15. A restaurant has one type of milk that has 2% fat and another that has 7% fat. how many quarts of each type does the restaurant need to make 40 quarts of a milk mixture that is 5% fat? a. 24 quarts of the 2% milk and 16 quarts of the 7% milk b. 16 quarts of the 2% milk and 24 quarts of the 7% milk c. 16 quarts of the 2% milk and 16 quarts of the 7 milk d. 24 quarts of the 2% milk and 24 quarts of the 7% milk
oh boy a mixture problem ...
@dtan5457
yeah o_o
lets call the amount of 7% milk x so the amount of 2% milk is \(40-x\) because the total is 40
\(5\% \) of \(40\) is \(2\) right?
that means \[0.07x+0.02\times (40-x)=2\] or to make it easier \[7x+2(40-x)=200\]
O_O I-I think
well this is how i did it.. x+y=40 2x+7y/40=5 2x+7y=200 x=40-y plug in to get 80-2y+7y=200 5y=200 y=24 x=16 b. 16 quarts of the 2% milk and 24 quarts of the 7% milk
yeah it is because ten percent of forty is four, so five percent is two
@satellite73 please check my work
wait nvm i already screwed up 5y=120
not 200..
o.o
at some point you are going to have to solve \[7x+2(40-x)=200\]
wow wow wow wow wow wow wow wow wow wow. now I gotta read all of this
yes! @Mimi_Styles TO understand IT!!!
if i just solve for y for 7% milk, isn't that enough?
ok we can start again if you like and do it slow
these problems are all more or less the same do one or two and you can do them all
<____>
i don't know if that is a yes or a no
i just wanna know if i my answer is correct so i can rest in peace
I made this face and it's called the confused face
ok first off, do you know that \(7\%\) means \(.07\)
yes.
and do you know that 7 percent of a number means that number times \(.07\) ?
if you do not, feel free to say "i know it now"
I know now... ._.
ok so for example 7% of 50 is \(.07\times 50\) and 7% of \(x\) is \(.07x\)
o.o
good to here?
ohh okay
if we are gong to solve one of these, we have to have a variable some people actually like two variables, i think that is more confusing, one will do and we have to say what it is i say "let \(x\) be the amount of 7% milk"
then in terms of \(x\) we know that the total number of gallons is 40 if \(x\) of them are 7% milk the rest, the 2% milk is \(40-x\) for example if there were 15 gallons of 7% milk there would have to be \(40-15=25\) gallon of 2% milk
since we now know that 7% of x is \(.07x\) and also that 2% of \(40-x\) is \(.02(40-x)\) we can make an equation
the total amount of fat or whatever is 5% of 40, which we decided is 2 that means \[.07x+.02(40-x)=2\]
o.o ohh
now that is somewhat annoying to solve
but not too bad if you multiply both sides by 100 and solve \[7x+2(40-x)=200\]
i am sure you can do that one once you get it, you will have the amount of 7% milk
uhhh D:
wait wait *lost*
what step
"but not too bad if you multiply both sides by 100 and solve 7x+2(40−x)=200 " <----
that one.
\[.07x+.02(40-x)=2\] is what needs to be solved and you can do it with the decimals in place if you like
but multiplying by 100 moved the decimal over two places, i.e. gets rid of them
if you multiply by 100 you have to multiply both decimals by 100 to get rid of the decimal, but you also have to multiply 2 by 100 to get 200 that is why \[.07x+.02(40-x)=2\] is the same as \[7x+2(40-x)=200\]
ohh~
ohhh! ^_^
so what im doing is...multiplying each sides by 100 ?
yes to get from the equation with decimals to the equation with out them you still have to solve it though
0.02 x + 0.07(40 -x ) =40(0.05) 2x + 280-7x = 200 -5x =-80 x = 16 fat = 40-16 = 24
ok looks reasonable
i am not sure how you got that one, but it is correct also
o.o
wait so 16 2% and 24 7%?
yes
@Mimi_Styles switched them
\[0.02 x + 0.07(40 -x ) =40(0.05) \] has \(x\) as the 2% milk
btw "whole milk" is about \(3.25\%\) or so not 5% and certainly not 7%
oooohhh
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