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Rate problem

by ru2008 » Mon Sep 06, 2010 5:54 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?
4 * (pi)- 1.6
4 * (pi) + 8.4
4 * (pi) + 10.4
2 * (pi)- 1.6
2 * (pi) - 0.8


Could someone explain the process?

Source: MGMAT
Last edited by ru2008 on Mon Sep 06, 2010 6:56 am, edited 1 time in total.
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by scorpionz » Mon Sep 06, 2010 6:12 am
Dude,

The options seem weird and all options are definitely incorrect...

Are you sure there is no pi value (3.14) in any of the options??
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by mj78ind » Mon Sep 06, 2010 7:16 am
Ok so the two cars together have to travel (2*pi*10 - 20) miles at the combined rate of (2 +3) mph. Also they then move apart by 12 miles so the total distance they have to travel = (2*pi*10 - 20) + 12 and they have to do this at 5 mph

Thus time of travel of cars = {(2*pi*10 - 20) + 12}/5 but wait there is another trick the car B had been traveling for 10 hours already before car A started. Thus total time for which car B has been traveling:

{(2*pi*10 - 20) + 12}/5 + 10 = 4*pi + 8.4
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by scorpionz » Mon Sep 06, 2010 7:31 am
ru2008 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?
4 * (pi)- 1.6
4 * (pi) + 8.4
4 * (pi) + 10.4
2 * (pi)- 1.6
2 * (pi) - 0.8


Could someone explain the process?

Source: MGMAT
Here's the process -

Total distance of track = 2*pi*r = 20pi
Initial time that B has run = 10 hrs --------------------------------------------> 1
Distance covered by B in 10 hrs = 20 miles
Hence distance between Starting point and B after 10 hours = 20pi - 20

Since A & B are approaching each other, the effective speed = speed of A + speed of B = 5 mph

Time taken for A & B to meet = (20pi - 20) / 5 = 4pi - 4 hrs ---------------------> 2

Time taken for A & B to separate 12 miles = Distance / Effective speed = 12 / 5 = 2.4 hrs ---------------> 3

Thus, the total time that B has run = sum of 1, 2 and 3 above
= 10 + 4pi - 4 + 2.4
= 4pi + 8.4 hrs
= Option B

Hope this helps!!

Cheers!!
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by Maciek » Mon Sep 06, 2010 7:50 am
Hi all!
Car B begins moving at 2 mph around a circular track with a radius of 10 miles.
Let X be the position of car B after 10 hours
t1 = 10 hours
X = 2*10 = 20 miles
the length of track is 2*(pi)*R = 2*(pi)*10 = 20*(pi)
Ten hours later, Car A leaves from the same point in the opposite direction, traveling at 3 mph.
the length of route between cars A and B is (20*(pi) - 20) miles
20*(pi) - 20 = sA + sB
tA = tB
tA = sA/vA = sA/3
tB = sB/vB = sB/2
sA/3 = sB/2
sA = sB*3/2
20*(pi) - 20 = sB*3/2 + sB
sB = (20*(pi) - 20)/(1 + 3/2)
sB = 2*20((pi) - 1)/5 = 8(PI - 1)
tB = sB/vB = 8((pi) - 1)/2 = 4(PI - 1)
When both cars pass each other, car B has travelled for t2 hours.
t2 = t1 + tB = 10 + 4((pi) - 1)
For how many hours will Car B have been traveling when car A has passed and has moved 12 miles beyond Car B?

12 = sA + sB
tA = tB
tA = sA/vA = sA/3
tB = sB/vB = sB/2
sA/3 = sB/2
sA = sB*3/2
12 = sB*3/2 + sB
sB = 12/(1 + 3/2)
sB = 2*12/5 = 24/5
tB = sB/vB = 24/(5*2) = 12/5 = 2.4
t3 = 10 + 4((pi) - 1) + tB = 6 + 4*(pi) + 2.4 = 4*(pi) + 8.4
Therefore, answer B is correct

Hope it helps!
Best,
Maciek
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