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Car A Car B,Time distance speed

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Car A Car B,Time distance speed

by rb90 » Fri Sep 10, 2010 2:33 am
Car B begins moving at 2 mph around a circular track with a radius of 10 miles. Ten hours later, Car A leaves from the same point in the opposite direction, traveling at 3 mph. For how many hours will Car B have been traveling when car A has passed and moved 12 miles beyond Car B?

(A) 4(PIE)- 1.6
(B) 4(PIE) + 8.4
(C) 4(PIE) + 10.4
(D) 2(PIE) - 1.6
(E) 2(PIE) - 0.8

The correct answer is B.
Please note that in every option,the first part is multiplied with pie i.e. 22/7(referred to as PIE)
Please help as i cant understand how to go about this sum.
Thanks!
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by kvcpk » Fri Sep 10, 2010 4:54 am
rb90 wrote:Car B begins moving at 2 mph around a circular track with a radius of 10 miles. Ten hours later, Car A leaves from the same point in the opposite direction, traveling at 3 mph. For how many hours will Car B have been traveling when car A has passed and moved 12 miles beyond Car B?
Radius is 10 miles.
Hence perimeter of track = 2*pi*10 = 20pi miles

Now, car B is travelling at 2 mph
in 10 hours, carb Travels 20 miles.
Distance remaining from start position for car B is 20pi-20 = 20(pi-1)

Now, Car A starts in opposite direction.
these are in opposite directions. hence speeds should be added.
Distance to be covered = 20(pi-1) + 12 [since car B has to be 12 miles away from car A]

Hence time = (20(pi-1)+12)/5 = 4(pi-1) +12/5 hours.
=4pi -4 +2.4
=4pi-1.6

But, we need to 10 hours to it. Because car A started after 10 hours.

4pi-1.6+10 = 4pi+8.4 hours.

Hope this helps!!
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by sanju09 » Fri Sep 10, 2010 5:14 am
rb90 wrote:Car B begins moving at 2 mph around a circular track with a radius of 10 miles. Ten hours later, Car A leaves from the same point in the opposite direction, traveling at 3 mph. For how many hours will Car B have been traveling when car A has passed and moved 12 miles beyond Car B?

(A) 4(PIE)- 1.6
(B) 4(PIE) + 8.4
(C) 4(PIE) + 10.4
(D) 2(PIE) - 1.6
(E) 2(PIE) - 0.8

The correct answer is B.
Please note that in every option,the first part is multiplied with pie i.e. 22/7(referred to as PIE)
Please help as i cant understand how to go about this sum.
Thanks!

On a loop stretch of 20 π miles, is a point P where the two cars are bound to start from. Ten hours later, Car B will be at a point Q, which means a 20 miles long arc stretch PQ, let's say Q in clock direction to P, then starts Car A from point P in counter-clock direction to P.

Remember, there remains a head to head arc stretch of 20 π - 20 miles between the two cars at the moment Car A starts from point P. Hence, cars need to cover this arc stretch of 20 π - 20 miles, which will be covered at a relative speed of 5 mph in 4 π - 4 hours plus 12/5 = 2.4 more hours in order to move 12 miles beyond Car B after passing.

Time taken by Car A to do so = (4 π - 4 + 2.4) = 4 π - 1.6 hours, and Car B has traveled 10 more hours than this by this time.

ANSWER = 10 + 4 π - 1.6 hours = [spoiler]4 π + 8.4 hours


B[/spoiler]
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by RCV » Fri Sep 10, 2010 6:23 am
sanju09 wrote:
rb90 wrote:Car B begins moving at 2 mph around a circular track with a radius of 10 miles. Ten hours later, Car A leaves from the same point in the opposite direction, traveling at 3 mph. For how many hours will Car B have been traveling when car A has passed and moved 12 miles beyond Car B?

(A) 4(PIE)- 1.6
(B) 4(PIE) + 8.4
(C) 4(PIE) + 10.4
(D) 2(PIE) - 1.6
(E) 2(PIE) - 0.8

The correct answer is B.
Please note that in every option,the first part is multiplied with pie i.e. 22/7(referred to as PIE)
Please help as i cant understand how to go about this sum.
Thanks!

On a loop stretch of 20 π miles, is a point P where the two cars are bound to start from. Ten hours later, Car B will be at a point Q, which means a 20 miles long arc stretch PQ, let's say Q in clock direction to P, then starts Car A from point P in counter-clock direction to P.

Remember, there remains a head to head arc stretch of 20 π - 20 miles between the two cars at the moment Car A starts from point P. Hence, cars need to cover this arc stretch of 20 π - 20 miles, which will be covered at a relative speed of 5 mph in 4 π - 4 hours plus 12/5 = 2.4 more hours in order to move 12 miles beyond Car B after passing.

Time taken by Car A to do so = (4 π - 4 + 2.4) = 4 π - 1.6 hours, and Car B has traveled 10 more hours than this by this time.

ANSWER = 10 + 4 π - 1.6 hours = [spoiler]4 π + 8.4 hours


B[/spoiler]
Awesome sanju09, do they test circular motion
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by GMATGuruNY » Fri Sep 10, 2010 9:32 am
rb90 wrote:Car B begins moving at 2 mph around a circular track with a radius of 10 miles. Ten hours later, Car A leaves from the same point in the opposite direction, traveling at 3 mph. For how many hours will Car B have been traveling when car A has passed and moved 12 miles beyond Car B?

(A) 4(PIE)- 1.6
(B) 4(PIE) + 8.4
(C) 4(PIE) + 10.4
(D) 2(PIE) - 1.6
(E) 2(PIE) - 0.8

The correct answer is B.
Please note that in every option,the first part is multiplied with pie i.e. 22/7(referred to as PIE)
Please help as i cant understand how to go about this sum.
Thanks!
Quickest approach is to ballpark. Pi = approximately 3.

Car B is traveling for more than 10 hours, so answer choices D and E are too small, and A is unlikely. The correct answer is either B or C.

Circumference of track = 20pi = 60 miles approximately.
In 10 hours, distance for B = 2*10 = 20 miles.
60-20 = 40 miles between A and B.
A and B now have to travel the 40 miles between them and then -- after they meet -- keep traveling in opposite directions until there is another 12 miles between them. So the total distance that they need to travel is 40+12 = 52 miles.
When things work together, we can add their rates. Combined rate for A+B = 3+2 = 5 miles/hour.
Time for A+B = Distance/Rate = 52/5 = 10.4 hours.
Since B traveled for 10 hours before A joined in, the total time for B = 10 + 10.4 = 20.4 hours.
Only answer choice B works: 4pi + 8.4 = 12 + 8.4 = 20.4 approximately.

The correct answer is B.
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by scorpionz » Fri Sep 10, 2010 9:54 pm
Good ol' BTG search to the rescue....

This question has already been discussed on multiple threads...

https://www.beatthegmat.com/rate-problem ... tml#294901

https://www.beatthegmat.com/rates-and-ci ... 45863.html

https://www.beatthegmat.com/a-rate-quest ... 51585.html

https://www.beatthegmat.com/hollyyy-diff ... 47861.html

https://www.beatthegmat.com/car-t60352.html
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