Asked by mohammed
A man travels from town a to b at an average speed of 90km/h.on his return journey he takes (one)1 hour less and his speed is 10km/h faster......1.how far apart are towns a and b, .............
..2.what Is his average speed from town a to b and back to a ..
..2.what Is his average speed from town a to b and back to a ..
Answers
Answered by
Henry
d1 = 90*t.
d2 = 100*(t-1).
1. d1 = d2.
90t = 100(t-1)
90t = 100t-100
90t-100t = -100
-10t = -100
t = 10 h.
d = 90*10 = 900 km
2. V = d/t=(2*900)km/(10+9)h=94.74 km/h.
d2 = 100*(t-1).
1. d1 = d2.
90t = 100(t-1)
90t = 100t-100
90t-100t = -100
-10t = -100
t = 10 h.
d = 90*10 = 900 km
2. V = d/t=(2*900)km/(10+9)h=94.74 km/h.
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