Respect Responsibility Readiness
1. A mixture containing 6% boric acid is to be mixed with 2 quarts of a mixture which is 15 % boric acid in order to obtain a solution which is 12% boric acid. How much (in quarts) of the 6% solution must be used?
The following three diagrams represent the three solutions.
The following show the percents in each original mixture and the combined mixture.
Find the one given amount.
Represent the other two amounts using unknowns. The question asks, "How much of the 6% solution must be used?" so that is what x represents. The total solution will have x quarts plus 2 quarts (or the total of the other two solutions).
Let x = quarts of 6% solution.
you are now ready to solve the problem. If you multiply the amount of solution by the percent of acid in the solution, you will find the amount of pure boric acid in each solution. The amount of pure acid in the final solution is equal to the amounts of pure acid in the two original solutions added. This equality gives you the equation information. Keep all amounts on top to avoid confusion. Show products of amount times percent in the lower part of the diagram. These products represent the amount of pure acid in each mixture.
0.06x + 2(0.15) = 012(x + 2)
It is safer to eliminate parentheses first and then eliminate decimals before proceeding to solve for x.
0.06x + 0.30 = 0.12x + 0.24
6x + 30 = 12x + 24
6 = 6x
x = 1
So 1 quart of the 6% solution must be used.
2. Penny Pincher wants to mix milk containing 3% butterfat with cream containing 30% butterfat to obtain 900 gallons (gal) of milk which is 8% butterfat. How much of each must he use?
Let x = gallons of 3% mixture
900 - x = gallons of 30% mixture
0.03x + 0.30(900 - x) = 0.08(900)
0.03x + 270 - 0.30x = 72
3x + 27,000 - 30x = 7,200
-27x + 27,000 = 7,200
-27x = -19,800
x = 733 1/3
900- x = 166 2/3
733 1/3 gallons of 3% mixture
166 2/3 gallons of 30% mixture
3. Cora Spondence wishes to mix two grades of coffee, one of which sells for $4.00 per pound and one of which sells for $6.00 per pound. She wants to sell the mixture for $5.50 per pound. If she has 25 pounds of the $4.00 coffee, how much of the $6.00 coffee must he add so the value of the final mixture is equal to the total value of the other two?
Let x = pounds (lb) of $6.00 coffee
25(4.00)+ 6.00x = 5.50(x + 25)
100 + 6.00x = 5.50x + 137.5
1000 + 60x = 55x + 1375
1000 + 5x = 1375
5x = 375
x = 75
75 pounds (lb) of $6.00 coffee must be added
4. Juan Ting has 20 pounds of
candy worth 80 cents per pound. It wishes to mix it with candy worth 50 cents
per pound so that the total mixture can be sold at 60 cents per pound without
any gain or loss. How much of the 50-cent candy must be used?
Let x = pounds of 50-cent candy
0.80(20) + 0.50x = 0.60(x + 20)
16 + 0.50x = 0.60x + 12
1600 + 50x = 60x + 1200
50x - 60x = 1200 - 1600
-10x = -400
x = 40
40 pounds of 50-cent candy
5. Miss Calculate purchased two lots of shoes. One lot she purchased for $32 per pair and the second lot she purchased for $40 per pair. There were 50 pairs in the first lot. How many pairs were in the second lot if she sold them all at $60 per pair and made a gain of $2800 on the entire transaction?
Let x = pairs in second lot.
32(50) + 40x + 2800 = 60(x + 50)
1600 + 40x + 2800 = 60x + 3000
-20x + 1600 = 3000
-20x = -1400
x = 70
70 pairs of shoes were sold in the second lot
6. Meg A. Bucks needs a solution of tannic acid that is 70% pure. How much
distilled water must she add to 5 gallons of acid which is 90% pure to obtain
the 70% solution?
6. Meg A. Bucks needs a solution of tannic acid that is 70% pure. How much distilled water must she add to 5 gallons of acid which is 90% pure to obtain the 70% solution?
Let x = gallons of water
0.90(5) + 0 = 0.70(x + 5)
4.5 + 0 = 0.70x + 3.5
45 = 7x + 35
-7x = 35 - 45
-7x = -10
x = 1 3/7
1 3/7 gallons of water must be added
7. How much pure alcohol must a nurse add to 10 cubic centimeters (cc) of a 60% alcohol solution to strengthen it to a 90% alcohol solution?
Let x = cubic centimeters of pure alcohol
0.60(10) + 1x = 0.90(x + 10)
6 + x = 0.90x + 9
60 + 10x = 9x + 90
x = 30
30 cubic centimeters of pure alcohol must be added
8. A 5-gallon radiator containing a mixture of water and antifreeze was supposed to
contain a 50% antifreeze solution. When tested, it was found to have only 40%
antifreeze. How much must be drained out and replaced with pure antifreeze so
that the radiator will then contain the desired 50% antifreeze solution?
8. A 5-gallon radiator containing a mixture of water and antifreeze was supposed to contain a 50% antifreeze solution. When tested, it was found to have only 40% antifreeze. How much must be drained out and replaced with pure antifreeze so that the radiator will then contain the desired 50% antifreeze solution?
Let x = gallons of solution drained and replaced
0.40(5) - 0.40x + 1x = 0.50(5)
2.00 - 0.40x + 1x = 2.50
200 - 40x + 100x = 250
60x = 50
x = 5/6
5/6 gallons of solution drained and replaced
9. Millie Watt has 300 grams of 20% hydrochloric acid solution. He wishes to drain
some off and replace it with an 80% solution so as to obtain a 25% solution. How
many grams must he drain and replace with the 80% solution?
9. Millie Watt has 300 grams of 20% hydrochloric acid solution. He wishes to drain some off and replace it with an 80% solution so as to obtain a 25% solution. How many grams must he drain and replace with the 80% solution?
Let x = grams of acid to drain and replace
0.20(300) - 0.20x + 0.80x = 0.25(300)
60 - 0.20x + 0.80x = 75
6000 - 20x + 80x = 7500
60x = 1500
x = 25
25 grams of acid must be drained and replaced
FOR LATER USE
Let x = liters of 10% solution.
0.60(40) + 0.10x = 1.20(40 + x)
24 + 0.10x = 8 + 0.20x
240 + x = 80 + 2x
-x + 240 = 80
-x = -160
x = 160
160 gallons of distilled water
Let x = grams of 40% solution.
20 - x = grams of 70% solution
0.40x + 0.70(20 - x) = 0.52(20)
0.40x + 14 - 0.70x = 10.4
4x + 140 - 7x = 104
-3x + 140 = 104
-3x = -36
x = 12
20 - x = 8
12 grams of 40% solution
8 grams of 70% solution
Let x = liters of distilled water (and solution drained off)
0.60(40) - 0.60x + 0(x) = 0.45(40)
24 - 0.60x = 18
240 - 6x = 180
-6x = -60
x = 10
Let x = ounces of 5% solution (and ounces of 15% solution drained off)
0.15(5) - 0.15x + 0.05x = 0.125(5)
0.75 - 0.15x + 0.05x = 0.60
75 - 15x + 5x = 60
-10x = -15
x = 1 1/2
1 1/2 grams
10. A florist wishes to make bouquets of mixed spring flowers. Each bouquet is to be made up of tulips at $30 a bunch and daffodils at $21 a bunch. How many bunches of tulips should she use to make 15 bunches which she can sell for $24 a bunch?
11. A farmer has 100 gallons of 70% pure disinfectant. He wishes to mix it with disinfectant which is 90% pure in order to obtain 75% pure disinfectant. How much of the 90% pure disinfectant must he use?
12. If an alloy containing 30% silver is mixed with a 55% silver alloy to get 800 pounds of 40% alloy, how much of the 30% silver must be used?
13. Richard discovers at the end of the summer that his radiator antifreeze solution has dropped below the safe level. If the radiator contains 4 gallons of a 25% solution, how many gallons of pure antifreeze must he add to bring it up to a desired 50% solution? (Assume there is room to add antifreeze without removing any solution.)
14. A store manager wishes to reduce the price on her fresh ground coffee by mixing two grades. If she has 50 pounds of coffee which sells for $10 per pound, how much coffee worth $6 per pound must she mix with it so that she can sell the final mixture for $8.50 per pound?
15. A hospital needs to dilute a 50% boric acid solution to a 10% solution. If it needs 25 liters of the 10% solution, how much water should it use?
16. Mary has 25 ounces of a 20% boric acid solution which she wishes to dilute to a 10% solution. How much water does she have to add in order to obtain the 10% solution?
17. Forty liters of a 60% disinfectant solution are to be mixed with a 10% solution to dilute it to a 20% solution. How much of the 10% solution must be used?
18. A doctor orders 20 grams of a 52% solution of a certain medicine. The pharmacist has only bottles of 40% and bottles of 70% solution. How much of the 40% solution must he use to obtain the 20 grams of the 52% solution?
19. Forty liters of a 60% salt solution are reduced to a 45% solution. How much must be drained off and replaced with distilled water so that the resulting solution will contain only 45% salt?
20. A pharmacist has to fill a prescription calling for 5 ounces of a 12% Argyrol solution. She has 5 ounces of a 15% solution and 5 ounces of a 5% solution. If she starts with the 5 ounces of 15%, how much must she draw off and replace with 5% in order to fill the prescription?
to You Shoppe sells cashews for $15 per kg and pecans for $10 per kg. How many
kilograms of each should be mixed in order to get 20 kg of a mixture worth $12
20 = total mixture
x = kg of cashews
20 - x = pecans
15x + 10(20-x) = 12(20)
15x + 200-10x = 240
5x + 200 = 240
5x = 40
x = 8
cashews = 8
pecans = (20 – x) or 12
So 8 kg of cashews at $15 ($120) and 12 kg of pecans at $10 ($120) = 20 total mixture at $12 ($240)
22. Coffee Grounds, Inc., has
two kinds of coffee. Coffee A costs $9 per kg and Coffee B costs $6 per kg. How
many kilograms of each should be combined to obtain 150 kg of a blend worth $8
23. C and Y Candy Company mixes
candy that costs $6.00 per kg with candy that costs $4.50 per kg. How many
kilograms of each are needed to make a 3 kg box that costs $15.00?
24. Trail Snax Corp. mixes
raisins that cost $5.00 per kg with peanuts that cost $3.80 a kg. How many
kilograms of raisins should be mixed with 10 kg of peanuts to obtain a mixture
worth $4.00 per kg?
25. Ground beef sells for $4.75
per kg and ground pork sells for $5.50 per kg. How many kilograms of ground pork
should be mixed with 8 kg of ground beef to make a mixture that sells for $5.10
26. Speed Seed Company mixes
bluegrass seed that Costs $7.60 per kilogram with ryegrass seed that costs $6.25
a kg. How many kilograms of bluegrass seed should be mixed with 200 kg of
ryegrass seed to make a mixture worth $7.00 per kg?
27. A card company mixes two
varieties of cards. Embossed Cards Cost $.65 each, and regular cards $.40 each.
How many cards of each type should be included in an assortment of 25 cards that
28. How many liters of water
must be added to 8 liters of a 4O% acid solution to obtain a 10% acid solution?
29. How many titers of water
must be added to 20 liters of a 70% antifreeze solution to produce a 50%
30. Bunson Beaker has 150 grams
of a 50% salt solution. How many grams of water must be added to obtain a 20%
31. How much water must be
added to 12 grams of a 90% iodine solution to produce a 25% iodine solution?
32. Moonshine has 50 liters of
a 70% alcohol solution. How many liters of pure alcohol must be added to obtain
an 80% alcohol solution?
33. How many kilograms of pure
salt must be added to 20 kilograms of a 10% salt solution to obtain a 25% salt
34. How much pure acid must be added to 6 milliliters of a 5% acid solution to produce a 40% acid solution?