My Math Forum Alligation or Mixture

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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!! 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.
 October 6th, 2014, 12:26 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,205 Thanks: 512 Math Focus: Calculus/ODEs As this is a question in algebra, and not an introduction, I have moved this thread appropriately. Thanks from saha
 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: $\displaystyle \left(\frac{3}{8}\right)\cdot x - \left(\frac{3}{8}\right)\cdot y$ = $\displaystyle \left(\frac{3}{8}\right)\cdot(x-y)$ 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. Thanks from saha

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