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Suppose there are three qualities of rice, A(1 dollar per Kg), b(2 dollar per Kg) and C(3 dollar per Kg). The salesmen want to mix these in a certain ratio a:b:c so as to make the price 2.5 dollar per kg.

How can we find the ratio by using methods of alligation and mixture?

A reference to the proof(or the proof itself if possible) of the method would also be helpful.

P.S: The formula for two types is $$ \dfrac{\text{Amount of A}}{\text{Amount of B}}=\dfrac{P_M-P_B}{P_A-P_M} $$ Where $P_M=$mean price, $P_A=$ Price of $A$ and $P_B=$ price of $B$.

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    $\begingroup$ There are infinitely many ratios among three items that will give the final value you want, at least if the final value is between the minimum and maximum of the three items. (Two items have only one ratio, but more than two changes that.) Do you want a parametrization of all of them? $\endgroup$ – Rory Daulton Jul 28 '15 at 20:01
  • $\begingroup$ @RoryDaulton In my book they are finding a unique solution with some trick. First they find ration of first and third varity of rice, and then ratio of 1st and 2nd variety and then somehow find a:b:c. $\endgroup$ – user103816 Jul 29 '15 at 8:08
  • $\begingroup$ If their are infinitely many solutions then I'd like to know how. $\endgroup$ – user103816 Jul 29 '15 at 8:09
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I am not familiar with the process of alligation, but here is how I would solve this problem. From what I can tell, this method is, according to Wikipedia, "A general formula that works for both alligation 'alternate' and alligation 'medial'".

Let $a$ be the number of kilograms of quality A rice at USD $1$ per kg, $b$ be the number of kilograms of quality B rice at USD $2$ per kg, and $c$ be the number of kilograms of quality C rice at USD $3$ per kg. Then

$$\text{Total Cost}=1a+2b+3c$$ $$\text{Total Mass}=a+b+c$$

We want

$$\text{Price}=\frac{\text{Total Cost}}{\text{Total Mass}}=2.50$$ $$\frac{a+2b+3c}{a+b+c}=\frac 52$$ $$2a+4b+6c=5a+5b+5c$$ $$c=3a+b$$

Since you want a ratio, not actual amounts, we'll just set $a=1$. This still does not decide the value of $b$, so we'll also set $b=r$ where $r$ is any arbitrary non-negative real number. Then the ratio of rice qualities A:B:C that gets us a price of USD $2.50$ is

$$1:r:(r+3)$$

This is easily checked. With those values,

$$\text{Total Cost}=1a+2b+3c=1\cdot 1+2\cdot r+3\cdot(r+3)=5r+10=5(r+2)$$ $$\text{Total Mass}=a+b+c=1+r+(r+3)=2r+4=2(r+2)$$ $$\text{Price}=\frac{5(r+2)}{2(r+2)}=\frac 52=2.50$$

You can see that there are infinitely many ratios of the three kinds of rice that will give you the desired price. This will be true in general for a mixture of three kinds of goods, as long as the final price is somewhere between the maximum and the minimum price of the three kinds of goods. (If the desired price is out of that range, we will get a negative number for at least one of the kinds of goods, which is unrealistic.)

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  • $\begingroup$ We have 3 variables and only one equation. This is why we are getting infinitely many solutions. But in case of two kinds of goods we get a specific solution -- this might be because we only want the ratio. $\endgroup$ – user103816 Jul 29 '15 at 11:36
  • $\begingroup$ @user103816: Yes, that is exactly correct. The "ratio" removes one of the dimensions in the answer, but only one. $\endgroup$ – Rory Daulton Jul 29 '15 at 11:37
  • $\begingroup$ My book is doing something quite different. Their prices are a=1.20, b=1.44 and c=1.74. The desired mean price is 1.41 -- all in dollar per Kg. They first picked type_A and type_C and then found their ratio which would make the average price of A-and-C=1.41. Then they pick typeA and typeB and found the ratio which makes price of A-and-B 1.41. At last they said (b/a)*(a/c)=b/c and found the ratio to be 11:77:7. $\endgroup$ – user103816 Jul 29 '15 at 11:45
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    $\begingroup$ @user103816: When I use those values, I get a final price of about USD $1.4343157895$, not $1.41$. In a moment I'll get my own answer to that problem. $\endgroup$ – Rory Daulton Jul 29 '15 at 11:49
  • $\begingroup$ @user103816: For that problem I get the ratios $$\frac{r+11}{7}:r:1$$ and that checks. Letting $r=11$ gives the ration $22:77:7$. Your book seems to be wrong. Either that or you mistyped the problem here. $\endgroup$ – Rory Daulton Jul 29 '15 at 11:56

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