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July 2nd, 2014, 03:08 AM   #1
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Mixture Problem

Q: Two vessels P and Q contains 'a' litres of petrol and 'b'litres of kerosine respectively. 'c' litres of petrol and same quantity of kerosine is taken out and transferred to Q and P respectively. The price litre become equal in both P and Q. What is the value of 'c' ? (Note the price of petrol and kerosine were different before the transfer )

Ans.The answer is always $c=\frac{ab}{a+b}$. My question why does this formula always works ? Are we equating money or are we equating quantity ? Why does the price per litre becomes same in both vessel ? What is the underlying logic ?
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July 2nd, 2014, 07:34 AM   #2
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Another formula that also works: Say a =140 and b=60 ,

Then, $\frac{(140X3)+(60X7)}{2(7+3)}$

Which is basically the same formula as above. So why does this formula works? Is there some proportions involved? Does both the vessels get same proportion of liquids ?
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July 2nd, 2014, 09:22 AM   #3
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Quote:
Originally Posted by Aman Verma View Post
The price litre become equal in both P and Q.
What do you mean by the "price litre"?
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July 3rd, 2014, 02:57 AM   #4
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Quote:
Originally Posted by Evgeny.Makarov View Post
What do you mean by the "price litre"?
It's Price per litre. Regret the typo !
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July 3rd, 2014, 03:57 AM   #5
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Let the prices of petrol and kerosene be $p$ and $k$, respectively. Then we have the following calculations.
\[
\begin{array}{r|c|c}&\text{Petrol} & \text{Kerosene}\\ \hline \text{Initial volume} & a & b\\ \text{Volume after taking out }c\text{ liters} & a-c & b-c\\ \text{Total price after taking out }c\text{ liters} & (a-c)p & (b-c)k\\ \text{Total price after putting in }c\text{ liters} & (a-c)p+ck & (b-c)k+cp\\ \text{Price per liter after putting in }c\text{ liters} & \dfrac{(a-c)p+ck}{a} & \dfrac{(b-c)k+cp}{b}\\\end{array}
\]
Equating the last two expressions and solving for $c$ gives $c=ab/(a+b)$.
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July 3rd, 2014, 04:12 AM   #6
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Thanks! That I do know how to derive. My question is that why this formula works ? Is it the weighted average ? Because this formula is similar to work/rate formula. Also one thing to be noted that the ratio of petrol and kerosine becomes same in both the vessels after the transfer. So is that the reason that the formula works? Is weighted average somehow related with the application of this formula ? Is it somehow related to the unitary method ? Has proportionality constant got anything to do with this formula ? How are the quantities 'a' and 'b' related to each-other that make this formula applicable?

Last edited by Aman Verma; July 3rd, 2014 at 04:17 AM.
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July 3rd, 2014, 04:30 AM   #7
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Quote:
Originally Posted by Aman Verma View Post
My question is that why this formula works ?
I am not sure what kind of explanation you want. My answer would be, because we derived it.

Quote:
Originally Posted by Aman Verma View Post
Also one thing to be noted that the ratio of petrol and kerosine becomes same in both the vessels after the transfer.
I don't agree. Why do you think so?

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How are the quantities 'a' and 'b' related to each-other that make this formula applicable?
$a$ and $b$ are not related; they are arbitrary positive values.
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July 3rd, 2014, 06:46 AM   #8
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Originally Posted by Evgeny.Makarov View Post

I don't agree. Why do you think so?
Check the values for a=140 and b=60. The value of c comes out to be 42.
Now, exchange 42 litres of petrol and kerosine between the two vessels as mentioned. Then check the ratio. Vessel 1. P:k=7:3 , Vessel 2, P:K=7:3. I think this has something to do with the fact why the fomula works. Also this might have something to do with weighted average because the formula resemble a lot with the weighted average and the proportion of a and b might might be equal to their weight in the value of c though I am not much sure. Advice needed in this regard from experts.
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July 3rd, 2014, 07:01 AM   #9
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Sorry, I calculated the ratio of petrol to kerosene in P and kerosene to petrol in Q; that's why they were not equal. Yes, the equation saying that the ratios after transfer are the same is
\[
\frac{a-c}{c}=\frac{c}{b-c}
\]
and it is equivalent to $c=ab/(a+b)$ for positive $a,b$.
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July 3rd, 2014, 08:27 AM   #10
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One last question: Has it got anything to do with weighted average ?
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