SOLUTION: Hello: please help me fiqure out this word problem for future math problems with the step to perform: i do recall this problems but it has been a while since i done this word mat

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Question 558198: Hello:
please help me fiqure out this word problem for future math problems with the step to perform: i do recall this problems but it has been a while since i done this word math problem
this is the problem: How much of an alloy that is 40% copper should be xixed with 500 ounce tha is 80% copper in order to get an alloy that is 60% copper.

Found 3 solutions by mananth, MathTherapy, josgarithmetic:
Answer by mananth(16949) About Me  (Show Source):
You can put this solution on YOUR website!
percent ---------------- quantity
Alloy type I 80 ---------------- 500 ounce
Alloy type II 40 ---------------- x ounce
Mixture 60 ---------------- 500 + x ounce

80*500+40x=60(500+x)
40000+40x=30000+60x
40x-60 x= 30000 -40000
-20 x = -10000
/ -20
x= 500 ounce of Alloy type II

Answer by MathTherapy(10801) About Me  (Show Source):
You can put this solution on YOUR website!
 Hello:
please help me fiqure out this word problem for future math problems with the step to perform: i do recall this problems
but it has been a while since i done this word math problem
this is the problem: How much of an alloy
that is 40% copper should be xixed with 500 ounce tha is 80% copper in order to get an alloy that is 60% copper.
*************************************
Adding a 40% substance to an existing 80% substance, to get a 60% substance results in an addition of the same amount of the
INITIAL substance. In other words, Percent: %28%2280%25%22+%2B+%2240%25%22%29%2F2+=+%22120%25%22%2F2+=+%2260%25%22, and AMOUNTS: %28500+%2B+500%29%2F2+=+%221%2C000%22%2F2+=+500.
So, amount of 40% copper to be added will be the INITIAL amount, 500 ounces.

The above is NOT STANDARD, so future mixture problems may definitley NOT be similar to this one. And, as you would like to
"fiqure out this word problem for future math problems with the step to perform...", you can do this problem, as follows:

Let the amount of 40% copper to be added, be C

Percent of copper in INITIAL amount of alloy: 80, or .8
Amount of copper in INITIAL amount of alloy: .8(500) = 400 oz

Percent of copper to be ADDED to INITIAL alloy: 40, or .4
Amount of copper to be ADDED: .4C

Percent of copper in RESULTANT alloy: 60, or .6
Amount of copper in RESULTANT alloy: .6(500 + C) = 300 + .6C

We then get: Initial amount of copper + ADDED amount of copper = RESULTANT amount uf copper, OR
                                           400                +                     .4C                   =                 300 + .6C
                                                                                                  .4C - .6C = 300 - 400
                                                                                                         - .2C = - 100
                                    Amount of copper to be ADDED to alloy, or 
 
Some folks tend to use a table, which could prove to be more organized and easier to follow, But, that's up to you.

Answer by josgarithmetic(39790) About Me  (Show Source):
You can put this solution on YOUR website!
PERCENT COPPER         QUANTITY          PURE COPPER

   40                     x                 0.4x
   80                    500 oz.            0.8*500

   60                    x+500              0.4x+0.8*500

highlight_green%28%280.4x%2B0.8%2A500%29%2F%28x%2B500%29=0.6%29

0.4x%2B0.8%2A500=0.6%28x%2B500%29
0.4x%2B0.8%2A500=0.6x%2B0.6%2A500
0.8%2A500-0.6%2A500=0.6x-0.4x
500%280.8-0.6%29=%280.6-0.4%29x

highlight%28x=500%28%280.8-0.6%29%2F%280.6-0.4%29%29%29
Compute this.