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YES. HE TOOK AS MUCH TIME FOR THE SECOND HALF OF HIS TRIP AS THE WHOLE TRIP WOULD HAVE TAKEN ON FOOT. SO NO MATTER HOW FAST THE TRAIN WAS, HE LOST EXACTLY AS MUCH TIME AS HE SPENT ON THE TRAIN. HE WOULD HAVE SAVED 1⁄30 OF THE TIME BY WALKING ALL THE WAY THE MESSENGER HAS TO HAVE TRAVELED 2 KM. IT DOESN'T MATTER WHAT SPEED THEY WALKED AT. AT THE BEGINNING OF THE PUZZLE, THE LINE IS 1 KM LONG. THE GENERAL IS THEREFORE 1 KM AHEAD OF HIM. THE MESSENGER MUST THEREFORE TRAVEL MORE THAN 1 KM TO REACH THE GENERAL. SINCE THE LINE MOVES 1 KM FORWARD, THE END IS WHERE THE BEGINNING WAS. EVEN IF HE WALKED 1.5 KM TO THE GENERAL, HE ONLY HAS TO WALK 0.5 KM TO GET BACK TO THE END OF THE LINE. IT GOES FASTER GOING BACK, BECAUSE NOW THEY ARE COMING TOWARDS HIM, AND NOT GOING AWAY AUDREY WILL REACH THE DESTINATION FIRST. SUPPOSE THEY COVER 12 MILES, BOTH WALKING AT A RATE OF 2 MILES PER HOUR AND RUNNING AT A RATE OF 6 MILES PER HOUR. USE THE FORMULA: RT = D (RATE X TIME = DISTANCE) TO FIND EACH PERSON'S TIME. NANCY (WALKS HALF THE DISTANCE AND RUNS HALF THE DISTANCE): 2T = 6 MILES, SO T = 3 HR. WALKING 6T = 6 MILES, SO T = 1 HR. RUNNING T = 4 HOURS TOTAL TIME AUDREY (WALKS HALF THE TIME AND RUNS HALF THE TIME): 2(0.5T) + 6(0.5T) = 12 MILES T + 3T = 12 4T = 12 T = 3 HOURS TOTAL TIME YES, THERE IS A POINT ALONG THE PATH THAT THE MONK OCCUPIES AT PRECISELY THE SAME TIME ON BOTH DAYS. IF YOU ASSUME THAT THE THERE ARE TWO MONKS, ONE DOES THE ROUTE FROM THE TOP, AND THE OTHER FROM THE BOTTOM. AT SOME POINT, THEY MUST MEET. THERE'S YOUR POINT