Word problem solving in Algebra: Mixture Problem
Mixture problem is one of the hardest among word problem. Unlike distance problem, you can't illustrate what happens to the mixture. We can draw beakers, flasks, etc but we don't know happen to the liquids. So this type of problem needs more practice as exposure to its nature. Usually percents of concentration and prices appears in mixture problems. How to deal with this is almost the same process. This is a bit complicated than other worded problem but we can handle the manner it goes.
Illustrative Example:
1. When we say " 25 quarts of a mixture containing 17% alcohol" means: 25(0.17) = 4.25 quarts pure alcohol.
2. How many percent of alcohol would a three quarts pure alcohol mixed into a mixture that produces 20 quarts? So;
Example problem 1:
"A solution containing 4% boric acid is to be mixed into a 1 quart mixture that is 17% boric acid to obtain 20% boric acid solution. Compute how much of the 4% solution is to be used"
Dealing with this problem need understand well what was happening because there's a confusion. A lot of percentages appears. Sometimes we don't know what and where to use. Remember that based on the problem we have three solutions; first mixture + second mixture = total mixture. Write it down.
So;
x = is the unknown solution.
first mixture = 4% = 0.04
second mixture = 17% =0.17
total mixture =20% = 0.20
Then:
Given: Boric acid, 4%, 1 quart mixture that is 17% boric acid, 20% boric acid solution
Required: Amount of 4% solution, x
Solution:
Let:
x = is the unknown solution.
first mixture + second mixture = total mixture
Do the addition:
first mixture + second mixture = total mixture (substitute their equivalents)
0.04 + 0.17 = 0.20
0.04 x + "1 quart mixture that is 0.17" = 0.20
0.04 x + 1(.17) = 0.20(x + 1)
0.04 x = (0.20 x + 0.20)- 1(.17)
0.04 x = 0.20 x + 0.20 - 0.17
0.04 x - 0.20 x = 0.03
0.02 x = 0.03
x = 1.5
Example problem 2:
"To obtain 500 gallons of milk, that is 5% butterfat, a mixture of milk (containing 2% butter fat ) and a cream (containing 20% butterfat) was mixed. How much of each is to be used?"
Given: x gal., 2%,
(500-x), 20%
Required: Amount of x and (500-x).
Solution: mix milk and butter
Let:
x = is the unknown solution.
face
Mixture problem is one of the hardest among word problem. Unlike distance problem, you can't illustrate what happens to the mixture. We can draw beakers, flasks, etc but we don't know happen to the liquids. So this type of problem needs more practice as exposure to its nature. Usually percents of concentration and prices appears in mixture problems. How to deal with this is almost the same process. This is a bit complicated than other worded problem but we can handle the manner it goes.
Illustrative Example:
1. When we say " 25 quarts of a mixture containing 17% alcohol" means: 25(0.17) = 4.25 quarts pure alcohol.
2. How many percent of alcohol would a three quarts pure alcohol mixed into a mixture that produces 20 quarts? So;
Example problem 1:
"A solution containing 4% boric acid is to be mixed into a 1 quart mixture that is 17% boric acid to obtain 20% boric acid solution. Compute how much of the 4% solution is to be used"
Dealing with this problem need understand well what was happening because there's a confusion. A lot of percentages appears. Sometimes we don't know what and where to use. Remember that based on the problem we have three solutions; first mixture + second mixture = total mixture. Write it down.
So;
x = is the unknown solution.
first mixture = 4% = 0.04
second mixture = 17% =0.17
total mixture =20% = 0.20
Then:
Given: Boric acid, 4%, 1 quart mixture that is 17% boric acid, 20% boric acid solution
Required: Amount of 4% solution, x
Solution:
Let:
x = is the unknown solution.
first mixture + second mixture = total mixture
Do the addition:
first mixture + second mixture = total mixture (substitute their equivalents)
0.04 + 0.17 = 0.20
0.04 x + "1 quart mixture that is 0.17" = 0.20
0.04 x + 1(.17) = 0.20(x + 1)
0.04 x = (0.20 x + 0.20)- 1(.17)
0.04 x = 0.20 x + 0.20 - 0.17
0.04 x - 0.20 x = 0.03
0.02 x = 0.03
x = 1.5
Example problem 2:
"To obtain 500 gallons of milk, that is 5% butterfat, a mixture of milk (containing 2% butter fat ) and a cream (containing 20% butterfat) was mixed. How much of each is to be used?"
Given: x gal., 2%,
(500-x), 20%
Required: Amount of x and (500-x).
Solution: mix milk and butter
Let:
x = is the unknown solution.
2% milk + 20%butterfat = 5%(500)
0.02 gal milk(x)+ 0.2(500 - x) = 0.05(500)
0.02 x + 0.2 (500 -x) = 0.05(500)
0.02 x + 100 - 0.2 x = 25
-0.18 x = 25 - 100
-0.18 x = -75
x =416.66, gallons
500 -x = 500 - 416.66 = 83.34, gallons
Example problem 3:
" A chemist wants to make an 80% alcohol solution. What is the amount of pure alcohol must she add to a 15 cc of a 65% alcohol?"
Given: x = volume of pure alcohol to be added (note that pure alcohol is 100%)
Required: Amount of x
Solution: add x to a 15cc of 65% alcohol
" A chemist wants to make an 80% alcohol solution. What is the amount of pure alcohol must she add to a 15 cc of a 65% alcohol?"
Given: x = volume of pure alcohol to be added (note that pure alcohol is 100%)
Required: Amount of x
Solution: add x to a 15cc of 65% alcohol
100% x + 65%(15cc) = 80%(x + 15)
1x + 0.65(15) = 0.8x + 12
x + 9.75 = 0.8x + 12
x - 0.8x = 12 - 9.75
0.2x = 2.25
x = 11.25 cc
Pop Quiz:
1. A 20% alloy is to be mixed in an alloy containing 60% silver to produce 600 lbs. of 35% alloy. What is the amount each is to be used?
Partial solution:
Let x = # of lbs. of 20% alloy
600-x = # of lbs. of 60% alloy
0.2 x + 0.6 (600 - x) = 0.35 (600)
2. There's a 250 grams of 17% hydrochloric acid solution. It needs to be drained some amounts down so that a 75% solution is to be add up, to produce a 20% solution. If you are the chemist, how many grams you needed to drain?
Partial solution:
Let x = # of grams of acid to be drained for replacement
17% (250) - 17% x + 75%x = 20 % (250)
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