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Re M0421
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15 Sep 2014, 23:23



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M0421
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Updated on: 07 Jul 2015, 06:20
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, 00:41.
Last edited by Raihanuddin on 07 Jul 2015, 06:20, edited 1 time in total.



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Re: M0421
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06 Jul 2015, 12:39
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, 00:18
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, 06:25
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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30 Jul 2016, 04:26
it would be really helpful to have some links to similar questions to make sure the concept has been internalized!



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Re: M0421
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26 Dec 2016, 18:56
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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20 Feb 2017, 23:22
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, 09:41
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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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, 06:43
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!



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Re M0421
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26 Jul 2017, 01:21
I think this is a poorquality question. The stem MUST specify "25% faster than" WHAT!!! Is it "25% faster than" NORMAL SPEED (the SPEED at which the soccer reaches the office ON TIME), or "25% faster than" 24 km/h ((the SPEED at which the soccer reaches the office 5 minutes LATE). Otherwise, it is confusing.
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Re: M0421
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13 Jan 2018, 00:28
leanhdung wrote: I think this is a poorquality question. The stem MUST specify "25% faster than" WHAT!!!
Is it "25% faster than" NORMAL SPEED (the SPEED at which the soccer reaches the office ON TIME), or "25% faster than" 24 km/h ((the SPEED at which the soccer reaches the office 5 minutes LATE).
I completely agree with you. It should have been clearly mentioned 25% faster than what, usual speed or 24kmph. It's creating ambiguity.
Otherwise, it is confusing.



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Re: M0421
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25 Dec 2018, 23:26
I agree that this a poor quality question as it is not explained what is the speed faster than? is it faster than 24kmph or the original speed x. as both will yield different answers



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21 Jan 2019, 12:29
Hi Bunuel, I attempted this question as follows but seem to be getting the wrong answer: Initial scenario
Rate = 24 km/hour = (6/15) km/min Assume time = (T+4) [because this scenario reaches 5 min late vs. faster scenario reaches 1 min late, therefore difference of 4 minutes] Distance = (6/15)*(T+4) Faster scenario (25% faster vs. original 24 km/hr)
Rate = (6/15)*(1+1/4) = 1/2 km/min Assume time = T Distance = (1/2)*T Equating both sides:
(6/15)*(T+4) = (1/2)*T ==> T = 16 min Thus, plugging T back into second equation: D = (1/2)*(16) = 8 km Where did I go wrong?










