Algebra Homework Help!?

Question

the octane number of gasoline can be thought of as the percent of the compound octane in the gasoline. For instance, 89-octane gas would be 89% octane and 11% something else. Suppose that a wholesaler has 84-octane gas and 91-octane gast. -----------A) the wholesaler mixes X gallons of 84-octane gas and y gallons of 91-octane gas. write an expression for the number of gallons of octane in the mixture.----------- B)evaluate the expression if 1000 gallons of 84 octane gas is mixed with 2000 gallons of 91-octane gas. -------C) what will be the octane number of the mixture in part b--------D) if the wholesaler recieves an order for 15,000 gallons of 89.7 octane gas, how much of each kind should be mixed to fill the order?

Answer 1

(a) The number of gallons of octane in the mixture is (84x+91y)/100.
(b) Put x = 1000gal and y = 2000gal. in the above formula:
(84*1000 + 91*2000) / 100
= 840 + 91*20
= 840 + 1820
= 2660 gal. of octane.
(c) There are 3000 gal. in the mixture.The octane number is therefore 2660*100/3000 = 266/3 = 88 2/3.
(d) Suppose he mixes x gal. of 84 octane with 15000 - x gal. of 91 octane.
Then 84x + 91(15000 - x) = 15000*89.7
(91 - 84)x = ( 91-89.7)*15000
7x = 1.3*15000
x = 1.3*15000/7 = 19500/7 gal.
15000 - x = (15000*7 - 19500)/7 = 85500/7 gal.
He therefore needs to mix 19500/7 gal of 84 octane with
85500/7 gal. of 91.7 octane.

Answer 2

A) .84x + .91y

B) .84x + .91y = .84(1000) + .91(2000) = 840 + 1820 = 2620 gallons

C) Octane number is 2620*100/(1000 + 2000) = 262/3 = 87.3

D)
.84x + .91y = (15,000).897
x + y = 15,000
x = 15,000 - y
.84(15000 - y) + .91y = 13455
12600 - .84y + .91y = 13455
.07y = 855
y = 12,214.286 gallons
x = 15000 - y = 2,785.714 gallons

Answer 3

you are in a jam with gosoline leak.

Answer 4

hmm

Answer 5

sry... that gave me a headache.......