432 algebra De mystif ieD
- A plane on a flight from Denver to Indianapolis flew with a 20-mph
tailwind. On the return flight, the plane flew into a 20-mph headwind.
The distance between Denver and Indianapolis is 1000 miles and the
plane was in the air a total of 5^12 hours. What would have been the
plane’s average speed without the wind? - A plane flew from Minneapolis to Atlanta, a distance of 900 miles,
against a 30 mph-headwind. On the return flight, the 30-mph wind
became a tailwind. The plane was in the air for a total of 5^12 hours. What
would the plane’s average speed have been without the wind?
✔SOLUTIONS
- Let r represent the plane’s average speed (in mph) with no wind. Then
the average speed from Dallas to Chicago (with the tailwind) is r + 40,
and the average speed from Chicago to Dallas is r − 40 (against the
headwind). The distance between Dallas and Chicago is 800 miles. The
time in the air from Dallas to Chicago plus the time in the air from Chi-
cago to Dallas is 5.08 hours. The time in the air from Dallas to Chicago is
800
r + 40
. The time in the air from Chicago to Dallas is^800
r – 40
. The equation
to solve is^800
40
800
40508
rr .
+ +
– = . The LCD is (r + 40)(r − 40).()()()() ()(rr
r
rr
rr+−
+++−
−=+40 40 800
40
40 40 800
4040 rr
rr rr−
−+ += −+40 508
800 40 800 40 508404)(.)
()().[()(0 0
800 32 000 800 32 000 508 1600
16002)]
rr,,.(r )
r−++=−
= 5508 8128
0508 1600 8128
1600 1602
2.
.
()(r
rrr−
=−−=−− ±−^0045088128
2508
1600 2 560 000 1651)(^2 .)()
(.)
,, ,−−= ±+^6 60 96
10 16
1600 2 725 160 96
10 161600 1650 80.
.
,,.
.= ± ≈ ±.^66155
10 16
320 1600 1650 806155
10 16.
.
.≈ − isnegativve