speed time related...

Question

a man increases his speed by 5km/h and reaches office 40 mintes early if he decreases his speed by 2km/h then reaches office 40 minutes late what is the distance?

Answer 1

distance = 280/27 = about 10.37 km

(x+5)(t-2/3) = d = (x- 2)(t+2/3)

x= current speed, t = time to arrive at x kph, 40 min = 2/3 hr

expand and combine like terms

xt +5t -2x/3 -10/3 = xt -2t +2x/3 -4/3

5t -4x/3 = -2t + 2

7t -4x/3 = 2

21t -4x = 6

d = xt = (x-2)(t+2/3) = xt -2t +2x/3 -10/3

0 =-2t +2x/3 -4/3

6t = 2x -4

2x = 6t +4

x = 3t +2

21t-4x = 6

21t -4(3t +2) = 6

21t -12t -8 = 6

9t = 14

t= 14/9 = 1 5/9 hours

x= 3t+2 = 42/9+18/9 = 60/9 = 6 2/3 hours

d= xt = 14/9)(60/3) = 280/27 = about 10.37 km