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A Good Mixture Problem

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by Ozlemg » Thu Jun 02, 2011 7:34 am
There are 2 bars of gold-silver alloy ;one pice has 2 parts of gold to 3 parts of silver and another has 3 parts of gold to 7 parts of silver. If both bars are melted into 8 kg bar with the final gold to silver ratio of 5:11. What was the weight of the first bar?

A.1 kg
B.3 kg
C.5 kg
D.6 kg
E.7 kg

OA is A
Last edited by Ozlemg on Thu Jun 02, 2011 8:10 am, edited 1 time in total.
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by Frankenstein » Thu Jun 02, 2011 7:41 am
Hi,
Fraction of gold in bar1 = 2/5
Fraction of gold in bar2 = 3/10
Fraction of gold in alloy = 5/16
Using the principle of allegations, bar1/bar2 = (5/16-3/10)/(2/5-5/16) = 1/7
Total weight of alloy is 8 kg.
So, weight of bar1 = 1kg
Hence, A
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by SoCan » Thu Jun 02, 2011 7:44 am
First gold bar is 4/10 gold, the second bar is 3/10 gold. We want to find the weights that give us an 8 pound bar with 5/16 gold.

(4/10)x + (3/10)(8-x) = (5/16)*8
x/10 + 24/10 = 5/2
x/10 = 1/10
x = 1
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by SoCan » Thu Jun 02, 2011 7:46 am
Frankenstein wrote:Hi,
Fraction of gold in bar1 = 2/5
Fraction of gold in bar2 = 3/10
Fraction of gold in alloy = 5/16
Using the principle of allegations, bar1/bar2 = (5/16-3/10)/(2/5-5/16) = 1/7
Total weight of alloy is 8 kg.
So, weight of bar1 = 1kg
Hence, A
I've never used alligation before in this way, but it seems to be quicker than algebra in many cases. I'll look into it for myself.
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by Frankenstein » Thu Jun 02, 2011 8:13 am
SoCan wrote:
Frankenstein wrote:Hi,
Fraction of gold in bar1 = 2/5
Fraction of gold in bar2 = 3/10
Fraction of gold in alloy = 5/16
Using the principle of allegations, bar1/bar2 = (5/16-3/10)/(2/5-5/16) = 1/7
Total weight of alloy is 8 kg.
So, weight of bar1 = 1kg
Hence, A
I've never used alligation before in this way, but it seems to be quicker than algebra in many cases. I'll look into it for myself.
Yes.. It is certainly quicker in many cases but your method is concise too in this case.
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by g.shankaran » Thu Jun 02, 2011 8:35 am
Frankenstein wrote:Hi,
Fraction of gold in bar1 = 2/5
Fraction of gold in bar2 = 3/10
Fraction of gold in alloy = 5/16
Using the principle of allegations, bar1/bar2 = (5/16-3/10)/(2/5-5/16) = 1/7
Total weight of alloy is 8 kg.
So, weight of bar1 = 1kg
Hence, A
If you don't mind can u explain a bit about the principle of allegations?
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by Frankenstein » Thu Jun 02, 2011 8:47 am
g.shankaran wrote:
Frankenstein wrote:Hi,
Fraction of gold in bar1 = 2/5
Fraction of gold in bar2 = 3/10
Fraction of gold in alloy = 5/16
Using the principle of allegations, bar1/bar2 = (5/16-3/10)/(2/5-5/16) = 1/7
Total weight of alloy is 8 kg.
So, weight of bar1 = 1kg
Hence, A
If you don't mind can u explain a bit about the principle of allegations?
Hi,
I have been typing allegations. It is actually alligations. Will rectify this typo hereafter
In a mixture of constituents 1 and 2, the ratio of the quantities of the two constituents is given by the following formula:
Q1/Q2 = (c2-c)/(c-c1)
where Q1,Q2 are the quantities of the constituents 1 and 2 respectively
c1,c2 are the concentrations of the constituents 1 and 2 respectively
c is the concentration of the mixture.
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by cans » Thu Jun 02, 2011 9:12 am
x be the weight of first bar. then (8-x) is weight of 2nd bar.
total gold = 2x/5 + 3*(8-x)/10 = (x+24)/10
total silver = 3x/5 + 7*(8-x)/10 = (56-x)/10
ratio of gold:silver = (x+24)/(56-x) = 5/11
11x + 264 = 280 - 5x
x=1
IMO A
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