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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, 04:23
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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 28 Sep 2018, 03:58
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hibobotamuss wrote:
VeritasKarishma chetan2u pls can someone explain this more clearly?



Hi...

Since we are working for %, we can assume some value for the initial alloy..

So let's take it 100 and B:G:S is 90:7:3
Now another alloy, A, is added to it and gold becomes 5% BUT the gold is not added at all..
So 7 becomes 5% of new total . .......5% of total = 7.... total = 7*100/5=140
Thus new alloy,A, added = 140-100=40

Now let's concentrate on the bronze..
Bronze is 85% of new total = 140*85/100=119
So increase in bronze=119-90=29=quantity of bronze in new alloy,A.
So % of bronze in A=100*29/40=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 03 Sep 2018, 04:54
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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, 12:40
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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%


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.
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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, 12:37
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, 13:07
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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%


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, 01:05
Can we solve this question using the allegation method?
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Re: an alloy of gold,silverand bronze contain 90% bronze, 7% gold and  [#permalink]

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New post 28 Sep 2018, 02:19
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VeritasKarishma chetan2u pls can someone explain this more clearly?
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Re: an alloy of gold,silverand bronze contain 90% bronze, 7% gold and  [#permalink]

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New post 28 Sep 2018, 07:37
yarawehbe1 wrote:
Can we solve this question using the allegation method?


consider the gold in alloy b = 0 and equate the ratios you get
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Re: an alloy of gold,silverand bronze contain 90% bronze, 7% gold and  [#permalink]

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New post 30 Sep 2018, 01:27
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 the original alloy consist of 100 units, hence it has bronze as 90 units, gold as 7 units & silver as 3 units.

Let the second alloy consist of x units of bronze & y units of silver.

Hence new alloy will have total 100 + (x+y) units, which includes, (90+x) units of bronze, 7 units of gold & (3+y) units of silver.

Given that new alloy contains 85% bronze, 5% gold & 10% silver. Hence we have,

5% of [100 + (x+y)] = 7, we get (x+y) = 40 units

also, 85% of [100 + (x+y)] = 90+x

or (85/100)*140 = 90+x, solving which we get x = 29 units

Therefore % of bronze in second alloy = (29/40)*100 = 72.5%


Answer B.


Thanks,
GyM
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Re: an alloy of gold,silverand bronze contain 90% bronze, 7% gold and &nbs [#permalink] 30 Sep 2018, 01:27
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