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Mathematics 16 Online
OpenStudy (anonymous):

Hard question Tell if u give up!!!!!!!!!!!!!!!! Owen made 100 sandwiches which she sold for exactly $100. She sold caviar sandwiches for $5.00 each, the bologna sandwiches for $2.00, and the liverwurst sandwiches for 10 cents. How many of each type of sandwich did she make?

OpenStudy (amistre64):

you have 2 equations with 3 unknowns ...

OpenStudy (anonymous):

yes! it is tricky try and solve it i have solved it and got it correct

OpenStudy (amistre64):

a+b+c = k 5a+2b+.1c=k the intersection of 2 planes is a line, not a point.

OpenStudy (anonymous):

so?

OpenStudy (inkyvoyd):

so, trolololol

OpenStudy (anonymous):

it has a solution!!

OpenStudy (inkyvoyd):

It has a lot of solutions.

OpenStudy (anonymous):

only 1 sollution i have solved it!!

OpenStudy (amistre64):

im sure it has a solution, but is that solution unique? as is: a = -33.333... + .6333... c b = 133.333... -1.6333... c if for some discrete c, a and b are nonnegative integers, then there may well be some solution we could also setup a grid to determine all the possible combinations

OpenStudy (anonymous):

they are whole numbers integers and ofcourse are positive integers

OpenStudy (amistre64):

0 >= -33.333... + .6333... c 33.333... --------- >= c 6.333... 0 >= 133.333... -1.6333... c 133.333... ---------- >= c 1.6333... this at least gives us a range for the interval of c

OpenStudy (anonymous):

shall i give sollution

OpenStudy (amistre64):

of course not

OpenStudy (anonymous):

ok!

OpenStudy (anonymous):

tell me the sollution by tommorow i have to go!

OpenStudy (amistre64):

should be .6333 on the top, mistyped the decimal :) got it

OpenStudy (asnaseer):

@amistre64 -- I get the same solution let c = number of caviar sandwiches b = number of bologna sandwiches l = number of liverwurst sandwiches then we get:\[\begin{align} c+b+l&=100\tag{1}\\ 5c+2b+0.1l&=100\tag{2}\\ \therefore 50c+20b+l&=1000\qquad\text{(2) x 10}\tag{3}\\ \therefore 49c+19b&=900\qquad\text{(3)-(1)}\tag{4}\\ \therefore 19b&=900-49c=19\cdot47+7-49c\\ \therefore 19(47-b)&=49c-7=7(7c-1) \end{align}\] Therefore \((7c-1)\) must be a multiple of 19 which leads to:\[c=\frac{19m+1}{7}\qquad\text{for m=0.1,...}\]Also, from (4) we can see that:\[c\le17\]Using this we get the only integer solution which is c=11. Thus c=11, b=19 and l=70

OpenStudy (amistre64):

:)

OpenStudy (anonymous):

@asnaseer ur great!!!

OpenStudy (asnaseer):

thank you :)

OpenStudy (asnaseer):

Did you use the same method to solve?

OpenStudy (anonymous):

yes!

OpenStudy (asnaseer):

then you are also great! :D

OpenStudy (anonymous):

lol! shall i ask one more question answer this one!!!

OpenStudy (asnaseer):

it is best to ask each question separelty

OpenStudy (asnaseer):

so better to close this and start a new one

OpenStudy (anonymous):

yes that's what i am doing

OpenStudy (asnaseer):

:)

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