Three brothers A, B, and C started simultaneously from a

AAPL wrote:3 brothers A, B, and C started simultaneously from a common point to go to a fair, which is 60 miles away from that point. A and B started on a bike that moves at a speed of 40 miles/hour, while C started walking at a constant speed of 10 miles/hour. B dropped A at a point X somewhere along the route and returned back on the bike to pick up C. B picked up C at point Y, and reached the fair at the same time A reached the fair. If B had traveled for 2 hours by the time he picked up C, find the distance that A had to walk?

A. 5
B. 10
C. 15
D. 20
E. 25

Let’s call the starting point S and the ending point E. We are given that SE (the distance from S to E) = 60 miles and we need to determine the distance A had to walk, which is XE (the distance from point X to E).

We are given that B had ridden the bike from S to X and then from X to Y (where Y is some point between S and X) and he had ridden the bike for 2 hours. Since his rate is 40 mph, he had travelled 2 x 40 = 80 miles. In other words, SX + XY = 80.

Since B had traveled for 2 hours by time time he picked up C at point Y, C also had traveled for 2 hours by walking. Since C’s rate is 10 mph, he had traveled 2 x 10 = 20 miles. In other words, SY = 20.

Notice that since Y is some point between S and X, we can break SX into two parts: SY and YX. Thus, we have:

SX + XY = 80 → SY + YX + XY = 80

Since SY = 20 and XY = YX, we have:

20 + YX + YX = 80

2(YX) = 60

YX = 30

Notice that Y and X are some points between S and E and we know that SE = 60, SY = 20 and YX = 30 and we need to determine XE. So we have:

SY + YX + XE = SE

20 + 30 + XE = 60

XE = 10

Alternate Solution:

Let the distance of the point where B drops A to the starting point be d. Since B rides his bike for two hours at 40 mph, he travels a total of 2 x 40 = 80 miles. Since C walks at 10 mph, he travels a total of 2 x 10 = 20 miles. We see that B first travels a distance of d miles to drop off A and then turns around and travels another d - 20 miles to pick up C (since C walked 20 miles in that time); thus traveling d + d - 20 = 2d - 20 miles in total. Since we know the total distance traveled by B is 80, we can create the equation:

2d - 20 = 80

2d = 100

d = 50

So, B dropped A off at a distance of 50 miles to the starting point; which means A has to walk another 60 - 50 = 10 miles.

Answer: B