George and Bill left town A for town B at the same time. George was riding a bicycle at a constant speed, while Bill was driving a car. Bill traveled 5 times faster than George. Bill's car broke halfway, and the rest of the way to the town B Bill traveled
by foot with a speed equal to half of the George's bicycle speed. Who got to B first: George or Bill?

## George and Bill left town A for town B at the same time.

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# 3 Answers

George Speed - X

Bills Speed - 5X

Bills speed by foot - X/2

d- distance from A-B

X = d/t1 5X + X/2 =d/t2

11X/2 = d/ t2

t1= d/X ---------------- time it takes Bill to go from A to B

t2 = d/ (11X/2)---------- Time it will take George to go from A to B

t1= d/X

t2=2d/11

t1 > t2 , Therefore Bill reaches earlier.

Note : The formula used was

Speed * ( time it would take) = distance traveled.

Let D be the distance between A and B. Let also George's speed be V. Then Bill's car speed was 5V, his walking speed was V/2.

Time it took George to get to B:

t

_{G}=L/Vtime it took Bill to get to B:

t

_{B}=(L/2)/(5V)+(L/2)/(V/2)=L/(10V)+L/V;It is immediately obvious that, since George's time is L/V, it took Bill longer to get to B. In fact, it took him 11/10*(L/V)=11/10*t

_{G},_{ }where t_{G}is the time it took George to get from A to B.So George got to B first.

The time George completed the trip: t = AB/v

The time Bill completed the trip: t(total) = t1 + t2

The time for the first half of the trip (car is not broken): t1 = (AB/2) / 5v = AB/10v

The time for the second half (by foot): t2 = (AB/2) / (v/2) = AB/v

Total time for Bill : T = t1 + t2 = AB/10v + AB/v. This time is longer than George's time ( AB/v)

Answer: George got to B first