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# How many grams of pure iron must be alloyed with 387 grams of pure cop

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How many grams of pure iron must be alloyed with 387 grams of pure cop  [#permalink]

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25 Jan 2017, 05:56
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Difficulty:

15% (low)

Question Stats:

80% (01:29) correct 20% (01:31) wrong based on 146 sessions

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How many grams of pure iron must be alloyed with 387 grams of pure copper to create an alloy that is 90% copper?

A. 38.7
B. 43
C. 129
D. 301
E. 430

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Re: How many grams of pure iron must be alloyed with 387 grams of pure cop  [#permalink]

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25 Jan 2017, 06:13
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4
Bunuel wrote:
How many grams of pure iron must be alloyed with 387 grams of pure copper to create an alloy that is 90% copper?

A. 38.7
B. 43
C. 129
D. 301
E. 430

387 grams of pure copper should serve the purpose of 90% of entire alloy

i.e. 387 = (90/100)*Total Alloy weight

i.e. Total Alloy Weight = 430 Grams

Amount of Iron to be added = 430 - 387 = 43 grams

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Re: How many grams of pure iron must be alloyed with 387 grams of pure cop  [#permalink]

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25 Jan 2017, 06:23
Bunuel wrote:
How many grams of pure iron must be alloyed with 387 grams of pure copper to create an alloy that is 90% copper?

A. 38.7
B. 43
C. 129
D. 301
E. 430

Hi,

We need option that gives 10% iron + 90% Copper
So by observation, you can eliminate C, D, & E
Now Coming to option A and B

Testing Option A
387+38.7=425.7(Total alloy)
Here if you subtract 425.7-42.5=383.15(less than 90%)

Hence option B
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Re: How many grams of pure iron must be alloyed with 387 grams of pure cop  [#permalink]

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25 Jan 2017, 10:25
@vertasprepkarishma

any way to get this done by weighted avg method
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Re: How many grams of pure iron must be alloyed with 387 grams of pure cop  [#permalink]

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26 Jan 2017, 11:32
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mohshu wrote:
@vertasprepkarishma

any way to get this done by weighted avg method

Hi Mohshu,

The algebraic approach is the best way to solve this problem as it requires a basic understand of percentages. If we keep the the weight of iron as 'x' grams, the total alloy will weigh '387+x' grams. Since copper makes up 90% of the alloy,

387/(387+x) = 90/100

Simplifying we get x = 43grams

The alternate approach is to make use of the alligation approach. To use the alligation approach there are two things that you need to do.

1. Draw the alligation diagram the basic framework of which is given below

Attachment:

Mixtures 1.png [ 8.12 KiB | Viewed 1188 times ]

2. Make sure that the values you substitute as the higher value, lower value and mean value are values associated with the word 'per' (percentages, ratios, averages...)

Now in this question we can form an alligation diagram using the percentages of Copper, we get the higher value as 100% and the lower value as 0%, since we have pure copper and pure iron respectively (the percentage of copper in pure iron is 0%)

Attachment:

Untitled.png [ 7.48 KiB | Viewed 1190 times ]

Since the ratio of copper to iron in the alloy is 9 : 1 and the weight of copper is 387 grams, then this 387 grams represents 9 parts of the ratio. So the weight of iron which is represented by 1 part in the ratio will be 387/9 = 43grams

Hope this helps!

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Re: How many grams of pure iron must be alloyed with 387 grams of pure cop  [#permalink]

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27 Jan 2017, 01:15
When looking at this problem, significant amount of the alloy is copper. So we can eliminate choices C, D & E.

When applying allegation method

Iron--------------Alloy---------Copper
0%----90-------90%---10---100%

C/I= 90/10=9............I=387/9..Do not need to divide.. 387 is divisible by 9 So only one answer fits.

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Re: How many grams of pure iron must be alloyed with 387 grams of pure cop  [#permalink]

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27 Jan 2017, 09:46
Bunuel wrote:
How many grams of pure iron must be alloyed with 387 grams of pure copper to create an alloy that is 90% copper?

A. 38.7
B. 43
C. 129
D. 301
E. 430

We need to create an alloy that is 90% copper. In other words:

copper/(iron + copper) = 90%

We can let i = the amount of pure iron in grams and create the following equation:

387/(i + 387) = 9/10

387(10) = 9(i + 387)

3870 = 9i + 3483

387 = 9i

i = 43

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How many grams of pure iron must be alloyed with 387 grams of pure cop  [#permalink]

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03 May 2017, 22:33
Let's say x grams of pure iron is added to form alloy.

New weight is (387+x).

The % of Iron in alloy = 10% = $$\frac{10}{100}$$

=> $$\frac{x}{(387+x)}$$ =$$\frac{10}{100}$$

=> 10x = 387 + x

=> x=$$\frac{387}{9}$$

=> x=43

Ans: B
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Re: How many grams of pure iron must be alloyed with 387 grams of pure cop  [#permalink]

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04 May 2017, 11:36
b
Bunuel wrote:
How many grams of pure iron must be alloyed with 387 grams of pure copper to create an alloy that is 90% copper?

A. 38.7
B. 43
C. 129
D. 301
E. 430

(387/.9)-387=43
B
Re: How many grams of pure iron must be alloyed with 387 grams of pure cop &nbs [#permalink] 04 May 2017, 11:36
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