explain it please.... thanks

Question

a man covered a certain distance at some speed. had he moved 3 kmph faster he would have taken 40 minutes less, if he had moved 2 kmph slower he would have taken 40 minutes more, the distance is?

Answer 1

d = r t

 

First note that 40 min is 2/3 hour

 

d = ( r + 3)(t - 2/3) <---  3 km per hour faster, takes 40 min less 

 

d = (r - 2) ( t  + 2/3) <-- 2 km per hour slower, takes 40 min longer

 

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Setting the first two equations equal:

 

rt = (r + 3)(t - 2/3)

 

rt = rt - (2/3)r + 3t - 2  <--- FOIL method , right side

 

0 = -(2/3)r + 3t - 2 <--- subtracts rt from both sides

 

(2/3)r =3t - 2 <--- adds (2/3)r to both sides

 

2r = 9t - 6 <--- multiplies everything by 3 to clear away the fractions

 

2r + 6 = 9t   <--- solves for t in terms of r

 

[ Please label this equation ALPHA]

 

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Setting the last two equations equal:

 

rt = (r-2)(t + 2/3)

 

rt = rt + (2/3)r - 2t - 4/3 <--- FOIL method right side

 

0 = (2/3)r - 2t - 4/3 <--- subtracts (2/3)r from both sides

 

0 = 2r - 6t - 4 <-- multiplies everything by 3

 

0 = r - 3t - 2  <--- divides everything by 2

 

3t + 2 = r <--- solves for r in terms of t 

 

[Please label this equation BETA]

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Substitutes equation BETA into equation ALPHA:

 

2(3t +2) + 6 = 9t

 

6t + 4 + 6 = 9t

 

6t + 10 = 9t

 

10  = 3t

 

t = 10/3  <--- the original time is 10/3 hours , or 3 and 1/3 hours = 3 hours and 20 minutes

 

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D = (r + 3)( 10/3 - 2/3) = (r + 3)(8/3) <--- plugging into the 2nd equation

 

D = (r - 2)(10/3 + 2/3) = (r-2)(12/3) = 4(r-2) <--- plugging into the 3rd equation

 

Setting them equal:

 

(r +3)(8/3) = 4(r-2)

 

8(r +3) = 12(r-2) <--- multiplies by 3

 

8r + 24 = 12r - 24 <--- distributive both sides

 

0 = 12r - 24 - 8r - 24 <--- everyone to the right side

 

0 = 4r - 48 <--- combines like terms

 

48 = 4r

 

r=12 <--- the original speed is 12 km per hour

 

the original distance is 12*(10/3) = 4*10 = 40 km

 

 

Increasing the speed by 3 km per hour, 40 km = (15 km/hr) * t

   

              t = 40/15 = 8/3 which is 2/3 less than 10/3

 

Decreasing the speed by 2 km per hour, 40 km = (10 km.hr) * t

 

                                              t = 40/10 = 4

 

  which is 4 = 12/3 = 10/3 + 2/3 , or 40 mins longer

 

40 km is the distance