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THERE IS NO WAY TO GET AN AVERAGE SPEED OF 30 MPH (MILES PER HOUR). IF YOUR AVERAGE SPEED FOR THE 2 MILES STRETCH MUST BE 30 MPH, YOU SHOULD DO IT IN 4 MINUTES. DURING THE ASCENT, YOU WOULD HAVE ALREADY TAKEN 4 MINUTES TO TRAVEL 1 MILE AT 15 MPH. SINCE YOU HAVE TAKEN 4 MINUTES TO GO UP THE HILL, YOU WILL NEED TO TAKE 0 SECONDS TO TRAVEL THE 1 MILE DESCENT IN ORDER TO GET AN AVERAGE SPEED OF 30 MPH. THIS IS NOT POSSIBLE THE TRAIN IS ABOUT TO LEAVE AND YOU WILL NOT BE ABLE TO CATCH THE TRAIN. THE AMOUNT OF TIME TAKEN TO TRAVEL TO THE HALFWAY POINT AT 15 MILES AN HOUR IS THE SAME AMOUNT OF TIME TAKEN TO REACH THE TRAIN STATION WHEN TRAVELLING TO THE RAILROAD STATION AT 30 MILES AN HOUR. FOR EXAMPLE, LET'S ASSUME THAT THE DISTANCE TO TRAVEL IS 30 MILES. WHEN TRAVELLING AT 30 MILES AN HOUR, IT TAKES 1 HOUR. WHEN TRAVELLING AT 15 MILES AN HOUR, IT WILL ALREADY TAKE 1 HOUR TO REACH THE HALFWAY POINT IF THE GIRLS HAD BEEN ON A STANDING TRAIN, THE FIRST GIRL'S CALCULATIONS WOULD HAVE BEEN CORRECT, BUT THEIR TRAIN WAS MOVING. IT TOOK 5 MINUTES TO MEET A SECOND TRAIN, BUT THEN IT TOOK THE SECOND TRAIN 5 MORE MINUTES TO REACH WHERE THE GIRLS MET THE FIRST TRAIN. SO THE TIME BETWEEN TRAINS IS 10 MINUTES, NOT 5, AND ONLY 6 TRAINS PER HOUR ARRIVE IN THE CITY ONE TRAIN WAS RUNNING TWICE AS FAST AS THE OTHER. LET: SPEED OF THE FAST TRAIN = F SPEED OF THE SLOW TRAIN = S TIME IT TAKES FOR THE TRAINS TO MEET (PASS EACH OTHER) = T SINCE BOTH TRAINS TRAVEL THE SAME TOTAL DISTANCE AND DISTANCE = TIME X SPEED: F(T+1) = S(T+4) WE'RE TRYING TO FIGURE OUT F/S WHICH IS EQUAL TO (T+4) / (T+1) FROM THE EQUATION ABOVE. SO WE NEED TO FIGURE OUT THE VALUE OF T. AFTER THEY MEET, THE FAST TRAIN TRAVELS ONE MORE HOUR AT SPEED F AND COVERS THE SAME DISTANCE THE SLOW TRAIN COVERED IN T HOURS: F1 = ST OR F = ST AFTER THEY MEET, THE SLOW TRAIN TRAVELS FOR 4 MORE HOURS AND COVERS THE SAME DISTANCE THE FAST TRAIN COVERED IN T HOURS: S4 = FT SUBSTITUTING ST FROM THE FIRST EQUATION IN FOR F IN THE 2ND EQUATION: 4S = STT 4 = TT 2 = T SUBSTITUTE 2 IN FOR T IN THE (T+4) / (T+1) EQUATION TO GET 6/3 OR 2. THE FAST TRAIN IS GOING TWICE AS FAST AS THE SLOW TRAIN