SOLUTION: Ann lived in Bristol and Amy lived in Fairvale. At 2:00 pm they left their respective towns and walked in opposite directions, with Amy walking twice as fast as Ann. By 4:00 pm, th

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: Ann lived in Bristol and Amy lived in Fairvale. At 2:00 pm they left their respective towns and walked in opposite directions, with Amy walking twice as fast as Ann. By 4:00 pm, th      Log On

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Question 24033: Ann lived in Bristol and Amy lived in Fairvale. At 2:00 pm they left their respective towns and walked in opposite directions, with Amy walking twice as fast as Ann. By 4:00 pm, they were 66 km apart. If, instead, they had walked from the towns toward each other, they would have been 12 km apart at 3:00 pm. How far apart are the towns?
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Ann lived in Bristol and Amy lived in Fairvale. At 2:00 pm they left their respective towns and walked in opposite directions, with Amy walking twice as fast as Ann. By 4:00 pm, they were 66 km apart. If, instead, they had walked from the towns toward each other, they would have been 12 km apart at 3:00 pm. How far apart are the towns?
LET THE 2 TOWNS BE X KM APART
LET ANNS SPEED BR Y . SO SPEED OF AMY IS TWICE Y OR =2Y
CASE 1 ..THEY ARE WALKING AWAY FROM EACH OTHER..
SO THEIR RELATIVE SPEED IS Y+2Y=3Y..THEY WALKED FROM 2=00 PM TO 4=00 PM THAT IS FOR 2 HRS .
IN 2 HRS TOGETHER THEY TRAVEL 2*3Y=6Y KM...SO THEIR DISTANCE OF SEPERATION =ORIGINAL DISTANCE OF X +TRAVELLED DISTANCE OF 6Y=X+6Y..THIS IS 66 KM .HENCE
X+6Y=66.................EQN.I
CASE 2....THEY ARE WALKING TOWARDS EACH OTHER..
SO THEIR RELATIVE SPEED IS Y+2Y=3Y..THEY WALKED FROM 2=00 PM TO 3=00 PM THAT IS FOR 1 HRS .
IN 1 HRS TOGETHER THEY TRAVEL 1*3Y=3Y KM...SO THEIR DISTANCE OF SEPERATION =ORIGINAL DISTANCE OF X -TRAVELLED DISTANCE OF 3Y=X-3Y..THGIS IS 12 KM.HENCE
X-3Y=12......................EQN.II
TO FIND X LET US ELIMINATE Y BY MULTIPLYING EQN.2 WITH 2 AND ADDING EQN.1 TO IT
2*EQN.2....2X-6Y=24..........III
EQN.I......X+6Y=66....ADDING
3X=90
X=90/2=30 KM.