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an alloy of gold,silverand bronze contain 90% bronze, 7% gold and

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an alloy of gold,silverand bronze contain 90% bronze, 7% gold and  [#permalink]

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New post 03 Sep 2018, 05:23
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Question Stats:

44% (04:02) correct 56% (02:48) wrong based on 45 sessions

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An alloy of gold,silverand bronze contain 90% bronze, 7% gold and3% silver.a second alloy of bronze and silver only is melted with the first and mixture contain 85% of bronze,5% of gold,10% of silver find the percentage of bronze in second alloy?
a. 75%
b. 72.5%
c. 70%
d. 67.5%
e. 65%
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Re: an alloy of gold,silverand bronze contain 90% bronze, 7% gold and  [#permalink]

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New post 03 Sep 2018, 05:54
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pratik2018 wrote:
An alloy of gold,silverand bronze contain 90% bronze, 7% gold and3% silver.a second alloy of bronze and silver only is melted with the first and mixture contain 85% of bronze,5% of gold,10% of silver find the percentage of bronze in second alloy?
a. 75%
b. 72.5%
c. 70%
d. 67.5%
e. 65%
Answer is B imho
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Re: an alloy of gold,silverand bronze contain 90% bronze, 7% gold and  [#permalink]

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New post 03 Sep 2018, 13:40
2
pratik2018 wrote:
An alloy of gold,silverand bronze contain 90% bronze, 7% gold and3% silver.a second alloy of bronze and silver only is melted with the first and mixture contain 85% of bronze,5% of gold,10% of silver find the percentage of bronze in second alloy?
a. 75%
b. 72.5%
c. 70%
d. 67.5%
e. 65%


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.

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fskilnik.
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an alloy of gold,silverand bronze contain 90% bronze, 7% gold and  [#permalink]

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New post 04 Sep 2018, 13:37
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
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an alloy of gold,silverand bronze contain 90% bronze, 7% gold and  [#permalink]

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New post 04 Sep 2018, 14:07
pratik2018 wrote:
An alloy of gold,silverand bronze contain 90% bronze, 7% gold and3% silver.a second alloy of bronze and silver only is melted with the first and mixture contain 85% of bronze,5% of gold,10% of silver find the percentage of bronze in second alloy?
a. 75%
b. 72.5%
c. 70%
d. 67.5%
e. 65%


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
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Re: an alloy of gold,silverand bronze contain 90% bronze, 7% gold and  [#permalink]

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New post 13 Sep 2018, 02:05
Can we solve this question using the allegation method?
Re: an alloy of gold,silverand bronze contain 90% bronze, 7% gold and &nbs [#permalink] 13 Sep 2018, 02:05
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