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A 480-pound alloy (which is just a mixture of metals) containing 30% silver was mixed with an alloy containing 55% silver to get an alloy containing 40% silver. How many pounds of the 55% alloy were used?

Sagot :

Answer:

"24 pounds" is the appropriate answer.

Step-by-step explanation:

According to the question, the equation will be:

⇒ x+y=60...(equation 1)

or,

⇒ \frac{55}{100}x+\frac{30}{100}y=\frac{40}{100}\times 60

     55x+30y=2400...(equation 2)

Now,

By multiplying (equation 1) by 55 and (equation 2) by 1, we get

⇒ 55x+55y=3300...(equation 3)

⇒ 55x+30y=2400...(equation 4)

From equation 3 and 4, we get

⇒ 25y=900

      y=\frac{900}{25}

         =36

By putting the values of "y" in equation 1, we get

⇒ x=60-36

      =24

Thus the value of x be "24 pounds".
Let x = amt of 55% alloy required:
Then
(800-x) = amt of 30% alloy required
:
Write a silver amt equation:
.55x + .30(800-x) = .40(800)
:
.55x + 240 - .3x = 320
:
.55x - .30x = 320 - 240
:
.25x = 80
:
x = 80/.25
:
x = 320 lb of 55% alloy
then
800 - 320 = 480 lb of 30% alloy
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