Question

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?

Answer 1

To find how much of each silver alloy to mix, we set up a system of linear equations based on the total weight and the percentage of silver in the resulting alloy. We then solve for the quantity of each alloy to get the desired concentration in the 800-pound mixture.

Step-by-step explanation:

To solve the problem of mixing two alloys to get an 800-pound 40% silver alloy, we need to use the concept of weighted averages to calculate the amount of each alloy required. Let's denote the amount of 30% silver alloy as x pounds and the amount of 55% silver alloy as y pounds. The total weight of the mixture is given as 800 pounds, so we have x + y = 800. We also know that the total amount of silver in the mixture should be 40% of 800 pounds, which is 320 pounds.

The amount of silver from the 30% alloy is 0.30x and from the 55% alloy is 0.55y. Combining these expressions, we get the equation 0.30x + 0.55y = 320. Now we have a system of equations with two variables:

1. x + y = 800
2. 0.30x + 0.55y = 320

We can solve this system by substitution or elimination. If we use substitution, we can express x as 800 - y and substitute into the second equation:

0.30(800 - y) + 0.55y = 320

After simplification, we find the value of y, which represents the amount of 55% alloy needed, and subsequently the value of x, which represents the amount of 30% alloy needed.

Answer 2

480 pounds of silver and 320 pounds of silver alloy

EXPLANATION

The alloy contains 30% silver and 55% silver alloy.

Let the amount of silver be s.

Let the amount of silver alloy be a.

So, we have that the total amount of alloy is 800 pounds.

This must mean that if we add the amount of silver and silver alloy, we will get 800:

s + a = 800 ____(1)

Also, 30% of s and 55% of a make up 40% of 800 pounds of alloy.

This means that:

30% * s + 55% * a = 800 * 40%

=> 0.3s + 0.55a = 800 * 0.4 (convert percentages to decimals)

=> 0.3s + 0.55a = 320 ____(2)

We now have two simultaneous equations:

s + a = 800 ____(1)

0.3s + 0.55a = 320 ____(2)

From (1):

a = 800 - s

Put that in (2):

0.3s + 0.55(800 - s) = 320

0.3s + 440 - 0.55s = 320

Collect like terms:

0.3s - 0.55s = 320 - 440

-0.25s = -120

Divide through by -0.25:

s = -120 / -0.25

s = 480 pounds

Recall:

a = 800 -s

=> a = 800 - 480

a = 320 pounds

Therefore, 480 pounds of silver and 320 pounds of silver alloy must be used.