Problem Solving

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

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??

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

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!!

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