Posted by mac on .
a motorist drives to a city in 4 hrs. and returns by another route which is 26 mi. longer. on the return trip he traveled 5 miles per hour faster and took 4 hours and 12 min. . find the length of the shorter route.
let d be the shorter distance.
d+26=(r+5)4.2 (4.2 hrs is 4hrs 12min)
two equations, two unknowns, solve for d.
length of shorter trip ---- x miles
speed for shorter trip ---- 4/x mph , since speed = distance/time
length of longer trip --- x+26
speed for longer trip --- 21/(5(x+26)) .....explanation : ( 21/5 hrs = 4hr. 12 min )
21/(5(x+26)) - 4/x = 5 , ... the difference is their speeds is 5 mph
multiply by 5x(x+26)
21x - 20(x+26) = 25x(x+26)
21x - 20x - 520 = 25x^2 + 650x
25x^2 + 649x + 520=0
x = (-649 ± 607.619)/50
this gave me two negative answers, but x has to be > 0
Unless I made some silly arithmetic error, this problem has no solution.
Go with bobpursley - math -
Never mind my solution, go with bobpursley
Found my "silly error" , just noticed I had my fractions upside - down. e.g. speed for shorter trip = x/4 , etc.
Can I blame it on the heat affecting my brain ?
could bob pursley finish the problem ? i still don't have the answer or could you, reiny, do it again without the error. i need to see the problem worked, thanks.
Yes, I could finish it, however, you don't want anyone thinking you are an answer grazer. You do the work, and I will critique it.
mr. pursley, Have you ever worked geo. & trig. etc with the heat on wheather you blew up the town or the target or the building collapased or was built in the right place. i'm not an answer graser. you don't know how long i worked on this problem. i'm trying to learn some different types of math now for fun. if you choose to not assist, please don't.