SOLUTION: Could somebody please explain to me *clearly* how to solve mixture problems? I've been trying to figure them out, but I just don't understand all of it. For example, how would you
Algebra ->
Customizable Word Problem Solvers
-> Mixtures
-> SOLUTION: Could somebody please explain to me *clearly* how to solve mixture problems? I've been trying to figure them out, but I just don't understand all of it. For example, how would you
Log On
Question 283890: Could somebody please explain to me *clearly* how to solve mixture problems? I've been trying to figure them out, but I just don't understand all of it. For example, how would you set up an equation for a problem like this?
2.) Brianna has 2 pounds of brand 176 nuts that sell for $3 a pound. She also has brand 391 nuts that sell for $1.74 a pound. If she wants a mixture of nuts that sell for $2.58 a pound, how many pounds of brand 391 should she add?
You can put this solution on YOUR website! First of all the brand names are only going to lead to confusion so let's change that.
let call brand 391 t for three
and brand 176 o for one
what do we know?
brand o sells for $3.00 lb
and she has 2 lbs of brand o so o=2
brand t sells for 1.74 lb
but we don't know how much she has of brand t . We will call it t
We want to sell a mix (o+t) for 2.58.
So how do we set up an equation?
the individual nuts have to be equal to the mix
What do we want to know?How many lbs of t are we mixing with o
to get o+t
so we are dealing with selling price.
the o selling price is
2*o where o =300
I am going to convert everything to cents so we don't deal with decimals.
2*300+t*174=(2+t)*258
solve for t
You can put this solution on YOUR website! Brianna has 2 pounds of brand 176 nuts that sell for $3 a pound. She also has brand 391 nuts that sell for $1.74 a pound. If she wants a mixture of nuts that sell for $2.58 a pound, how many pounds of brand 391 should she add?
How do you come to that conclusion?
------------------------
B1 costs 300/pound
B3 costs 174/pound
She has 2 pounds of B1
x = pounds of B3 to add
------------------------
The cost of the mix is:
B1*300 + x*174
The amount of the mix is B1 + x
and the cost of the mix is (B1 + x)*258
-----------------
There are 2 equations for the price of the mix, and they're equal
B1*300 + x*174 = (B1 + x)*258
B1 is 2
600 + 174x = 516 + 258x
x = 1 pound
email via the Thank You note if that's not clear, or to Moral Loophole@aol.com
Let represent the number of pounds of the inexpensive brand.
Then the value of the expensive brand is 2 lbs times $3.00 per pound is $6.00. The value of the inexpensive brand is . The total weight of the mixture when all is said and done is lbs., and the total value of the mixture has to be . The sum of the values of the constituent parts of the mixture has to be equal to the total value of the mixture, so:
