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Two groups of hikers take two different routes from Camp A to Camp B. The first group leaves Camp A at 9:00 AM and takes a route that is 30 miles longer than the second group's route. The second group leaves Camp A at 1:00 PM. Their hiking speed is 2 mph slower than the first group's speed. Both groups reach Camp B at 4:00 PM. How fast does each group hike?

Respuesta :

Answer: first groups hiking was 6mph, second groups was 4mph.

Step-by-step explanation:

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Speed of first group = 6 mph

Speed of second group = 4 mph

  • We are told that first group leaves Camp A at 9:00 AM.

If the normal distance between camp A and camp B is x, then the total distance this first group covers will be;

(x + 30) miles

  • This first group reached camp B at 4pm. This is 7 hours from when they left camp A.

Thus, their speed is;

v1 = (x + 30)/7 mph

  • We are told that;

Second group left by 1pm and got to camp B by 4pm. This is 3 hours.

  • We are also told that their speed was 2mph slower than the first group.

Thus;

v2 = ((x + 30)/7) - 2 mph

  • Since normal distance from camp A to camp B is x. Then this second group will have;

x = ((x + 30)/7) - 2) × 3

Since;

distance = speed × time

Divide both sides by 3 to get;

x/3 = ((x + 30)/7) - 2)

Multiply through by 21 to get;

7x = 3(x + 30) - 42

7x = 3x + 90 - 42

7x - 3x = 48

4x = 48

x = 48/4

x = 12 miles

Thus, putting 12 for x in V1 equation gives;

v1 = (12 + 30)/7

v1 = 42/7

v1 = 6 mph

Similarly;

v2 = ((12 + 30)/7) - 2

v2 = (42/7) - 2

v2 = 6 - 2

v2 = 4 mph

Read more at; brainly.com/question/18642827

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