Question 1195325: An alloy is made by melting and combining two or more metals. A metalsmith has two alloys, each containing different amounts of silver, that will be melted and combined to form another alloy. Every 10 grams of alloy A contains 2 grams of silver, and every 10 grams of alloy B contains 7 grams of silver. To obtain 100 grams of an alloy that contains 50 grams of silver, how many grams of alloy A should be combined with alloy B?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website! .
An alloy is made by melting and combining two or more metals.
A metalsmith has two alloys, each containing different amounts of silver,
that will be melted and combined to form another alloy.
Every 10 grams of alloy A contains 2 grams of silver,
and every 10 grams of alloy B contains 7 grams of silver.
To obtain 100 grams of an alloy that contains 50 grams of silver,
how many grams of alloy A should be combined with alloy B?
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Take x times 10 grams of alloy A and y times 10 grams of alloy B.
Then the total mass equation is
10x + 10y = 100 grams (1)
while the pure silver mass equation is
2x + 7y = 50 grams. (2)
So, you have this system of 2 equations for x and y
x + y = 10, (1')
2x + 7y = 50. (2)
From eq(1'), express x = 10 - y and substitute it into equation (2). You will get
a single equation for y
2*(10-y) + 7y = 50
20 - 2y + 7y = 50
5y = 50 - 20
5y = 30
y = 30/5 = 6.
So, you should take 6*10 = 60 grams of alloy B and the rest, 100 - 60 = 40 grams of alloy A. ANSWER
CHECK. We check the content of silver. It is 7*6 + 4*2 = 42 + 8 = 50 grams. ! Correct !
Solved.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
The solution from tutor @ikleyn shows a typical standard algebraic solution. You should understand that method and be able to use it to solve similar problems.
You can also view the problem as a "mixture" problem.
You are mixing an alloy that is 20% silver (2 grams silver per 10 grams alloy) with another alloy that is 70% silver, producing an alloy that is 50% silver.
A non-algebraic way to solve mixture problems is to view the three percentages on a number line; the ratio in which the two original alloys need to be mixed to produce the new alloy depends on where the percentage of the new alloy lies between the percentages for the two original alloys.
I will show the calculations for solving the problem with a minimum number of words so you can see how easy the method is to use.
The three percentages are 20, 50, and 70....
70-20 = 50;
50-20 = 30;
30/50 = 3/5
50% is 3/5 of the way from 20% to 70%; that means 3/5 of the mixture needs to be the 70% alloy.
ANSWER: 3/5 of 100 grams, or 60 grams, of 70% silver; the other 40 grams of 20% silver
CHECK:
.70(60)+.20(40) = 42+8 = 50
.50(100) = 50
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