A goldsmith has two gold alloys. The first alloy is 20% gold; the second alloy is 60% gold. How many grams of each should be mixed to produce 40 grams of an alloy that is 30% gold?

Question

amount of 20% gold_____g
Amount of 60% gold_____g

Answer 1

grams 20 % = a
grams 60% = b

total gold = .2 a + .6 b
total mass = a + b = 40 so b = 40-a

so
.2 a + .6 b = .3 (40) = 12
.2 a + .6 (40-a) = 12
-.4 a + 24 = 12
.4 a = 12
a = 30
so
b = 10

Answer 2

It can be difficult to come up with formulas from word problems.

Your going to add some weight of the first alloy and some weight of the second allow to get 40 grams.
let x be the weight used from the first alloy and y the weight from the second.
so x+y=40

How much gold is in the 40 grams?

Well it's 30% gold, so
40*(.3)=12g of gold

likewise the amount of gold present from the first alloy would be 0.2x

the amount in the 60% sample is 0.6y

those two amounts add together to be the 12g of gold.

0.2x+0.6y=12

so two formulas, two variables.

if x+y=40 then
x=40-y

plug it in
0.2(40-y)+0.6(y)=12
8-0.2y+0.6y=12
0.4y=4
y=10

therefore x=30

check that it works to be safe
0.2(30)+0.6(10)=12
6+6=12
excellent.

hope this helps!

Answer 3

There is a rule for this class of problems of finding proportions of mixtures. Most of the time this can be done mentally, for example by chemists or nurses.

We have one ingredient A at 20%, and B at 60%. The required mixture is 30% (must be between those of A & B).

Take the difference between the ingredients, namely 60-20=40%.

The target percentage lies at 10% from A and 30% from B, or in the ratio 1:3

The proportions required for A and B will be 3:1, namely 30g of A and 10g of B.

Answer 4

Interesting approach Mathmate. Hadn't seen that before.