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October 6th, 2014, 10:32 AM   #1
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Alligation or Mixture

Hello Friends,
Please help me in 3rd and 4th steps of the solution!!

Que-A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?

Sol-
Suppose the vessel initially contains 8 litres of liquid.

Let x litres of this liquid be replaced with water.
Quantity of water in new mixture =[3 -3x/8+ x] litres

Quantity of syrup in new mixture =[5 -5x/8+x] litres
3 -3x/8+ x =5 - 5x/8
5x + 24 = 40 - 5x
10x = 16
x=8/5
So, part of the mixture replaced -> 8/5x1/8=1/5

In this solution, please explain the 3rd and 4th step.
Quantity of water in new mixture =[3 -3x/8+ x] litres and
Quantity of syrup in new mixture =[5 -5x/8+x] litres


Or, can explain the another way to solve the same question as well.






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October 6th, 2014, 11:26 AM   #2
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Math Focus: Calculus/ODEs
As this is a question in algebra, and not an introduction, I have moved this thread appropriately.
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October 6th, 2014, 07:18 PM   #3
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Hello, saha.
Let the volume of the vessel = x litres.
As $\displaystyle \frac{3}{8}$th part of it is full of water, so volume of water = $\displaystyle \left(\frac{3}{8}\right)\cdot x$ litres.
As $\displaystyle \frac{5}{8}$th part of it is full of syrup, so volume of syrup = $\displaystyle \left(\frac{5}{8}\right)\cdot x$ litres.

Let us suppose that the volume of mixture drawn = y litres = volume of water added.
In y litres of mixture drawn, there is $\displaystyle \left(\frac{3}{8}\right)\cdot y$ litres of water and $\displaystyle \left(\frac{5}{8}\right)\cdot y$ litres of syrup.
So after drawing y litres of mixture, the mixture in the vessel contains:
$\displaystyle \left(\frac{3}{8}\right)\cdot x - \left(\frac{3}{8}\right)\cdot y$ = $\displaystyle \left(\frac{3}{8}\right)\cdot(x-y)$ litres of water and
$\displaystyle \left(\frac{5}{8}\right)\cdot x - \left(\frac{5}{8}\right)\cdot y$ = $\displaystyle \left(\frac{5}{8}\right)\cdot(x-y)$ litres of syrup.

Now, y litres of water is added.
So, the mixture in the vessel contains:

$\displaystyle \frac{3\cdot(x-y)}{8}+y = \frac{3x+5y}{8}$ litres of water and $\displaystyle \frac{5 \cdot (x-y)}{8}$ litres of syrup.

[Check: Total volume of mixture finally in the vessel =$\displaystyle \frac{5 \cdot(x-y)}{8}+\frac{3x+5y}{8}$ = x litres = volume of vessel. This happens because the volume of mixture taken out from the vessel has been substituted by an equal volume of water.]

$\displaystyle \frac{5\cdot(x-y)}{8}+\frac{3x+5y}{8}$ or $\displaystyle 5(x-y)=3x+5y$ or $\displaystyle 2x=10y$ or $\displaystyle y=\left(\frac{1}{5}\right)\cdot x$

Thus, volume of mixture to be taken out and subsequently replaced by water is $\displaystyle \frac{1}{5}$th part of the volume of the vessel.
In other words $\displaystyle \frac{1}{5}$th of the mixture in the vessel is to be replaced by water.
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