 700 Check alternative approaches here: m04q21soccercoachbikeride70736.htmlAs for time management: practice should help. Want to speed up? Check this: Timing Strategies on the GMATOther discussions dedicated to this issue:
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Re: M0421
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20 Jul 2015, 07:25
I solved this problem in a very prevaricate way. I got the right answer but I'm just concerned I'm not solving these the fastest way. It took me almost three minutes and involved some calculation.
The overall approach was to create two equations, set them equal to each other and after finding t, and plugging t back into the original equation to find d.
Let d equal distance to office. Let t equal scheduled time to office.
\(d=rt\)
\(d=24 km/h*(t + 5)\) \(d=30 km/h*(t  4)\)
\(30(t  4) = 24(t+5)\) \(30t  120 = 24t + 120\) \(6t=240\) \(t=40\)
Plugging t into the first equation: \(d=24(40+5) = 1080 km/h\)
But that's per hour. We need to find the distance, so we must divide that by 60.
\(\frac{1080}{60} = 18 km\)
Answer choice A.
Any tips on how I can see different ways to solve this? I'm getting closer to solving these harder problems, but I'm not quite there yet to doing it flawlessly.



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Re: M0421
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26 Dec 2016, 19:56
I did this similar to gmatser1, but a bit different (in hours).
24 km/hr > 1/12 hrs late 30 km/hr > 1/15 hrs early
D=24*(t+1/12)=30*(t1/15) solve for t=2/3 hours (how long it should take at the "typical" speed  not given)
go back to the equation above to find D: D=24*(2/3+1/12)=24*(9/12)= 18km



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Re: M0421
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21 Feb 2017, 00:22
Answer: A Initial speed = 24 km/hr Later speed = (5/4) *24 km/hr = 30 km/hr Now, Difference of time in hrs, D/ 24  D/30 = 9/60 Therefore D = 18
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Re: M0421
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18 Jul 2017, 10:41
Can someone explain why this reasoning is wrong:
I assumed for path 1, time = t
for path 2, time = t  9/60
Using this method doesn't yield the same results, what is incorrect in my underlying assumption?
Thanks,



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18 Jul 2017, 20:08
omullick wrote: Can someone explain why this reasoning is wrong:
I assumed for path 1, time = t
for path 2, time = t  9/60
Using this method doesn't yield the same results, what is incorrect in my underlying assumption?
Thanks, Path 1 : speed : 24km/hr ; time: t hrs Path 2 : speed : 30km/hr ; time: t 9/60 hrs therefore, Distance= 24*t = 30* (t9/60) t = 3/4 hrs d = 24*t = 18
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Re: M0421
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19 Jul 2017, 07:43
Spartan85 wrote: omullick wrote: Can someone explain why this reasoning is wrong:
I assumed for path 1, time = t
for path 2, time = t  9/60
Using this method doesn't yield the same results, what is incorrect in my underlying assumption?
Thanks, Path 1 : speed : 24km/hr ; time: t hrs Path 2 : speed : 30km/hr ; time: t 9/60 hrs therefore, Distance= 24*t = 30* (t9/60) t = 3/4 hrs d = 24*t = 18 Wow, thanks for your help. I did this question 3 times and each time I said D=rate/time instead of D=rate*times.... don't do math at 2am. Thanks again!










