Question 199079
Let {{{t[1]}}} = time of express train and {{{t[2]}}} = time of local train


Also, let {{{r[1]}}} = speed of express train and {{{r[2]}}} = speed of local train



Since the "express train travels twice as fast as the local train", this means that {{{r[1]=2*r[2]}}}



Also, because "the express train arrives 2 hours ahead of the local train", this tells us {{{t[1]=t[2]-2}}}




Now let's set up the formula for the local train:



{{{d=rt}}} Start with the distance rate time formula



{{{60=r[2]*t[2]}}} Plug in {{{d=60}}}, {{{r=r[2]}}} and {{{t=t[2]}}}



{{{60/t[2]=r[2]}}} Divide both sides by {{{t[2]}}}.



So after isolating the second speed, we get {{{r[2]=60/t[2]}}}




Now let's set up the formula for the express train:



{{{d=rt}}} Start with the distance rate time formula



{{{60=r[1]*t[1]}}} Plug in {{{d=60}}}, {{{r=r[1]}}} and {{{t=t[1]}}}



{{{60=(2*r[2])*(t[2]-2)}}} Plug in {{{r[1]=2*r[2]}}} and {{{t[1]=t[2]-2}}}



{{{60=(2*(60/t[2]))*(t[2]-2)}}} Plug in {{{r[2]=60/t[2]}}}



{{{60=(120/t[2])*(t[2]-2)}}} Multiply



{{{60t[2]=120(t[2]-2)}}} Multiply both sides by {{{t[2]}}}.



{{{60t[2]=120t[2]-240}}} Distribute



{{{60t[2]-120t[2]=-240}}} Subtract {{{120t[2]}}} from both sides.



{{{-60t[2]=-240}}} Combine like terms.



{{{t[2]=-240/(-60)}}} Divide both sides by -60 to isolate {{{t[2]}}}.



{{{t[2]=4}}} Divide



So it takes 4 hours for the local train



{{{60=r[2]*t[2]}}} Go back to the local train formula



{{{60=r[2]*4}}} Plug in {{{t[2]=4}}}



{{{60/4=r[2]}}} Divide both sides by 4.



{{{15=r[2]}}} Divide.



So the speed of the local train is 15 mph


Now simply double the speed of the local train to get {{{15*2=30}}}


So the speed of the local train is 15 mph and the speed of the express train is 30 mph. 




Note: nowhere in the solution above is a mention of 12:00 noon. This information is unnecessary and it is put in there to throw you off.