SOLUTION: I haven't done too bad with the word problems but this one has stumped me: The typical cruising speed for a Boeing 747-400 airplane is 30 mph faster than that for a Boeing 737-8

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Question 67165This question is from textbook Intermediate Algebra with Applications
: I haven't done too bad with the word problems but this one has stumped me:
The typical cruising speed for a Boeing 747-400 airplane is 30 mph faster than that for a Boeing 737-800. If a Boeing 747-400 can make a trip of 1680 mi in 3 h, how long would it take a Boeing 737-800 to make the same trip? Round to the nearest tenth.
I know the equation to work with is rate x time = distance and have made a table solving for 30r x 3 = 1680. I have gotten r = 18.67 mph which I checked with the equation for the Boeing 747-400. When I try to take 18.67 x t = 1680 I'm getting 89.98 which is completely different from the answer in the book. I need to know how to set up the rest of the problem to solve for the correct answer.
This question is from textbook Intermediate Algebra with Applications

Found 2 solutions by ankor@dixie-net.com, astein76:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
The typical cruising speed for a Boeing 747-400 airplane is 30 mph faster than that for a Boeing 737-800. If a Boeing 747-400 can make a trip of 1680 mi in 3 h, how long would it take a Boeing 737-800 to make the same trip? Round to the nearest tenth.
I know the equation to work with is rate x time = distance and have made a table solving for 30r x 3 = 1680. I have gotten r = 18.67 mph which I checked with the equation for the Boeing 747-400. When I try to take 18.67 x t = 1680 I'm getting 89.98 which is completely different from the answer in the book. I need to know how to set up the rest of the problem to solve for the correct answer.
:
Let s = the speed of the 737; then (s+30) = the speed of the 747
:
Find the value of s and divide that into 1680, to get the time of the 737:
3(s+30) = 1680

Answer by astein76(1) About Me  (Show Source):