Question

Brad has two alloys containing copper and silver. Alloy A contains 50% pure silver, and alloy B contains 75% pure silver. He wants to make 10 ounces of a new alloy containing 70% pure silver. How much alloy of each type does he need to mix?

Answer 1

Answer:

I think the answer would be

Alloy A 30%

Alloy B 40%

The reason I say that is because Alloy B has a higher amount of pure silver so yeah

I hope this helps mark me brainliest and tell me if I'm wrong please so Then I could fix it.

P.S. Wegnerkolmp2741o is gonna answer so I would go with his

Answer 2

Answer:

There are 2 ounces of 50% silver

There are 8 ounces of 75% silver

Step-by-step explanation:

The percentage of silver times the amount of silver + the percentage of silver times the amount of silver = The percentage of silver times the  total amount of silver

We will put A in the first spot and B in the second spot  and The total on the right

We will take x ounces of A at 50%

We need 10 ounces of the total

10-x  is the amount we need for B  (  The amount in A Plus the amount in B has to equal 10)

We have (10-x) at 75% for B

We have 10 ounces at 70% for The total

 .50 *x + .75*(10-x) = .70*10

.5x +  .75*(10-x) = 7

Distribute

.5x +7.5 - .75x =7

Combine like terms

-.25x + 7.5 = 7

Subtract 7.5 from each side

-.25x +7.5-7.5 =7-7.5

-.25x =-.5

Divide by -.25 on each side

-.25x/-.25 = -.5/-.25

x =2

There are 2 ounces of 50% silver

To find how many ounces of 75% silver

10-x = 10-2 = 8

There are 8 ounces of 75% silver