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Help requested for mixture question

by cyrusthegreat » Tue Apr 13, 2010 2:30 pm
An alloy contains copper and zinc in the ratio of 5:3 and another contains copper and tin in the ratio of 8:5. If equal weights of the two are melted together to form a new alloy, find the weight of tin per kg in the new alloy.

A. 42/126 Kg

B. 5/28 Kg

C. 5/13 Kg

D. 5/26Kg

E. None of these


I came up with 25/129 and so chose e but it was incorrect.

This is how I got to my answer:

40c:24z 40c:25t 40+24+40+25 = 129 t=25 25/129 equals the proportion of a kg that is tin if all parts are melted down and recast as a new alloy.

Anyone know where I went wrong?

Thanks in advance
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by kevincanspain » Tue Apr 13, 2010 2:45 pm
Alloy 1
Copper = 5x Zinc =3x
Total 8x


Alloy 2
Copper = 8y, Tin =5y
Total 13y =8x (the weights of the alloys are equal)


Total weight 26y, of which 5y is tin: Answer: D

You mistakenly equated the weights of copper in the two alloys
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by cyrusthegreat » Tue Apr 13, 2010 3:25 pm
Thanks Kevin. I have another one that I'd appreciate some help with as well:

Two liquids are mixed in the proportion of 3:2 and the mixture is sold at $11 per litre at a 10% profit. If the 1st liquid costs $2 more per litre than the second, what does it cost per litre?

A. $11

B. $10.80

C. $10.40

D. $11.75

Thanks in advance
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by cyrusthegreat » Tue Apr 13, 2010 3:27 pm
Does anyone know at what level of difficulty these 2 questions would be considered?
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by dxgamez » Tue Apr 13, 2010 5:01 pm
Mixture original price would be at $10, as it is sold at 10% profit, which costs $11.

Let cost of 1st liquid be $x per litre and cost of 2nd liquid be $y per litre

$x = $y + $2

(3/5)*(x) + (2/5)*(y) = $10 --> everything is $ per litre and the proportions are stated in the qn.

sub x = y + 2 into above

(3/5)*(y+2) + (2/5)*y = 10 and multiplying both sides by 5 gives:-

3(y+2) + 2y = 50
y = 44/5 = $8.8 per litre

thus, x = $8.8 + $2 = $10.80...Answer is B.

Kevin, your take pls, for a clearer explanation?
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by cyrusthegreat » Tue Apr 13, 2010 6:30 pm
Thanks dxgames. Anyone know the difficulty level of these questions?
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by eaakbari » Wed Apr 14, 2010 10:41 am
kevincanspain wrote:Alloy 1
Copper = 5x Zinc =3x
Total 8x


Alloy 2
Copper = 8y, Tin =5y
Total 13y =8x (the weights of the alloys are equal)


Total weight 26y, of which 5y is tin: Answer: D

You mistakenly equated the weights of copper in the two alloys
Hey Kevin

Dont you mean 21 y in the bold part as 13 + 8 = 21
I got the answer as 5/21
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by harshavardhanc » Wed Apr 14, 2010 11:09 am
eaakbari wrote:
kevincanspain wrote:Alloy 1
Copper = 5x Zinc =3x
Total 8x


Alloy 2
Copper = 8y, Tin =5y
Total 13y =8x (the weights of the alloys are equal)


Total weight 26y, of which 5y is tin: Answer: D

You mistakenly equated the weights of copper in the two alloys
Hey Kevin

Dont you mean 21 y in the bold part as 13 + 8 = 21
I got the answer as 5/21
it will be 5/26.

See it this way : you take 1 Kg each of both the mixtures.

We have (5/13) * 1 kg of tin in 2 Kgs of the total mixture. so tin is (5/26) per Kg.
Regards,
Harsha
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