Submitted by: Submitted by leaapostol
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Category: Business and Industry
Date Submitted: 03/20/2011 02:29 AM
Mixture Word Problems:
Word problem is a part of algebra in which the problem is given in a statement form, from which the equations has to be figured out and then they have to be solved to achieve the answers. In mixture problems, a certain mixture of some concentration is combined with a certain mixture of other concentration asking to find out the the new mixture’s exact concentration. Word problems require some very good understanding ability because the equation has to be formed.
Example on Mixture Problems:
Ex 1 : Two solutions of 90% and 97% purity are mixed resulting in 21liters of mixture of 94% purity. How much is the quantity of the first solution in the resulting mixture?
Sol : Let x be the first solution of the two,
So, 90% of x +97 % of other solution gives 21litres of 94% new solution,
Therefore the equation is as follows,
x (0.9) + 0.97 (21-x) =21 (0.94)
Solving it for x 0.9x + 20.37 – 0.97x = 19.74
Dividing by 0.07 on both sides 0.07x = 0.63
X= 9
Ex 2: A mixture of 20ltrs of brandy and water contains 10% water. How much water should be added to it to increase the percentage of water to 25%?
Sol : 10 % of 20 liters is 2 liters
So 2 liters of water + 18 liters of brandy = 20 liters.
18 liters is 0.75
So, how much is? = 0.25,
18 (1/3) = 6,
Already there is 2 liters
So add 4 liters to it, 2+4 = 6 liters.
http://www.tutorvista.com/math/algebra-mixture-problems
Problem 1: How many liters of 20% alcohol solution should be added to 40 liters of a 50% alcohol solution to make a 30% solution?
Solution to Problem 1:
Let x be the quantity of the 20% alcohol solution to be added to the 40 liters of a 50% alcohol. Let y be the quantity of the final 30% solution. Hence
x + 40 = y
We shall now express mathematically that the quantity of alcohol in x liters plus the quantity of alcohol in the 40 liters is equal to the quantity of alcohol in y liters. But remember the alcohol is measured in percentage...