Question 1195325
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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.<br>
You can also view the problem as a "mixture" problem.<br>
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.<br>
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.<br>
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.<br>
The three percentages are 20, 50, and 70....<br>
70-20 = 50;
50-20 = 30;
30/50 = 3/5<br>
50% is 3/5 of the way from 20% to 70%; that means 3/5 of the mixture needs to be the 70% alloy.<br>
ANSWER: 3/5 of 100 grams, or 60 grams, of 70% silver; the other 40 grams of 20% silver<br>
CHECK:
.70(60)+.20(40) = 42+8 = 50
.50(100) = 50<br>