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# M04-21

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Math Expert
Joined: 02 Sep 2009
Posts: 52344

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15 Sep 2014, 23:23
1
9
00:00

Difficulty:

75% (hard)

Question Stats:

68% (01:58) correct 32% (02:25) wrong based on 179 sessions

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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

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Joined: 02 Sep 2009
Posts: 52344

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15 Sep 2014, 23:23
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:
$$\frac{D}{24} = x + \frac{1}{12}$$
$$\frac{D}{30} = x - \frac{1}{15}$$
$$D = 24x + 2$$
$$D = 30x - 2$$
$$6x = 4$$
$$x = \frac{2}{3}$$
$$D = 24*\frac{2}{3} + 2 = 18$$

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Joined: 11 Sep 2013
Posts: 148
Concentration: Finance, Finance

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Updated on: 07 Jul 2015, 06:20
2
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.
Intern
Joined: 24 Jun 2015
Posts: 46

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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:
$$\frac{D}{24} = x + \frac{1}{12}$$
$$\frac{D}{30} = x - \frac{1}{15}$$
$$D = 24x + 2$$
$$D = 30x - 2$$
$$6x = 4$$
$$x = \frac{2}{3}$$
$$D = 24*\frac{2}{3} + 2 = 18$$

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
Math Expert
Joined: 02 Sep 2009
Posts: 52344

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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:
$$\frac{D}{24} = x + \frac{1}{12}$$
$$\frac{D}{30} = x - \frac{1}{15}$$
$$D = 24x + 2$$
$$D = 30x - 2$$
$$6x = 4$$
$$x = \frac{2}{3}$$
$$D = 24*\frac{2}{3} + 2 = 18$$

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: m04-q21-soccer-coach-bike-ride-70736.html

As for time management: practice should help.

Want to speed up? Check this: Timing Strategies on the GMAT
Other discussions dedicated to this issue:

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Manager
Joined: 07 Feb 2015
Posts: 64

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20 Jul 2015, 06:25
1
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.

$$\frac{1080}{60} = 18 km$$

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.
Current Student
Joined: 30 Dec 2015
Posts: 188
Location: United States
Concentration: Strategy, Organizational Behavior
GPA: 3.88
WE: Business Development (Hospitality and Tourism)

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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!
Current Student
Joined: 23 Nov 2016
Posts: 75
Location: United States (MN)
GMAT 1: 760 Q50 V42
GPA: 3.51

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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*(t-1/15)
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
Manager
Joined: 31 Jan 2017
Posts: 58
Location: India
GMAT 1: 680 Q49 V34
GPA: 4
WE: Project Management (Energy and Utilities)

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20 Feb 2017, 23:22

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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Intern
Joined: 13 Apr 2017
Posts: 2

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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,
Manager
Joined: 31 Jan 2017
Posts: 58
Location: India
GMAT 1: 680 Q49 V34
GPA: 4
WE: Project Management (Energy and Utilities)

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18 Jul 2017, 19:08
1
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* (t-9/60)
t = 3/4 hrs
d = 24*t = 18
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Intern
Joined: 13 Apr 2017
Posts: 2

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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* (t-9/60)
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!
Manager
Joined: 04 Oct 2015
Posts: 248
Location: Viet Nam
Concentration: Finance, Economics
GMAT 1: 730 Q51 V36
GPA: 3.56

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26 Jul 2017, 01:21
1
I think this is a poor-quality 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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Do not pray for an easy life, pray for the strength to endure a difficult one - Bruce Lee

Intern
Joined: 05 Sep 2016
Posts: 20

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13 Jan 2018, 00:28
leanhdung wrote:
I think this is a poor-quality 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.
Intern
Joined: 03 Mar 2017
Posts: 9
Location: India
Schools: ISB '20

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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
Intern
Joined: 19 Jan 2019
Posts: 1

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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?
Re: M04-21 &nbs [#permalink] 21 Jan 2019, 12:29
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# M04-21

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