let the rate of the slower plane be x mph
let the rate of the faster plance be x+100 mph
let the time it took be t hours
distance covered by first plane = tx miles
distance covered by second plane = t(x+100)
tx + t(x+100) = 2000
2tx + 100t = 2000
t(2x + 100) = 2000
t = 2000/(2x+100)
= 1000/(x+50)
there is no unique solution, but there are many solutions.
Let's assume a reasonable speed for a plane
e.g. let x = 250 mph
then t = 1000/300) = 10/3 hours
let x = 200 mph
then t = 1000/250 = 4 hours
I will check the last answer:
slow plane = 200 mph
it went 4(200) or 800 miles
faster plane = 300 mph
it went 4(300) = 1200 miles
total distance between them = 800+1200 = 2000 miles
Hello, could someone help me solve/show the following algebra problem? If there's not enough info given, indicate that no solution can be found.
'Two planes leave chicago - one traveling due east and the other due west. The first travels 100 miles per hour faster than the second. How long will it be until they are 2000 miles apart?'
Methinks that there is no soultion.
1 answer
2 answers