Please help with mixture word problem
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blamocur
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Is it?
Copper : (zinc and tin)
18 : 7
total proportional parts =25
[math]\text{one proportional parts} =\frac{660}{25}= 26.4[/math]weight of 660 kg of a new alloy containing equal weight of A and B = [math]26.4 \times 18= 475.2[/math]Am I right?
The Highlander
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No, I'm afraid you are not "right".
Firstly: "weight of 660 kg of a new alloy containing equal weight of A and B =" 660 kg not 475.2 (bananas?) though I believe you intended to write: "weight of copper in 660 kg of a new alloy containing equal weight of A and B = 475.2 kg" so we can forgive that little mistake.
But you cannot just combine the ratios of the two alloys in the fashion you did.
The correct way to approach this problem is to calculate the weight of Copper in each of the alloys (A & B) before they are combined to form the new alloy.
So, A has Cu:Zn in the ratio 7: 3 which means that 330 kg of A contain \(\displaystyle \frac{330}{10}\times 7 = 231\) kg of Copper.
While B has Cu:Sn in the ratio 11:4 which means that 330 kg of B contain \(\displaystyle \frac{330}{15}\times 11 = 242\) kg of Copper.
So how much Copper will there be in 660 kg of the new alloy (if equal weights of A & B have been combined to make it)?
Hope that helps.
I would attempt it the following way:
Assume the new alloy is 30 kg of A and 30 kg of B
From A we get 21 kg of copper and 9 kg of Zinc
From B we get 22 kg of copper and 8 kg of Zinc
So in 60 kg of new alloy 43 kg of Copper and 17 kg of Zinc
You decide!!
The Highlander
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Why change the quantities when I have already illustrated this using the specified weights?
Simpler arithmetic operations - without calculator.
Where did you get 330kg from or is an assumption? Because 330 kg is half of 660 kg. One meaning I can make from 330 kg you are using is that both alloy A and B are having equal weights as stated in problem.
Okay I now understand. 330 kg are equal weights of copper.
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The Highlander
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No, they are not "equal weights of copper". They are equal weights of the alloys A & B (both of which contain Copper in the ratios specified).
But, yes, the question specifies that equal weights of alloys A & B are used to make 660 kg of the new alloy and so 330 kg of each is used.
So you determine the Copper content of each those (330 kg of) alloys A & B first and then add those weights of Copper to get the amount of Copper present in (660 kg of) the new alloy.
I now understand. The total weights of copper in the new alloy is
[math]231 + 242 = 473~kg[/math] Thank you.
Sorry that is oversight statement.
Alternative method.
From the condition above since [imath]Cu[/imath] and [imath]Zn[/imath] are in ratio [imath]7:3[/imath], by weight in the first mixture, each kg of the first mixture contains [imath]7/10[/imath] kg of [imath]Cu[/imath]and [imath]3/10[/imath] kg of [imath]Zn[/imath]. Similarly, [imath]1[/imath] kg of the second mixture contains [imath]11/15[/imath] kg of [imath]Cu[/imath] and [imath]4/15[/imath] kg of [imath]Sn[/imath]. If we take [imath]x[/imath] kg of the first mixture and [imath]y[/imath] kg of the second and mix them, we get [imath]x+y[/imath] kg of the new mixture (alloy), which will contain [imath]7x/10+11y/15[/imath] kg of [imath]Cu[/imath], [imath]3x/10[/imath] kg of [imath]Zn[/imath] and [imath]4y/15[/imath] kg of [imath]Sn[/imath]. But we know that we take equal weight of each mixture. That means [imath]x=y=330[/imath] kg. From this, it follows that the new alloy contains [imath]473[/imath] kg of [imath]Cu[/imath].
From the condition above since [imath]Cu[/imath] and [imath]Zn[/imath] are in ratio [imath]7:3[/imath], by weight in the first mixture, each kg of the first mixture contains [imath]7/10[/imath] kg of [imath]Cu[/imath]and [imath]3/10[/imath] kg of [imath]Zn[/imath]. Similarly, [imath]1[/imath] kg of the second mixture contains [imath]11/15[/imath] kg of [imath]Cu[/imath] and [imath]4/15[/imath] kg of [imath]Sn[/imath]. If we take [imath]x[/imath] kg of the first mixture and [imath]y[/imath] kg of the second and mix them, we get [imath]x+y[/imath] kg of the new mixture (alloy), which will contain [imath]7x/10+11y/15[/imath] kg of [imath]Cu[/imath], [imath]3x/10[/imath] kg of [imath]Zn[/imath] and [imath]4y/15[/imath] kg of [imath]Sn[/imath]. But we know that we take equal weight of each mixture. That means [imath]x=y=330[/imath] kg. From this, it follows that the new alloy contains [imath]473[/imath] kg of [imath]Cu[/imath].
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