Question

how much mg of a metal containing 4% silver must be combined with 11 mg of a metal containing 38% silver to form an alloy containing 26% silver

Answer 1

Answer:

6 mg of the metal needs to be added.

Step-by-step explanation:

Let the amount (in mg) of metal that needs to be added by y.

Therefore, the amount of silver in the above metal is 0.04y.

Prior to mixing, 11 mg of a metal contained 38% of silver (Given).

Therefore, the amount of silver before= $\frac{38}{100}*11$= 4.18 mg

The total amount of silver after mixing, 4.18 + 0.04y mg

The total amount of metal after mixing, 11 + y mg

New percentage of silver = 26% .

Thus, $\frac{4.18+0.04y}{11+y}*100= 26$

$(4.18 +0.04 y) * 100 = 26 *(11 +y)$

$418+4y=286+26y$

$132=22y$

y=6 mg

Therefore, the amount of metal that needs to be added is 6 mg.