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Question 1170861: Hello i need ASAP help can someone please answer
Steve is a jeweller who wants to create rings of a certain gold purity. 24-kt (karat) is considered
pure gold. Steve has a stock of 18-kt and 10-kt on hand. He wants to have 200g of 16-kt gold. How much
18-kt and 10-kt does he need to have a 200g sample?
thank you
Found 3 solutions by math_tutor2020, josgarithmetic, greenestamps:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
kt = karat
24 kt = 100% pure gold
18 kt = (18/24)*100% = (0.75)*100% = 75%
This indicates the 18 kt stock is 75% pure gold
Similarly,
10 kt = (10/24)*100% = (0.41667)*100% = 41.667%
The 10 kt stock is roughly 41.667% pure gold.
And,
16 kt = (16/24)*100% = 66.667%
The 16 kt stock is roughly 66.667% pure gold
I'll use the fractional form of each decimal value to avoid rounding errors.
In other words,
18 kt = (18/24) of pure gold = (3/4) of pure gold
10 kt = (10/24) of pure gold = (5/12) of pure gold
16 kt = (16/24) of pure gold = (2/3) of pure gold
Fractions can be a pain to work with, but they provide exact values.
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Let
x = amount of the 18 kt gold stock
y = amount of the 10 kt gold stock
amounts are in grams
If steve has x grams of 18 kt gold, then he has (3/4)x grams of pure gold. Recall that in the previous section 3/4 = 75% was the amount of pure gold from this stock. Effectively we have (3/4)x grams of pure gold (24 kt gold) mixed in with some other metal, likely copper or iron perhaps. This means the other 1/4 is the other metal.
If steve has y grams of 10 kt gold, then he has (10/24)y grams of pure gold. The fraction 10/24 reduces to 5/12. So (10/24)y = (5/12)y
The two amounts of pure gold add to
(3/4)x+(5/12)y
Steve wants to have 200 grams of 16 kt gold, so he wants 200*(2/3) = 400/3 grams of pure gold.
The 2/3 is from reducing 16/24.
Set the result 400/3 equal to the previous sum we got, and we end up with this equation
(3/4)x+(5/12)y = 400/3
This equation represents summing the pure gold amounts of each stock, getting a final amount of pure gold for the mixture.
To make life easier, multiply both sides by the LCD 12 to clear out the fractions
(3/4)x+(5/12)y = 400/3
12 * [(3/4)x+(5/12)y] = 12*400/3
9x+5y = 1600
Now everything is a whole number which is fairly nice.
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Since he wants the final amount to be 200 grams, this must mean the subtotals of x and y have to add to 200
We can then solve for y getting
x+y = 200
x+y-x = 200-x
y = 200-x
Plug this into the equation we found at the conclusion of the last section. Then solve for x.
9x+5y = 1600
9x+5(y) = 1600
9x+5(200-x) = 1600
9x+1000-5x = 1600
4x+1000 = 1600
4x+1000-1000 = 1600-1000
4x = 600
4x/4 = 600/4
x = 150
He needs 150 grams of the 18 kt gold stock.
Use this to find y
y = 200-x
y = 200-150
y = 50
He needs 50 grams of the 10 kt gold stock
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Answers:
150 grams of the 18 kt gold stock.
50 grams of the 10 kt gold stock.
Answer by josgarithmetic(39618) (Show Source):
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
You have received, to this point, two formal algebraic solutions. One uses a typical approach and gets to an answer relatively quickly. The other uses a very roundabout approach, eventually reaching an answer.
If a formal algebraic solution method is not required, here is the fastest and easiest way to get the answer.
You are mixing 10-kt and 18-kt gold to get 16-kt gold.
Consider the three numbers on a number line. 16 is 3/4 of the way from 10 to 18.
That means 3/4 of the mixture needs to be the 18-kt gold.
ANSWER: 3/4 of 200g, or 150g, is the 18-kt gold; the other 50g is the 10-kt gold.
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