A goldsmith has two alloys of gold, the first being 80 % pure gold, the second is 50 % pure gold.?

Question

How many onces of each must he take to make 75 onces of an alloy which shall be 72 % pure gold?

Answer 1

x=80% alloy, y=50% alloy
x+y = 75, y = 75-x
75*.72 = 54 = .8x + .5*(75 - x), or .3x + 37.5 = 54, so
.3x = 16.5, and x = 55, and y = 20

Answer 2

This is a problem requiring two equations.

Your first equation is the total weight of the alloys X and Y,
as follows:

x + y = 75 oz

The second equation is the required percent formulation:

80%(x) +50% (y) = 72%(75)
0.8X + 0.5y = 0.72(75)
0.8x + .5y = 54

Using the first equation, we can solve for x:

x + y = 75
x = 75-y

Substituting this x value in equation no. 2, we get:

0.8(75-y) + 0.5y = 54
60 - .0.8y + 0.5y = 54
0.3y = 6
y = 6/0.3 = 20 oz

Solving for x:
x = 75- 20 = 55 oz.

Therefore, we need 55 oz of X (80% pure gold) and 20 oz of Y (50% pure gold), to form an alloy of 75 oz of (72% pure gold).

I hope that was a golden solution for you. You have a great, blessed evening! A golden one.

Answer 3

Hi,

20 ounces of the 50% pure gold and 55 ounces of 80% pure gold is the answer.

You can get it by making 2 equations:

ounces of 50% alloy + ounces of 80% alloy = total ounces
x + y = 75

ounces of pure gold from 50% alloy + ounces of pure gold from 80% alloy = total ounces of pure gold in 72% alloy

.50x + .80y = .72(75)

Solve this system by multiplying the second equation by -2 and then adding them.

x + y = 75
.50x + .80y = 54

x + y = 75
-2(.50x + .80y = 54)

x + y = 75
-x - 1.60y = -108
------------------------
-.6y = -33
y = 55, so 55 ounces of 80% pure gold alloy are needed.

75 - 55 = 20, so 20 ounces of 50% pure gold alloy are needed.



There is a second way to do mixture problems like this. Put the percentages along a number line and figure out how far apart they are.

50%............72%..80%
-+----------------+-----+
.<------22------><-8->
.<--------------30.----->

50 is 22/30 or 11/15 of the distance from 72.
80 is 8/30 or 4/15 of the distance from 72.

Now switch these fractions, so 4/15 goes with 50% and 11/15 goes with 80%. This means that 4/15 of the 75 ounces must come from the 50% alloy and 11/15 of the 75 ounces must come from the 80% alloy.

4/15 x 75 = 20 ounces of 50%
11/15 x 75 = 55 ounces of 80%



I hope those help!! :-)

Answer 4

Hello

Let e = amt of 80% added
Let f = amt of 50% added

Then we have two equations

e+f = 75
80%e + 50%f = 72%*75

Solve the first one for e giving us e = 75-f and plug it into the second one

Since there are % on both sides we can divide all by 100 and eliminate them we have

80(75-f) + 50 f = 72*75
so f = 20 and e = 55



Hope This Helps!

Answer 5

.5X+.8y=.72*75=54

75=X+Y
75-Y=X

.5(75-Y) + .8Y = 54
Y= 55 is the 80% pure
X=20 is the 50% pure