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Lets say t = meeting time, obviously cars won't meet in the starting point right away so they will definitely do some rounds first, lets say a = amount of rounds first car does, b - second car. We get an equation.
\(t = \frac{2*pi*6*a}{6} = \frac{2*pi*6*b}{8}\)
after solving the 2nd equation I get the relation between a and b: a/b = 3/4. So basically their first meeting at starting point happens when first car does 3 laps and 2nd car does 4 laps, therefore after we put a into our equation for t:
\(t = \frac{2*pi*6*3}{6} = 6*pi\) hours. At this point I'm not sure what I'm doing wrong.
edit: with the amendments to the task, result is 6, answer is D
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Bunuel
Two cars run in opposite directions on a circular track. Car A travels at a rate of 6? miles per hour and Car B runs at a rate of 8? miles per hour. If the track has a radius of 6 miles and the cars both start from Point S at the same time, how long, in hours, after the cars depart will they again meet at Point S?

(A) 6/7 hrs
(B) 12/7 hrs
(C) 4 hrs
(D) 6 hrs
(E) 12 hrs


Kudos for a correct solution.

This is case when we can't use sum of speeds because we need know when cars meet at the start of the circle distance.
So we should not sum speeds but subtract it 8pi - 6pi = 2pi
Circumference of circle equal to 2piR = 12pi
12pi / 2pi = 6 hours

Answer is D
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could you plz explain why finding the difference of speeds helps us to reach to the solution? And in what cases do we need to find it?
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marinab
could you plz explain why finding the difference of speeds helps us to reach to the solution? And in what cases do we need to find it?


when you are asked that

1) after how many hours the two cars will meet again. Sum the speeds .

2) after how many hours the two cars will meet at the starting point again . Subtract the speeds.
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could you plz explain why finding the difference of speeds helps us to reach to the solution? And in what cases do we need to find it?

Our cars drive by circle. One clockwise and another counterclock wise (actually it doesnt' matter).
And we need to find when this two cars intersect on the start.

First car driving with speed 6pi and another car with speed 8pi
Circle equal to 12 pi

Hand calculating approach (for better understanding what's going on):
After two hours, first car (6pi) will be on start but another car will drive 16 pi because it have speed 8pi and by two hours it'll be 16pi
After the next hours first will drive 24pi and will be again on the start but second car will be on 32pi so it's not the start position
And finally after six hours both cars will be simultaneously on the start 6pi * 6 = 36 (3 times 12) and 8pi * 6 = 48 (4 times 12)

Formulas approach:
Why we should subtract speeds? Second car drives faster. So we want to know when second car make on 1 circle more than first car which drives slowly.
We should find how much second car faster than first car. 8pi - 6pi = 2pi
And for now we should divide distance 12pi on 2pi and we get 6 hours.

So we know that after 6 hours second car will drive on 1 circle more.
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Bunuel
Two cars run in opposite directions on a circular track. Car A travels at a rate of \(6\pi\) miles per hour and Car B runs at a rate of \(8\pi\) miles per hour. If the track has a radius of 6 miles and the cars both start from Point S at the same time, how long, in hours, after the cars depart will they again meet at Point S?

(A) 6/7 hrs
(B) 12/7 hrs
(C) 4 hrs
(D) 6 hrs
(E) 12 hrs


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Circumference of the track - \(12\pi\)
Car A, with a speed of \(6\pi\), will get back to point S in two hours.
Car B, with a speed of \(8\pi\), will get back to point S in 1h3m.

so let's say they start at 12:00 PM...
1:30PM car B full lap
2:00PM car A full lap

3:00PM car B full 2 lap
4:00PM car A full 2 lap

4:30PM car B full 3 lap
6:00PM car A full 3 lap

6:00PM car B full 4 lap
8:00PM car A full 4 lap

so..we have some overlaps here..
when Car A finishes 3 laps, car B finishes 4 laps, and both cars are at the same point..S...
therefore, it takes 6 hours for them to meet again at the point S.

the answer is D.
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Since the track is circular so if the cars are driving in opposite direction it can be considered they meet twice
So to calculate the hours taken there can be two scenarios
1) after how many hours the two cars will meet again. Find the sum of the speeds
2) after how many hours the two cars will meet at the starting point again . Find the difference of the speeds.
Now that we know the circumference is 12*pi.
Every hour, car B that runs at a speed of 8π mph covers 2π miles more compared with the distance covered by car A in that time (which runs at a speed of 6ππ mph). So car B to cover 12π miles (a full circle) more than car A should be given by 12π/2π hrs = 6 hrs.

Option D.
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LCM of (L/a, ;L/b) is answer. L- Length of track; a- speed of car A, b- speed of car B

LCM (12pi/6pi, 12pi/8pi) = LCM (2, 1.5) = 6 (ANSWER)
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This is a very simple problem so need to make it unnecessarily complicated.

Let the speeds of Car A and Car B be Va and Vb respectively. The ratio of their speeds is:
Va/Vb=6pi/8pi=3/4 which means that B completes 4 revolutions in the time that A completes 3 revs after which they meet at their starting point S.
Time for A to complete 3 revs = 12pi*3/6pi = 6 hours.
Or, time for B to complete 4 revs = 12pi*4/8pi = (also) 6 hours. ANS: D
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Bunuel
Two cars run in opposite directions on a circular track. Car A travels at a rate of \(6\pi\) miles per hour and Car B runs at a rate of \(8\pi\) miles per hour. If the track has a radius of 6 miles and the cars both start from Point S at the same time, how long, in hours, after the cars depart will they again meet at Point S?

(A) 6/7 hrs
(B) 12/7 hrs
(C) 4 hrs
(D) 6 hrs
(E) 12 hrs

Circumference of the track = 2πr = 2*π*6 = 12π miles.
Implication:
For a car to complete a loop and return to S, the distance traveled must be a MULTIPLE OF 12π MILES.
Since Car A travels 6π miles each hour, it takes 2 hours for Car A to travel a multiple of 12π miles and return to S:
2*6π = 12π miles
Since Car B travels 8π miles each hour, it takes 3 hours for Car B to travel a multiple of 12π miles and return to S:
3*8π = 24π miles
Thus, for both cars to return to S simultaneously, the time must be a MULTIPLE OF 2 HOURS AND 3 HOURS.
The smallest answer choice divisible by 2 hours and 3 hours is D.

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can you please explain this one ?
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Pranjal3107
ScottTargetTestPrep
can you please explain this one ?
Solution:

Notice that the question is not asking when the two cars will meet again, but instead is asking for when the two cars will meet again at point S.

Since the radius of the track is 6 miles, its circumference is 12𝜋. Car A will again be at point P after traveling 12𝜋 miles, 24𝜋 miles, 36𝜋 miles, 48𝜋 miles etc. Thus, car A will be again at point P after 12𝜋/6𝜋 = 2 hours, 24𝜋/6𝜋 = 4 hours, 36𝜋/6𝜋 = 6 hours, 48𝜋/6𝜋 = 8 hours and so on.

Similarly, car B will again be at point P after traveling the same number of miles, which correspond to 12𝜋/8𝜋 = 3/2 hours, 24𝜋/8𝜋 = 3 hours, 36𝜋/8𝜋 = 9/2 hours, 48𝜋/8𝜋 = 6 hours and so on.

We see that the two cars are again together at point P after 6 hours.

Solution: D
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Circumference of the track = 2π∗r=2π∗6=12π or 12*pi

Time needed by Car A to complete 1 circle = t=12∗pi / 6∗pi=2hrs.

Time needed by Car B to complete 1 circle = t=12∗pi / 8∗pi=1.5hrs.

LCM of 1.5 and 2 = 6 (i.e LCM of 15 and 20 = 60).

This means that at 6 hrs from starting time, both Car A & Car B will be at starting point S.
Answer: D

Alternate way to understand:
Car A : At every {0, 2, 4, 6, 8, 10, .....} hrs , Car A will be at point S.
Car B : At every {0, 1.5, 3, 4.5, 6, 7.5 ......} hrs , Car B will be at point S.

So, at 6th hour both Car A & Car B will be at point S.
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King407
Bunuel
Two cars run in opposite directions on a circular track. Car A travels at a rate of \(6\pi\) miles per hour and Car B runs at a rate of \(8\pi\) miles per hour. If the track has a radius of 6 miles and the cars both start from Point S at the same time, how long, in hours, after the cars depart will they again meet at Point S?

(A) 6/7 hrs
(B) 12/7 hrs
(C) 4 hrs
(D) 6 hrs
(E) 12 hrs


Kudos for a correct solution.


Circumference of the track = \(2\pi*r = 2\pi*6 = 12\pi\)

Time needed by Car A to complete 1 circle = \(t = \frac{12*pi}{6*pi} = 2 hrs.\)

Time needed by Car B to complete 1 circle = \(t = \frac{12*pi}{8*pi} = 1.5 hrs.\)

LCM of 1.5 and 2 = 6 (i.e LCM of 15 and 20 = 60).

Hence D is the correct answer
Much more intuitive method IMO. Also love that DP!
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