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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 KOLKHOZ IS 120 MILES AWAY FROM THE CITY AND THE TRUCK SHOULD TRAVEL AT 24 MILES AN HOUR. AT 30 MILES PER HOUR A TRUCK TRAVELS A MILE IN 2 MINUTES; AT 20 MILES PER HOUR, IN 3 MINUTES. AT THE LATTER SPEED THE TRUCK IS 1 MINUTE SLOWER PER MILE. TO LOSE 2 HOURS, OR 120 MINUTES, TAKES 120 MILES, WHICH IS HOW FAR THE KOLHOZ IS FROM THE CITY. AT 30 MPH, THE TRUCK WOULD COVER 120 MILES IN 4 HOURS. THE TRIP IS TO TAKE 1 HOUR LONGER, OR TO ARRIVE AT 11:00 A.M., AND CALLS FOR A SPEED OF 24 (120/5) MILES PER HOUR 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