1. Ave(city) = 15 mi/gal,
Ave(exp.) = x mi/gal,
Ave = (15 + x) / 2 = 25 mi/gal,
Multiply both sides by 2 and get:
15 + x = 50,
x = 50 - 15,
x = 35 mi/gal, express way.
2. d = 2 * 35 km = 70 km,
Speed = 70 km / 4.8h = 14.58 km/h in
still water,
C = 15 - 14.58 = 0.417 km/h Current
speed.
3. x = old membership size,
x + 8 = new membership size,
1200 / x = assessment under old membership,
1200 / (x + 8) = assessment under new
membership,
1200 / (x + 8) = ( 1200 / x ) - 7.50,
1200 / (x + 8) - 1200 / x = -7.50,
common denominator = x(x + 8),
(1200x - 1200(x + 8)) / x(x + 8) = -7.5
(1200x - 1200x - 9600) / x(x + 8) = -7.50,
-9600 / x(x + 8) = -7.50,
Cross multiply:
-7.50 * x(x + 8) = -9600,
Divide both sides by -7.50 and get:
x(x + 8) = -9600 / -7.50,
x^2 + 8x = 1280,
x^2 +8x - 1280 = 0,
Factor Eq and get:
(x - 32) (x + 40) = 0,
x - 32 = 0,
x = 32,
x + 40 = 0,
x = -40.
Solution set: x = 32, and x = -40.
select pos. value: x = 32.
x + 8 = 32 + 8 = 40 = new membership.
1. During 60mi of city driving, Jenna averaged 15mi/gal. She then drove 140mi on an expressway and averaged 25mi/gal for the entire 200mi. Find the average fuel consumption on the expressway.
2. The excursion boat Holiday travels 35km upstream and then back again in 4h 48min. If the speed of the Holiday in still water is 15km/h, what is the speed of the current?
3. Members of the Computer Club were assessed equal amounts to raise $1200 to buy some software. When 8 new members joined, the per-member assessment was reduced by $7.50. What was the new size of the club?
2 answers
Correction:
14.58 km/h is NOT still wqater.
14.58 km/h is NOT still wqater.