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October 6th, 2014, 11:32 AM  #1 
Newbie Joined: Oct 2014 From: usa Posts: 10 Thanks: 0  Alligation or Mixture Hello Friends, Please help me in 3rd and 4th steps of the solution!! QueA 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. 
October 6th, 2014, 12:26 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs 
As this is a question in algebra, and not an introduction, I have moved this thread appropriately.

October 6th, 2014, 08:18 PM  #3 
Senior Member Joined: Jul 2014 From: भारत Posts: 1,178 Thanks: 230  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: $\displaystyle \left(\frac{3}{8}\right)\cdot x  \left(\frac{3}{8}\right)\cdot y$ = $\displaystyle \left(\frac{3}{8}\right)\cdot(xy)$ litres of water and $\displaystyle \left(\frac{5}{8}\right)\cdot x  \left(\frac{5}{8}\right)\cdot y$ = $\displaystyle \left(\frac{5}{8}\right)\cdot(xy)$ litres of syrup. Now, y litres of water is added. So, the mixture in the vessel contains: $\displaystyle \frac{3\cdot(xy)}{8}+y = \frac{3x+5y}{8}$ litres of water and $\displaystyle \frac{5 \cdot (xy)}{8}$ litres of syrup. [Check: Total volume of mixture finally in the vessel =$\displaystyle \frac{5 \cdot(xy)}{8}+\frac{3x+5y}{8}$ = 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.] $\displaystyle \frac{5\cdot(xy)}{8}+\frac{3x+5y}{8}$ or $\displaystyle 5(xy)=3x+5y$ 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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