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C _ T Y Clue
THE FLY WOULD HAVE TRAVELLED 90 MILES. THE TRAINS ARE 100 MILES APART, AND THE TRAINS ARE TRAVELLING TOWARD EACH OTHER AT 40 AND 60 MPH, THE TRAINS WILL MEET IN ONE HOUR. THE FLY WILL HAVE BEEN FLYING FOR AN HOUR AT 90 MPH AT THAT POINT, SO THE FLY WILL HAVE TRAVELLED 90 MILES FOR THE TRAIN TO PASS COMPLETELY THROUGH THE TUNNEL, IT MUST TRAVEL 2 MILES. AFTER 1 MILE TRAVEL, THE TRAIN WOULD BE COMPLETELY IN THE TUNNEL, AND AFTER ANOTHER MILE IT WOULD BE COMPLETELY OUT AND SINCE THE TRAIN IS TRAVELING AT 1 MILE A MINUTE, IT WILL TAKE 2 MINUTES TO PASS THROUGH THE TUNNEL 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 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