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# M04 Q21 - soccer coach bike ride

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M04 Q21 - soccer coach bike ride [#permalink]

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25 Sep 2008, 17:23
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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?

(A) 18 km
(B) 24 km
(C) 36 km
(D) 40 km
(E) 72 km

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

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
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Re: M04 Q21 - soccer coach bike ride [#permalink]

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26 Sep 2008, 17:35
1
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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.

***** You can solve this using alegbra too

24km/hour = 24km/60 minutes = 2/5 km/min=.4km/min
30 km/hour= 30km/60 minutes = 1/2 km/min=.5km/min

Distance is unknown X
Desired time is unknown y

1) X/2/5= y +5

2) X/1/2= y -4

Solve for y using either equation

(using no. 2) y= X/1/2 + 4

Substitute for y, X/2/5= y + 5, X/2/5= X/1/2 + 4 + 5
Convert Fractions 5x/2 (or 2.5) = 2X + 9
Subtract 2x from both sides = 2.5X-2X = 9, .5x=9
Multiply both sides by 2,
x=18km
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Re: M04 Q21 - soccer coach bike ride [#permalink]

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21 May 2010, 08:32
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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
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Re: M04 Q21 - soccer coach bike ride [#permalink]

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13 Oct 2010, 05:54
A (18 km) will be the answer.
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Re: M04 Q21 - soccer coach bike ride [#permalink]

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13 Oct 2010, 06:52
Yeah..the answer got to be A.
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Re: M04 Q21 - soccer coach bike ride [#permalink]

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13 Oct 2010, 07:38
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Let t be the time

25% faster than 24 km/h = 30 km/h

(t+(5/60))24 = (t-(4/60))30
(t+ (1/12))24 = ( t - (1/15))30
(12t+1)2 = (15t-1)2
t = 2/3 hours

Distance can be calculated by substituting the time in any one of the equations
(t+(5/60))24 OR (t-(4/60))30

Distance = 18 km Answer A

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Re: M04 Q21 - soccer coach bike ride [#permalink]

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13 Oct 2010, 11:18
3
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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

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Re: M04 Q21 - soccer coach bike ride [#permalink]

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13 Oct 2010, 12:12
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Very Easy And Alternate Way

let Speed S1 =24 km/Hr then time taken T1 & Distance =D
where T1 =T+5 (says reaches 5 min late)

Now if 25% faster
Then S2= 125%*24= 30 Km/hr then time taken T2 & Distance=D (distance is same)
where T2 =T-4 (says reaches 4 min early)

by speed formula:

D=S1*T1 ....i)

D=S2*T2......ii)

Taking taking equation i) &ii)

S1*T1=S2*T2

24*(T+5)=30(T-4)
24T +120=30T -120
6T=240
T=240/6=40 (remember this is in min)

so time taken when speed is 24 Km is 45 min...

45 min = 3/4 hrs
Plugging it in any eq say i)
24=D/(3/4)
24*3=4D
D=18 Km

Hope this helps.... kindly give ur feedback about the method...

Karan
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Re: M04 Q21 - soccer coach bike ride [#permalink]

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13 Oct 2010, 20:20
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I have another approach which is quicker.

T1 is the time the coach needs when travel with speed v1 = 25km/h
T2 is the time he needs when travel with speed v2 = 1.25v1 = 30km/h

Compare to the scheduled time, T1 is 5 mins later, T2 is 4 mins earlier. We have:

T1 - T2 = 9 mins = 9/60 hour

Therefore, we have:

d/24 - d/30 = 9/60
=> d = 18 km. The answer is A.
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Re: M04 Q21 - soccer coach bike ride [#permalink]

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13 Oct 2010, 22:43
thutran wrote:
I have another approach which is quicker.

T1 is the time the coach needs when travel with speed v1 = 25km/h
T2 is the time he needs when travel with speed v2 = 1.25v1 = 30km/h

Compare to the scheduled time, T1 is 5 mins later, T2 is 4 mins earlier. We have:

T1 - T2 = 9 mins = 9/60 hour

Therefore, we have:

d/24 - d/30 = 9/60
=> d = 18 km. The answer is A.

This is the best and shortest explaination
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Re: M04 Q21 - soccer coach bike ride [#permalink]

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14 Oct 2010, 07:05
(t+5)24/60=(t-4)30/60
t=40 min
40-4/2=18km
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Re: M04 Q21 - soccer coach bike ride [#permalink]

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14 Oct 2010, 13:11
A

key is find T and then to find D and not stop at T.
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Re: M04 Q21 - soccer coach bike ride [#permalink]

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28 Sep 2011, 07:01
I solved that way

24*H=R
(24*1,12)*(H-9/60)=R, where 9/60 is 5+4 minutes
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Re: M04 Q21 - soccer coach bike ride [#permalink]

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17 Oct 2011, 06:52
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
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Re: M04 Q21 - soccer coach bike ride [#permalink]

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17 Oct 2011, 11:28
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Re: M04 Q21 - soccer coach bike ride [#permalink]

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17 Oct 2011, 15:18
Another fast approach to solve this problem:

At 25% faster, speed is 30 Km/hr.
At 30km/hr, lets say time taken is T.

Distance still to be covered @ 24 km/hr speed (this is after time T) =
24 * 9/60 = 3.6 km

At 6 km/hr (which is the speed difference) in time T, Distance covered =
3.6 km (calculated above)

So at 30 km/hr distance covered in time T =
3.6 * 5 = 18kms.
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Re: M04 Q21 - soccer coach bike ride [#permalink]

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17 Nov 2011, 23:19
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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.

Maybe 30 sec.
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Re: M04 Q21 - soccer coach bike ride [#permalink]

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17 Oct 2012, 23:43
I didn't check view all threads . however I'll provide a method with which you can solve practically any problem whether speed or time is given.

24km/hr --- 5 mins late
30km/hr--- 4 mins early
Diff in time = 9mins.

(had he reached at 8:00am ideal time; 24km/hr --- reaches at 8:05
30km/hr --- reaches at 7:56.)

lcm ( 24 and 30) =120..
hence if the distance was 120km..
24km/hr would take 5hrs
30km/hr would take 4hrs.
Diff in time = 1 hr..

apply Unitary method

dist............time diff.
120...........1hr(60min)
? ............. 9mins.

(120*9)/60= 18km

** Note-- if only the time taken is given , you can use the same method to solve . Only replace speed with time for the lcm..

Cheers
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Re: M04 Q21 - soccer coach bike ride [#permalink]

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18 Oct 2012, 00:59
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?

(A) 18 km
(B) 24 km
(C) 36 km
(D) 40 km
(E) 72 km

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

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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Re: M04 Q21 - soccer coach bike ride [#permalink]

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14 Dec 2012, 12:23
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.
Re: M04 Q21 - soccer coach bike ride   [#permalink] 14 Dec 2012, 12:23

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# M04 Q21 - soccer coach bike ride

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