Question 918210
When a man travels equal distance at speeds v1 and v2 km/hr, his average speed is 4 km/hr.
 But when he travels at these speeds for equal times his average speed is 4.5 km/hr.
 Find the difference of two speeds
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let 2d = the total distance
"When a man travels equal distance at speeds v1 and v2 km/hr, his average speed is 4 km/hr."
A time equation; time = dist/speed
{{{d/(v1)}}} + {{{d/(v2)}}} = {{{(2d)/4}}}
reduce fraction
{{{d/(v1)}}} + {{{d/(v2)}}} = {{{d/2}}}
divide thru by d
{{{1/(v1)}}} + {{{1/(v2)}}} = {{{1/2}}}
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"when he travels at these speeds for equal times, his average speed is 4.5 km/hr"
let 2t = total time
A distance equation; dist = speed * time
v1*t + v2*t = 4.5(2t)
v1*t + v2*t = 9t
divide by t
v1 + v2 = 9
v2 = (9-v1)
Replace v2 in the 1st simplified equation
{{{1/(v1)}}} + {{{1/(9-v1)}}} = {{{1/2}}}
multiply by 2(v1)(9-v1), cancel the denominators, you have
2(9-v1) + 2(v1) = v1(9-v1)
18 - 2(v1) + 2(v1) = 9(v1) - (v1)^2
arrange to forma quadratic equation on the left
(v1)^2 - 9(v1) + 18 = 0
Factors to
(v1 - 3)(v1 - 6) = 0
Two solutions
v1 = 3; then v2 = 6
v1 = 6; then v2 = 3
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Obvious the difference is 3 in the two speeds
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There should be an easier way to do this, but just can't come up with it this evening!