An alloy of gold,silverand bronze contain 90% bronze, 7% gold and 3%

Without loss of generality (to be understood later) we may assume we have 100 grams of the first alloy, therefore \(\,\,100\,\,{\text{g}}\,\,\,\left\{ \begin{gathered}\\
\,\boxed{90\,\,{\text{g}}\,\,{\text{bronze}}} \hfill \\\\
\,7\,\,{\text{g}}\,\,{\text{gold}} \hfill \\\\
\,3\,\,{\text{g}}\,\,{\text{silver}} \hfill \\ \\
\end{gathered} \right.\,\,\)

The second alloy does not have gold, and when the alloys are combined, we have 5% gold, therefore we may conclude (in grams of gold) that:

\(\frac{5}{{100}}\left( {{\text{total}}\,\,{\text{combined}}} \right)\,\, = \,\,\,7\,\,\,\,\,\, \Rightarrow \,\,\,\,{\text{total}}\,\,{\text{combined}} = \,\,\,140\,\,{\text{g}}\,\,\left\{ \begin{gathered}\\
\,\,\left( {\frac{{85}}{{100}}} \right)140 = \boxed{119\,\,{\text{g}}\,\,{\text{bronze}}}\,\, \hfill \\\\
\,\,\left( {\frac{5}{{100}}} \right)140\,\,{\text{g}}\,\,{\text{gold}} \hfill \\\\
\,\,\left( {\frac{{10}}{{100}}} \right)140\,\,{\text{g}}\,\,{\text{silver}} \hfill \\ \\
\end{gathered} \right.\,\,\,\,\,\,{\text{AND}}\,\,\,\,\,\,\,\)

\({\text{second}}\,\,{\text{alloy}} = \,\,\,40\,\,{\text{g}}\,\,\,\left\{ \begin{gathered}\\
\boxed{x \cdot 40\,\,{\text{g}}\,\,{\text{bronze}}} \hfill \\\\
\left( {1 - x} \right) \cdot 40\,\,{\text{g}}\,\,{\text{silver}} \hfill \\ \\
\end{gathered} \right.\,\,\,\,\,\,\,\,\,{\text{where}}\,\,\,\,? = x\,\,\,\left( {0 < x < 1} \right)\)

The equation (in grams of bronze) obtained using the "frames" presented above ends our solution:

\(90 + x \cdot 40 = 119\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = x = \left( {\frac{{29}}{{40}}} \right) \cdot 100\% = \,\,\underleftrightarrow {\frac{5}{2}\left( {28 + 1} \right)\% } = \left( {70 + 2.5} \right)\%\)


The above follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.

let A and B=first and second alloy weights respectively
x=% of bronze in alloy B
because B contains no gold,
.07A=.05(A+B)➡
B=.4A
substituting,
.9A+.4Ax=.85(A+.4A)➡
x=.725=72.5%
B

Not sure if the method is correct, but answer is right. Took long time to arrive on it

Mixture 1: Assume x gram totally, B=0.9x G=0.07x S=0.03x
Mixture 2: Assume y gram totally, B=ay S=by (a is what we need to find)
Mixture 3: Mixture 1 + 2: Total x + y gram, B=0.85(x+y) S=0.1(x+y) G=0.05(x+y)

See gold is not present in mixture 2, so gold in mixture 1 and 3 should be same
0.07x = 0.05(x+y)
x=5y/2 --> (1)

Now we need to find the amount of bronze in mixture 2:
Mixture 1 bronze + Mixture 2 bronze = Mixture 3 bronze
0.9x + ay = 0.85(x + y)
0.9x - 0.85x = 0.85y - ay
0.05x = (0.85 - a)y
sub eq (1)
(0.85 - a)y = 0.05 * 5y/2
a = 1.45/2 = 0.725

a is 72.5%

Answer: B

VeritasKarishma chetan2u pls can someone explain this more clearly?

consider the gold in alloy b = 0 and equate the ratios you get

W1 ---> weight of alloy :- gold, silver and bronze contain \(90\)% bronze, \(7\)% gold and \(3\)% silver
W2 ---> weight of alloy :- a second alloy of bronze and silver only (\(0\)% Gold)

Mixture becomes ---> \(85\)% of bronze, \(5\)% of gold, \(10\)% of silver

\(\frac{W1}{W2}\) = \(\frac{0-5}{5-7}\) =\( \frac{5}{2}\)

\(\frac{W1}{W2}\) = \(\frac{b-85}{85-90}\) = \(\frac{b-85}{-5}\)

\(\frac{5}{2}\) =\( \frac{b-85}{-5}\)

Solve for \(b = 72.5\)

Hello!

The way the question is phrased is a little confusing. Here it is rewritten a little more clearly with the same meaning

An alloy of gold, silver, and bronze contain 90% bronze, 7% gold and 3% silver. A second alloy consisting only of bronze and silver is melted with the first alloy and the new mixture contains 85% bronze, 5% gold, and 10% silver. What was the percentage of bronze in the second alloy?

a. 75%
b. 72.5%
c. 70%
d. 67.5%
e. 65%


When a number is unknown, such as the total weight of the first alloy, and especially when working with percents, it is very helpful to use numbers that are easy to work with, like 100.

Let's say the first alloy had a mass of 100 grams. Using the percents given to us, we know therefore that in the first alloy was 90g of bronze, 7g of gold, and 3g of silver.

We are told that the second alloy contains no gold, yet the resulting mixture of the two alloys is 5% gold.

The total amount of gold in the new mixture cannot increase as there was no gold in the second alloy, so it is still the same 7 grams of gold there was in the first alloy.

We know therefore that the new mixture contains 7g of gold, and that that is 5% of its total mass

To find the total mass of the new mixture, set up an equation

\(\frac{5}{100} = \frac{7}{x}\\
\)

5x = 700

x = 140

The new mixture has a mass of 140 grams.

That means that the second alloy had a total mass of 40 grams, since we said that the first one had a mass of 100 grams.

100 + 40 = 140

This new mixture is 10% silver. 10% of 140g = 14g

There are 14 grams of silver in this new mixture, and there were 3 grams of silver in the first alloy.

14 - 3 = 11

That means the second alloy contained 11 grams of silver.

If the total mass of the second alloy was 40 grams and 11 grams of it was silver, the rest must have been bronze

40 - 11 = 29

There were 29 grams of bronze in the second alloy

The question wants to know what percent of the second alloy was bronze. Again we set up an equation

\(\frac{29}{40} = \frac{x}{100}\)

2900 = 40x

Chop off a 0 from both sides (equivalent to dividing both sides of the equation by 10)

290 = 4x

With some quick long division 290/4 = 72.5

The answer is (B), 72.5% of the second alloy was bronze

Can we solve this question using the allegation method?

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