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Donovan and Michael are racing around a circular 400-meter track. If D

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Donovan and Michael are racing around a circular 400-meter track. If D  [#permalink]

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New post 16 Feb 2015, 06:59
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Donovan and Michael are racing around a circular 400-meter track. If Donovan runs each lap in 45 seconds and Michael runs each lap in 40 seconds, how many laps will Michael have to complete in order to pass Donovan, assuming they start at the same time?

A. 8
B. 9
C. 10
D. 11
E. 12


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Re: Donovan and Michael are racing around a circular 400-meter track. If D  [#permalink]

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New post 16 Feb 2015, 21:08
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@
Bunuel wrote:
Donovan and Michael are racing around a circular 400-meter track. If Donovan runs each lap in 45 seconds and Michael runs each lap in 40 seconds, how many laps will Michael have to complete in order to pass Donovan, assuming they start at the same time?

A. 8
B. 9
C. 10
D. 11
E. 12


Kudos for a correct solution.


Simply take the LCM of 40 and 45 - you get 360. This is the time they will take to meet for the first time. In this time, Michael would have covered 360/40 = 9 laps.
Answer (B)

The reason for this:

Judging from the question, we know that Michael and Donovan are running in the same direction. When will Michael overtake Donovan? Michael will always be a bit ahead of Donovan and will keep increasing this distance between them till the time he completes one full lap more than Donovan. That is when they will both again be at the starting point together. The time that should have passed at this point should be a multiple of both 40 and 45 so that they are at the starting point. So we find the LCM of 40 and 45 which is 360. In this time, Donovan completes 360/45 = 8 laps and Michael completes 360/40 = 9 laps.
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Re: Donovan and Michael are racing around a circular 400-meter track. If D  [#permalink]

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New post 16 Feb 2015, 07:36
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Bunuel wrote:
Donovan and Michael are racing around a circular 400-meter track. If Donovan runs each lap in 45 seconds and Michael runs each lap in 40 seconds, how many laps will Michael have to complete in order to pass Donovan, assuming they start at the same time?

A. 8
B. 9
C. 10
D. 11
E. 12


Kudos for a correct solution.


As per options given, Assuming that the two runners are running along the track in the SAME DIRECTION

Speed of Michael = Distance/Time = 400/40 = 10 m/Sec
Speed of Donovan = Distance/Time = 400/45 = 80/9 m/Sec

Relative Speed for the two objects moving in the same direction = Difference between their speed

Note: Relative Speed = The rate at which the distance between the two moving objects increases or decreases
In same Direction, The Distance Increases at lower rate, and the Rate = S1 - S2
In Opposite Direction, The Distance Increases at Higher rate, and the Rate = S1 + S2


Hence Relative Speed of Michael and Donovan = 10 - (80/9) = 10/9 m/sec

Relative Distance to be covered by Michael to pass Donovan = Length of complete Track = 400 mt

Time Taken by Michael to pass Donovan = Relative Distance / Relative Speed = 400 / (10/9) = 360 Secs

No. of Laps completed by Michael in 360 Secs = Total Time to pass Donovan / Time of Michael per Lap = 360/40 = 9

Answer: Option
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Re: Donovan and Michael are racing around a circular 400-meter track. If D  [#permalink]

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New post 16 Feb 2015, 07:51
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Bunuel wrote:
Donovan and Michael are racing around a circular 400-meter track. If Donovan runs each lap in 45 seconds and Michael runs each lap in 40 seconds, how many laps will Michael have to complete in order to pass Donovan, assuming they start at the same time?

A. 8
B. 9
C. 10
D. 11
E. 12


Kudos for a correct solution.


ALTERNATE METHOD:

Since Time taken by Michael to cover track < Time taken by Donovan to cover track
therefore, Speed of Michael > Speed of Donovan

Now, Let's say that Michael passes Donovan in 't seconds' time

Then Michael will have to cover the distance of 400 mt more than Donovan's distance in order to pass him running in the same direction.

Therefore if Distance covered by Donovan = D mt
then Distance covered by Michael = (D+400) mt

Speed of Michael = Distance/Time = 400/40 = 10 m/Sec
Speed of Donovan = Distance/Time = 400/45 = 80/9 m/Sec

For Constant Time
S1 / S2 = D1 / D2


10 / (80/9) = (D+400) / D

i.e. 9 D = 8 D + 3200
i.e. D = 3200 mt.

Laps Completed by Michael = Total Distance Covered by Michael / Distance per Lap = (3200+400) / 400 = 9

Answer: Option
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Re: Donovan and Michael are racing around a circular 400-meter track. If D  [#permalink]

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New post 16 Feb 2015, 07:52
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Bunuel wrote:
Donovan and Michael are racing around a circular 400-meter track. If Donovan runs each lap in 45 seconds and Michael runs each lap in 40 seconds, how many laps will Michael have to complete in order to pass Donovan, assuming they start at the same time?

A. 8
B. 9
C. 10
D. 11
E. 12


Kudos for a correct solution.


Speed of D=400/45=8.88(approx)m/s
Speed of M=400/40=10m/s
Rel speed =1.2m/s
So M gets a lead of 1.2 m in one second
so in order to get a lead of 400m he will take=400/1.2=333.33 sec and it equivalents to 333.33/40=8.33 rounds and hence ans is B.
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Re: Donovan and Michael are racing around a circular 400-meter track. If D  [#permalink]

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New post 16 Feb 2015, 19:23
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Hi All,

Since we have the "lap speeds" of each of the two runners, the length of the lap is IRRELEVANT. The "math" behind this question is the same as a "Chase Down"/Combined Rate question.

Donovan runs each lap in 45 seconds.
Michael runs each lap in 40 seconds.

The question asks how long it will take Michael to "pass" (re: "lap") Donovan.

Since the difference in their two times is 45-40 = 5 seconds, Michael will "catch up" 5 seconds with every lap. Donovan spends 45 seconds per lap, so Michael will need 45/5 = 9 laps to catch Donovan.

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Re: Donovan and Michael are racing around a circular 400-meter track. If D  [#permalink]

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New post 16 Feb 2015, 20:41
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jmo: the easiest way to nail down this question s to plug the number in
for 8: 400*8 = 3200, both 3200/45 and 3200/40 are not integer, kill
for 9: 400*9 = 3600, both 3600/45 and 3600/40 are integer, so they overlap.
-> Correct answer is B in 30 seconds.
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Re: Donovan and Michael are racing around a circular 400-meter track. If D  [#permalink]

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New post 17 Feb 2015, 01:24
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Answer = B. 9

LCM of 40 & 45 = 360

360 is the point wherein both will "meet" after starting up of the race.

Laps required by Michael \(= \frac{360}{40} = 9\)
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Re: Donovan and Michael are racing around a circular 400-meter track. If D  [#permalink]

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New post 17 Feb 2015, 02:17
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[quote="Bunuel"]Donovan and Michael are racing around a circular 400-meter track. If Donovan runs each lap in 45 seconds and Michael runs each lap in 40 seconds, how many laps will Michael have to complete in order to pass Donovan, assuming they start at the same time?

A. 8
B. 9
C. 10
D. 11
E. 12


Kudos for a correct solution.[/quote

One way of approaching this question is by Relative speed method
1. Speed/ Rate of Donovan = Distance/ time => 400/45 =>80/9
2. Speed/ Rate of Michael = Distance/ time => 400/40 => 10

Relative Speed between them = 10 - 80/9 => 10/9 (We subtract the Rates if moving in the same direction and add the rates if moving in the opposite direction)

In order to pass Donovan-
Distance to be covered = 400, Relative Rate = 10/9

Total Time taken by Micheal to surpass Donovan = Distance / rate => 400*9/10 => 3600/10 => 360

No. of laps taken by Michael = Total time / Michael's rate => 360/40 => 9

Hence correct answer is 9 Laps.
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Re: Donovan and Michael are racing around a circular 400-meter track. If D  [#permalink]

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New post 22 Feb 2015, 11:25
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Bunuel wrote:
Donovan and Michael are racing around a circular 400-meter track. If Donovan runs each lap in 45 seconds and Michael runs each lap in 40 seconds, how many laps will Michael have to complete in order to pass Donovan, assuming they start at the same time?

A. 8
B. 9
C. 10
D. 11
E. 12


Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION

Solution: B.

This problem can be tricky with algebra, but with logic it’s much simpler. On each lap that Donovan runs, Michael runs an additional 1/8 of a lap with the additional 5 seconds. In order to get a full lap ahead, Michael will need Donovan to have run 8 laps.

The “trick” comes here – Donovan will have run 8 laps, but Michael will have completed his ninth. So the correct answer is B – when Donovan completes his 8th lap, Michael will be completing his ninth.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Donovan and Michael are racing around a circular 400-meter track. If D  [#permalink]

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New post 15 Feb 2016, 16:24
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M's rate is 1/40 lap per second
D's rate is 1/45 lap per second
(1/40)/(1/45)=9/8
M runs 9 laps for every 8 laps D runs
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Re: Donovan and Michael are racing around a circular 400-meter track. If D  [#permalink]

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New post 21 Mar 2017, 06:08
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Bunuel wrote:
Donovan and Michael are racing around a circular 400-meter track. If Donovan runs each lap in 45 seconds and Michael runs each lap in 40 seconds, how many laps will Michael have to complete in order to pass Donovan, assuming they start at the same time?

A. 8
B. 9
C. 10
D. 11
E. 12


We are given that Donovan runs each lap in 45 seconds and Michael runs each lap in 40 seconds. Thus, their respective speeds are 400/45 = 80/9 meters per second and 400/40 = 10 meters per second.

If Michael passes Donovan t seconds after they start running, he will have traveled exactly 400 meters more than Donovan (because 400 meters is equal to one lap). Since distance = rate x time, we can say that in t seconds, Donovan covers a distance of 80t/9 meters and Michael covers a distance of 10t meters. Since the difference between these distances has to equal 400, we have the following:

10t - 80t/9 = 400

90t/9 - 80t/9 = 400

10t/9 = 400

10t = 3600

t = 360 seconds

In 360 seconds, Michael completes 360/40 = 9 laps.

Answer: B
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Re: Donovan and Michael are racing around a circular 400-meter track. If D  [#permalink]

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New post 23 Mar 2017, 12:44
Michael makes a lap in 40 sec. He makes x laps. Time spent on road = 40x
When Michael reaches Donovan, Donovan has made exactly 1 lap less. It takes him 45 sec to make a lap. Total time spent on road = 45(x-1)
40x=45(x-1)
5x=45
x=9 laps (B)
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Re: Donovan and Michael are racing around a circular 400-meter track. If D  [#permalink]

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New post 03 Apr 2018, 10:04
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Bunuel wrote:
Donovan and Michael are racing around a circular 400-meter track. If Donovan runs each lap in 45 seconds and Michael runs each lap in 40 seconds, how many laps will Michael have to complete in order to pass Donovan, assuming they start at the same time?

A. 8
B. 9
C. 10
D. 11
E. 12


Kudos for a correct solution.


If Michael passes Donovan, then Donovan completes ONE LAP LESS than Michael completes IN THE SAME AMOUNT OF TIME.
So, if x = # of laps Michael completes, then....
x - 1 = # of laps Donovan completes.

So, we can write the following WORD EQUATION:
(time for Michael to complete x laps) = (time for Donovan to complete x - 1 laps)
It takes 40 seconds for Michael to complete EACH lap, and it takes 45 seconds for Donovan to complete EACH lap.
So, we get: (40)(x) = (45)(x - 1)
Expand: 40x = 45x - 45
Solve to get x = 9

So, Michael must complete 9 laps

Answer: B

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Re: Donovan and Michael are racing around a circular 400-meter track. If D &nbs [#permalink] 03 Apr 2018, 10:04
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