A chemist has two alloys, one of which is 15% gold and 20% lead and the other which is 30% gold and 50% lead. How many grams of each of the two alloys should be used to make an alloy that contains 82.5 g of gold and 113 g of lead?

490 grams of the first alloy should be used.

30 grams of the second alloy should be used.

Step-by-step explanation:

Let x represent the weight of the first alloy in grams that should be used.

Let y represent the weight of the second alloy in grams that should be used.

A chemist has two alloys, one of which is 15% gold and 20% lead. This means that the amount of gold and lead in the first alloy is

0.15x and 0.2x

The second alloy contains 30% gold and 50% lead. This means that the amount of gold and lead in the second alloy is

0.3y and 0.5y

If the alloy to be made contains 82.5 g of gold, it means that

0.15x + 0.3y = 82.5 - - - - - - - - - - - -1

The second alloy would also contain 113 g of lead. This means that

0.2x + 0.5y = 113 - - - - - - - - - - - - -2

Multiplying equation 1 by 0.2 and equation 2 by 0.15, it becomes

0.03x + 0.06y = 16.5

0.03x + 0.075y = 16.95

Subtracting, it becomes

- 0.015y = - 0.45

y = - 0.45/- 0.015

y = 30

Substituting y = 30 into equation 1, it becomes

0.15x + 0.3 × 30 = 82.5

0.15x + 9 = 82.5

0.15x = 82.5 - 9 = 73.5

x = 73.5/0.15

x = 490

sqrt 10 = 3.16

sqrt 15 = 3.87

So the difference would be 0.7

I hope this helps! :)

~Ayden