Question 25868: A man with a scooter and two friends had to make a journey of 52 miles. The man decided he would take the first friend a certain distance while the second walked. He would drop the first friend off to walk the remainder of the 52 miles, return to meet the second friend, pick him up, and proceed to the end of the journey finishing just as the first friend finished. If the scooter travels 20 mph, the first friend 4 mph, and the second friend 5 mph, how should they arrange the journey?
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! A man with a scooter and two friends had to make a journey of 52 miles. The man decided he would take the first friend a certain distance while the second walked. He would drop the first friend off to walk the remainder of the 52 miles, return to meet the second friend, pick him up, and proceed to the end of the journey finishing just as the first friend finished. If the scooter travels 20 mph, the first friend 4 mph, and the second friend 5 mph, how should they arrange the journey?
TOTAL DISTANCE = D= 52 MILES
SCOOTER SPEED = 20 MPH
FIRST FRIEND (FF)SPEED = 4 MPH
SECOND FRIEND (SF) SPEED = 5 MPH.
I PART OF JOURNEY.SAY T HRS.
SCOOTER WILL GO 20T MILES.
BY THIS TIME SF WILL GO 5T MILES.
HENCE DISTANCE BETWEEN SCOOTER AND SF AT THAT TIME =20T-5T=15T
II PART OF JOURNEY
DISTANCE OF SEPERATION AT THE BEGINING =15T
RELATIVE SPEED WHILE GOING IN OPPOSITE DIRECTIONS = 20+5 MPH
TIME NEEDED TO MEET =15T/25=3T/5
BY THIS TIME SCOOTER WOULD GO BACK =20*3T/5=12T SO HE IS NOW AT 20T-12T=8T MILES FROM THE START.
III PART OF JOURNEY
DISTANCE TO BE COVERED = 52 - 8T
TIME TAKEN BY SCOOTER TO GO THIS DISTANCE
=(52-8T)/20
FINAL MEETING
DURING THIS PERIOD OF SECOND TRAVEL OF3T/5 + FINAL TRAVEL (52-8T)/20.....THE FF REACHED DESTINATION IN (52-20T)/4=13-5T HRS.HENCE
3T/5+(52-8T)/20=13-5T
12T+52-8T=20(13-5T)=260-100T
104T=260-52=208
T=208/104=2 HRS. SO THE SCOOTERIST SHOULD DRIVE FOR 2 HRS TO DROP THE FF BEFORE RETURNING TO PICKUP THE SF
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