SOLUTION: Evan went on a bike ride of 120 miles. He realized that if he had gone 18 mph faster, he would have arrived 15 hours sooner. How fast did he actually ride?

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Question 1139188: Evan went on a bike ride of 120 miles. He realized that if he had gone 18 mph faster, he would have arrived 15 hours sooner. How fast did he actually ride?
Found 4 solutions by josgarithmetic, ikleyn, josmiceli, greenestamps:
Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
              SPEED        TIME(hours)          DISTANCE

ACTUAL           r         120%2Fr               120

IF              r+18       120%2F%28r%2B18%29          120

120%2Fr-120%2F%28r%2B18%29=15

-
8%2Fr-8%2F%28r%2B18%29=1
8%28r%2B18%29-8r=r%28r%2B18%29
8r%2B8%2A18-8r=r%5E2%2B18r
8%2A18=r%5E2%2B18r
r%28r%2B18%29=144
system%28r=6%2Cand%2Cr%2B18=24%29

highlight%28r=6%29

Answer by ikleyn(52884) About Me  (Show Source):
You can put this solution on YOUR website!
.

            It is a standard problem to be solved using the "time" equation.

            It was solved tens (if not hundred) times at this forum, so I will be short. 


Let "r" be the actual speed, in miles per hour.

Then the hypothetical speed is (r+18) mph.


The "time" equation is


    120%2Fr - 120%2F%28r%2B18%29 = 15   hours.


Cancel the factor 15 in both sides 


    8%2Fr - 8%2F%28r%2B18%29 = 1.


Now multiply both sides by r*(r+18), simplify and solve for "r"


   8*(r+18) - 8r = r*(r+18).

    8*18 = r*(r+18)

    r^2 + 18r - 144 = 0

    (r-6)*(r+24) = 0.


There are two roots, 6 and -24, and only positive value of r= 6 is the solution to the problem.


ANSWER.  Actual speed was  6 mph.    (Not much . . .)

Solved.

----------------

Using "time" equation is the STANDARD method of solving such problems.
From my post,  learn on how to write,  how to use and how to solve a  "time"  equation.

To see many other similar solved problems,  look into the lessons
    - Had a car move faster it would arrive sooner
    - How far do you live from school?
    - Earthquake waves
    - Time equation: HOW TO use, HOW TO write and HOW TO solve it
in this site.


Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +s+ = the speed he actually went
(1) +d+=+%28+s+%2B+18+%29%2A%28+t+-+15+%29+
(2) +d+=+s%2At+
-------------------------------------
(2) +s+=+d%2Ft+
(1) +d+=+%28+d%2Ft+%2B+18+%29%2A%28+t+-+15+%29+
(1) +120+=+%28+120%2Ft+%2B+18+%29%2A%28+t+-+15+%29+
(1) +120+=+120+%2B+18t+-+1800%2Ft+-+270+
(1) +270+=+18t+-+1800%2Ft+
(1) +270t+=+18t%5E2+-+1800+
(1) +18t%5E2+-+270t+-+1800+=+0+
(1) +t%5E2+-+15t+-+100+=+0+
(1) +%28+t+-+20+%29%2A%28+t+%2B+5+%29+=+0+
+t+=+20+
and
+120+=+s%2A20+
+s+=+6+
He rode 6 mi/hr
---------------------
check:
(1) +120+=+%28+120%2Ft+%2B+18+%29%2A%28+t+-+15+%29+
(1) +120+=+%28+120%2F20+%2B+18+%29%2A%28+20+-+15+%29+
(1) +120+=+%28+6+%2B+18+%29%2A5+
(1) +120+=+24%2A5+
(1) +120+=+120+

Answer by greenestamps(13209) About Me  (Show Source):
You can put this solution on YOUR website!


An algebraic solution....

let x be the speed he rode at
Then x+18 is the speed he could have ridden at

The times to go 120 miles at the two speeds are 120/x and 120/(x+18).

The problem says the time would be 15 hours less at the higher speed:

120%2Fx-120%2F%28x%2B18%29+=+15
120%28x%2B18%29-120x+=+15x%28x%2B18%29
8%28x%2B18%29-8x+=+x%28x%2B18%29
144+=+x%5E2%2B18x
x%5E2%2B18x-144+=+0
x%2B24%29%28x-6%29+=+0
x = -24 (nonsense) or x = 6

His actual speed was 6mph. (A very slow bicycle speed!)

CHECK:
120/6 = 20
120/(6+18) = 120/24 = 5
20-5 = 15

An informal solution -- if algebra is not required....

Make a list of pairs of numbers whose product is 120 and find two pairs that meet the conditions of the problem.

1*120
2*60
3*40
4*30
5*24
6*20
8*15
10*12

The pairs 5*24 and 6*20 satisfy the given conditions: 6+18 = 24; 20-5 = 15.