Call the speeds j and c.
j - c = 100 -> j = 100 + c
(100 + c) = 5c / 4 (from the "25% faster")
Solve for c.
The average speed of a jet is 100km/h faster than a cargo plane. To travel 1400km, the cargo plane requires 25% more time than the jet. Find the average speed of the jet.
This is the formula I figured out so far.
V(average)=V(jet)-V(cargo
100km/h= d/t - d/t
100= [1400km/x] - [1400/(.25x+x)]
But from here I don't know what to do..
2 answers
Creating a chart makes these kind of problems real easy
3 columns: Distance, Rate , Time and remember D = RT
2 rows: jet and cargo plane
-----D-----R------T
Jet 1400 -- x+100 -- 1400/(x+100)
Cargo 1400 ----x -- 1400/x
(hope the spacing comes out halfdecent.)
It said the time of the cargo plane is 25% more than the time of the jet, so
1400/x = 1.25(1400)/(x+100)
1400/x = 1750/(x+100)
cross-multiply
1750x = 1400x + 140000
x = 400
So the cargo's speed is 400 km/h and the jet goes 500 km/h
check:
time of jet = 1400/500 = 2.8 hours
time of cargo = 1400/400 = 3.5
What is 25% of 2.8 added to 2.8 ?
YUP!!!!
3 columns: Distance, Rate , Time and remember D = RT
2 rows: jet and cargo plane
-----D-----R------T
Jet 1400 -- x+100 -- 1400/(x+100)
Cargo 1400 ----x -- 1400/x
(hope the spacing comes out halfdecent.)
It said the time of the cargo plane is 25% more than the time of the jet, so
1400/x = 1.25(1400)/(x+100)
1400/x = 1750/(x+100)
cross-multiply
1750x = 1400x + 140000
x = 400
So the cargo's speed is 400 km/h and the jet goes 500 km/h
check:
time of jet = 1400/500 = 2.8 hours
time of cargo = 1400/400 = 3.5
What is 25% of 2.8 added to 2.8 ?
YUP!!!!
1 answer