What is the answer and how?

1.If an alloy containing 30% silver is mixed with a 55% silver alloy to get 800 pounds of 40% alloy, how much of each must be used?

2.A hospital needs to dilute a 50% boric acid solution to a 10% solution. If it needs 25 liters of the 10% solution, how much water should it use

Top 1 Answers

A few days ago

hayharbr

Favorite Answer

Choose a variable for the pounds of one alloy, say the 30%; I’ll use x. Then since the total is 800 lb, the 55% quantity is (800 – x)

Then write expressions for % times amount for each alloy, and add them to get % times amount of final alloy

30x + 55(800-x) = 40(800)

30x + 44000 – 55x = 32000

-25x + 44000 = 32000

-25x = -12000

x = 480

So 480 lb of 30% and 800-480 lb of 55%

On the other, let x = amount of water added. Then (25 -x) will be the amount of 50% .So as in the above,

50(25-x) + 0(x) = 10(25) [since 0% of the water is acid]

1250 – 50x = 250

-50x = -1000

etc

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