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Half an hour after car A started traveling from Newtown to Oldtown, a distance of 62 miles, car B started traveling along the same road from Oldtown to Newtown. The cars met each other on the road 15 minutes after car B started its trip. If car A traveled at a constant rate that was 8 miles per hour greater than car B’s constant rate, how many miles had car B driven when they met?

Re: Half an hour after car A started traveling from Newtown to [#permalink]
29 Oct 2017, 01:56

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Let's see which information we have. Car A has traveled 45 minutes at a speed 8m/h greater than car B speed (x). Car B traveled 15 minutes at speed x m/h.

Since they meet along a road 62 km long, it must be that the distance the two cars traveled sum to 62. Thus, we can proceed by plugging-in the choices we have and check if they do the job. Let's start from the middle one, C = 10 m.

If car B would have traveled 10 m, its speed would have been 10/(1/4) = 40 m/h (15 minutes is 1/4 of hour). Thus, car A speed would have been 48 m/h and it would have been traveled for 48*(3/4) = 24 m. Then, 10m+24m = 34m < 62 m. Thus, choice C is not our answers. We have to look for something higher.

If we proceed with this same kind of check we get that if car B traveled for 14m, then its speed is 14/(1/4) = 56 m/h so that car B speed is 64 m/h and it traveled for 48m, which summed to 14 gives us 62, the length of the road.

Re: Half an hour after car A started traveling from Newtown to [#permalink]
03 Nov 2017, 13:19

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Total traveling time for car A = 45 min or 3/4 hour. Let the speed of car B be x miles/hour. So, car's A speed = (8+x) miles/hour Total distance covered by car A and car B is 62 miles.

Therefore, the equation is 3/4*(8+x) + 1/4*x = 62 x = 56 Distance traveled by car B = 1/4*x = 1/4 *56 = 14