A _ D R _ Y
W _ L L
R _ _ C H
T H _
D _ S T _ N _ T _ _ N
F _ R S T .
S _ P P _ S _
T H _ Y
C _ V _ R
1 2
M _ L _ S ,
B _ T H
W _ L K _ N G
_ T
_
R _ T _
_ F
2
M _ L _ S
P _ R
H _ _ R
_ N D
R _ N N _ N G
_ T
_
R _ T _
_ F
6
M _ L _ S
P _ R
H _ _ R .
U S _
T H _
F _ R M _ L _ :
R T
=
D
( R _ T _
X
T _ M _
=
D _ S T _ N C _ )
T _
F _ N D
_ _ C H
P _ R S _ N ' S
T _ M _ .
N _ N C Y
( W _ L K S
H _ L F
T H _
D _ S T _ N C _
_ N D
R _ N S
H _ L F
T H _
D _ S T _ N C _ ) :
2 T
=
6
M _ L _ S ,
S _
T
=
3
H R .
W _ L K _ N G
6 T
=
6
M _ L _ S ,
S _
T
=
1
H R .
R _ N N _ N G
T
=
4
H _ _ R S
T _ T _ L
T _ M _
A _ D R _ Y
( W _ L K S
H _ L F
T H _
T _ M _
_ N D
R _ N S
H _ L F
T H _
T _ M _ ) :
2 ( 0 . 5 T )
+
6 ( 0 . 5 T )
=
1 2
M _ L _ S
T
+
3 T
=
1 2
4 T
=
1 2
T
=
3
H _ _ R S
T _ T _ L
T _ M _ Clue
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 THE SON WILL CATCH UP WITH HIS FATHER IN 10 MINUTES. IN 10 MINUTES, THE SON WOULD HAVE TRAVELED HALF THE DISTANCE, SINCE HIS TOTAL TRAVELING TIME IS 20 MINUTES. THE FATHER STARTED 5 MINUTES EARLIER THAN THE SON, SO HE WOULD HAVE TRAVELED FOR A TOTAL OF 15 MINUTES WHEN HIS SON REACHED HIM. THEREFORE, THE FATHER WOULD HAVE ALSO TRAVELED HALF THE DISTANCE IN 15 MINUTES