Question 306138: How many mg of a metal containing 45% nickel must be combined with 6mg of pure nickel to form an alloy containing 78% nickel? Found 3 solutions by josmiceli, nospmoht, HydroKid:Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website! Let = mg of metal containing 45% nickel to be added = nickel content of added metal
In words:
(mg of nickel in final alloy)/(mg of final alloy) = 78%
4 mg of 45% nickel metal must be added
check:
OK
You can put this solution on YOUR website! Aight, first you have to sort your information. You have:
.45x (represents an unknown amount of 45% nickel)
1(6) (represents 6 mg of 100% nickel)
.78(6+x) (represents 6 mg plus an unkown amount of 78% nickel)
Then just make an equation. In 'mixture' word problems, the outcome or mixture will be on one side and the two substances being mixed will be on the other. This is how the equation for the problem would look:
.45x +1(6)=.78(6+x)
Solve for x:
.45x +1(6)=.78(6+x)
.45x +6 =4.68 + .78x
1.32 = .33x
4 = x
Then you plug it into the unknown information (shown at the top. So:
.45x now is .45(4), which equals 1.8 mgs
Answer: 1.8 mgs of 45% nickel mixed with 6 mgs of 100% nickel will result in 7.8 mgs of 78% nickel.
You can put this solution on YOUR website! .45x+6/x+6=.78
Solve the given equation step by step.
This is also equal to this system
.45x+6=.78(x+6)
You will get the final answer as
x=4
Therefore,
4 mg of a metal containing 45% nickel must be combined with 6 mg of pure nickel to form an alloy containing 78% nickel.
