## anonymous 5 years ago last year you could get a hamburger,fries, and coloa for $2.70. since the price of hamburger increased 20%, the prices of fries has increased 10%, and cola 60%. The same meal now costs 3.46$ if the price of a cola is now 20 cents more than that of a hamburger, what was the price of each item last year?

1. amistre64

c = h+.20 h + c + f = 2.70 (then) h(.2) + c(.6) + f(.1) = 3.46 (now)

2. anonymous

so for then its would be 20 cents 60 cents 10 cents?

3. amistre64

dont know yet :) im thinknig I should figure the percentage of the price as well....

4. amistre64

if we test out that they all cost .30 then; we can see where our guess leads us...

5. amistre64

.30(3) = 2.70 so that much would be good; .3(.2) + .3(.6) + .3(.1) .06 +.18 +.03 = .27 is not right, so we are missing something; we went lower in price instead of higher... they should be. the percentages should be add to 1 to account for the base price and the rise in percentage...

6. amistre64

h(1.2) + c(1.6) + f(1.1) = 3.46 is more accurate equation..

7. amistre64

at .30 cents then we would get: .36 +.48 +.33 = .36 + .81 = 1.17 is better; given that my .30 was off to begin with; shoulda been .90 :)

8. amistre64

h(1.2) = c(1.6) - .20 : is good for this the total price increased by .76. 76/270 = percent changed. which is an awful decimal so we will keep it as a fraction to make it easier on us. ................................................ h(1.2) + c(1.6) + f(1.1) = 2.70(3.46/2.70) (c(1.6) - .20)(1.2) + c(1.6) + f(1.1) = 3.46 c(1.92) - .24 + c(1.6) + f(1.1) = 3.46 c(3.52) + f(1.1)=3.70 is what I get so far...

9. amistre64

h(1.2) + c(1.6) + f(1.1) = 2.70(3.46/2.70) c(1.6) - .20 + c(1.6) + f(1.1) = 3.46 <-- corrected for my stupidity :) c(3.2) + f(1.1) = 3.70 is what we are looking at; not that it helps me much at the moment :)

10. amistre64

ack!!..... c(3.2) + f(1.1) = 3.66 ...and thats should be right lol... we really need someone smarter then me on this ;)

11. anonymous

lol thanks for all your help ;)

12. anonymous

so just so were absolutely clear..lol... soda is \$3.20 fries are 1.10 and burger is 1.20?

13. amistre64

f = 2.70 - h - c; if re can get f in terms of c this would be delightful... f(1.1) = (3.46 -h(1.2) - c(1.6)) f = [3.46 -h(1.2) - c(1.6)]/1.1 h = [c(1.6) - .20]/1.2 now we can test this theory of mine out and see where it ends up ;) we got everything in terms of "c" c + h + f = 2.70 c + [c(1.6) - .20]/1.2 + [3.46 -{[c(1.6) - .20]/1.2}(1.2) - c(1.6)]/1.1 = 2.70 the equation editor should be able to clean that up for us: $c + \frac{[c(1.6) - .20]}{1.2} + \frac{[3.66 -c(3.2)]}{1.1} = 2.70$

14. amistre64

get common denominators: (1.2)(1.1) c(1.32) +(1.1)[c(1.6) - .20] + (1.2)[3.66 -c(3.2)] = 2.70(1.32) c(1.32) + c(1.76) - .22 + 4.392 - c(3.84) = 3.564 c(1.32 + 1.76 - 3.84) = 3.564 +.22 -4.392

15. amistre64

c(-.76) = -.608 c = .80 cents.... with alot of luck ;)

16. amistre64

lets get some confidence and try to work out the rest :) c(1.6) = price of cola today .8(1.6) = 1.28 h(1.2) = 1.28 - .20 = 1.18 is the cost of a hamburger today. f(1.1) = 3.46 - 1.18 - 1.28 f(1.1) = cost of fries today = 1.00 corssing fingers and typing is not a good combination :) but im almost there lol

17. amistre64

f = 1/1.1 = 0.9090909090909091 = .91 h = 1.18/1.2 = 0.9833333333333333 = .98, or .99 c = 1.28/1.6 = .80 lets add them and see if we are close :) .80 + .91 + .98 = 2.70 2.69 = 2.70 ; so Id go with h= .99

18. amistre64

h = .99; c=.80; f=.91 last year