Question 726240
Let {{{ d }}} = distance from Swansea to London
Let {{{ s }}} = his speed for the 1st hour
{{{ d[1] = s*1 }}}
{{{ d[1] = s }}}
He now has a distance of {{{ d - s }}} to go 
He travels at {{{ (3/5)*s = .6s }}}
for time = {{{ t[1] }}}
{{{ d - s = .6s*t[1] }}}
{{{ t[1] = ( d - s ) / ( .6s ) }}}
He eventually arrived in London two hours later than he'd expected
he would have got there when he expected if his speed was {{{ s }}}
for the whole distance {{{ d }}}
{{{ d = s*t[2] }}}
{{{ t[2] = d/s }}}
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I can say:
{{{ ( d - s ) / ( .6s ) - d/s = 2 }}}
Multiply both sides by {{{ .6s }}}
{{{ d - s - .6d = 2*( .6s ) }}}
{{{ .4d = 1.2s + s }}}
{{{ .4d = 2.2s }}}
{{{ d = 2.2s / .4 }}}
{{{ d = 5.5s }}}
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Meeting the traffic 50 further on means that he
would travel at a speed of {{{ s }}} for
a distance = {{{ s*1 + 50 }}}
This is a time of {{{ t[3] = ( s + 50 ) / s }}}
After the traffic, his time is {{{ t[4] = ( d - ( s + 50 ) ) / ( .6s ) }}}
{{{ t[3] + t[4] = ( s + 50 ) / s + ( d - s - 50 ) / ( .6s ) }}}
{{{ t[3] + t[4] = ( s + 50 ) / s + ( 5.5s - s - 50 ) / ( .6s ) }}}
{{{ t[3] + t[4] = ( s + 50 ) / s + ( 4.5s - 50 ) / ( .6s ) }}}
I am given that
{{{ t[1] - ( t[3] + t[4] ) = 2/3 }}} ( 40 min is 2/3 hr ) 
{{{ ( d-s )/(.6s) - ( s+50 ) / s - ( 4.5s - 50 ) / (.6s)  = 2/3 }}}
{{{ ( 5.5s-s )/(.6s) - ( s+50 ) / s - ( 4.5s - 50 ) / (.6s)  = 2/3 }}}
{{{ ( 4.5s )/(.6s) - ( s+50 ) / s - ( 4.5s - 50 ) / (.6s)  = 2/3 }}}
Multiply both sides by {{{ .6s }}}
{{{ 4.5s - .6*( s + 50 ) - 4.5s + 50 = (2/3)*( .6s ) }}}
Multiply both sides by {{{ 3 }}}
{{{ 13.5s - 1.8s - 90 - 13.5s + 150 = 1.2s }}}
{{{ 1.8s + 1.2s = 150 - 90 }}}
{{{ 3s = 60 }}}
{{{ s = 20 }}}
since {{{ d = 5.5s }}},
{{{ d = 5.5*20 }}}
{{{ d = 110 }}} answer
There are no units given, so I can't fill these in. ( km? )
Hope I got it