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Mixture Problems
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Mixture Problems: Difficult Problems with Solutions
Problem 1
Some alcohol solution is mixed with [tex]24[/tex] gallons of [tex]65\%[/tex] alcohol to get [tex]60[/tex] gallons of [tex]59\%[/tex] alcohol. What is the concentration of alcohol in the first solution (in %)?
Solution:
The second solution is [tex]60-24=36[/tex] gallons. If its concentration is [tex]d\%[/tex], the pure alcohol gives:
[tex]0.65\times 24+\dfrac{36d}{100}=0.59\times 60[/tex]
[tex]15.6+0.36d=35.4 \Rightarrow 0.36d=19.8 \Rightarrow d=55[/tex]
Problem 2
Brass is an alloy of copper and zinc in which the weights of copper and zinc are in the ratio [tex]3:2[/tex]. How many pounds of zinc must be melted together with a brass block weighing [tex]4.4[/tex] lb to get a new alloy containing [tex]60\%[/tex] zinc?
[tex]1.76[/tex]
[tex]4.4[/tex]
[tex]2.2[/tex]
[tex]2.64[/tex]
Solution:
The block contains [tex]\frac{2}{5}\times 4.4=1.76[/tex] lb of zinc and [tex]2.64[/tex] lb of copper. Add [tex]x[/tex] lb of zinc:
[tex]1.76+x=0.6(4.4+x) \Rightarrow 0.4x=0.88 \Rightarrow x=2.2[/tex]
Check: [tex]\dfrac{1.76+2.2}{4.4+2.2}=\dfrac{3.96}{6.6}=0.6[/tex]
Problem 3
A jeweler has [tex]1.4[/tex] ounces of pure gold. How many ounces of copper must be added to it to get [tex]14[/tex]-karat gold? (Pure gold is [tex]24[/tex]-karat.)
Solution:
[tex]14[/tex]-karat gold is [tex]\frac{14}{24}=\frac{7}{12}[/tex] gold. If the new alloy weighs [tex]m[/tex] ounces:
[tex]\dfrac{7}{12}m=1.4 \Rightarrow m=2.4[/tex]
[tex]2.4-1.4=1[/tex]
Problem 4
An alloy of copper and gold contains [tex]60\%[/tex] copper. After [tex]100[/tex] ounces of gold were added, the copper content dropped to [tex]40\%[/tex]. How many ounces of copper and how many ounces of gold does the new alloy contain?
[tex]120, 180[/tex]
[tex]180, 120[/tex]
[tex]120, 80[/tex]
[tex]80, 220[/tex]
Solution:
Let the original alloy weigh [tex]x[/tex]; it contains [tex]0.6x[/tex] of copper, and the copper does not change:
[tex]0.6x=0.4(x+100) \Rightarrow 0.2x=40 \Rightarrow x=200[/tex]
The new alloy weighs [tex]300[/tex] oz: copper [tex]0.6\times 200=120[/tex] oz and gold [tex]80+100=180[/tex] oz.
Problem 5
A [tex]10[/tex]-gallon container is full of a [tex]40\%[/tex] acid solution. How many gallons must be poured out so that, after refilling the container with a [tex]30\%[/tex] solution of the same acid, the result is a [tex]37\%[/tex] solution?
Solution:
Pour out [tex]x[/tex] gallons. The acid in the container then is:
[tex]0.4(10-x)+0.3x=0.37\times 10 \Rightarrow 4-0.1x=3.7 \Rightarrow x=3[/tex]
Problem 6
A warehouse holds [tex]100[/tex] tons of cucumbers that are [tex]98\%[/tex] water. After some time the water content dropped to [tex]96\%[/tex]. How many tons of water evaporated?
Solution:
The dry matter is [tex]2\%[/tex] of [tex]100[/tex] tons, i.e. [tex]2[/tex] tons. It does not change and is now [tex]4\%[/tex] of the total:
[tex]2\div 0.04=50[/tex]
[tex]100-50=50[/tex] tons of water evaporated — half of the weight!
Problem 7
Freshly threshed wheat usually has a moisture content of [tex]16\%[/tex]. After some drying, [tex]200[/tex] lb of grain became [tex]20[/tex] lb lighter. What is the moisture content of the grain after drying?
[tex]4\%[/tex]
[tex]6\frac{2}{3}\%[/tex]
[tex]10\%[/tex]
[tex]12\%[/tex]
Solution:
Water: [tex]0.16\times 200=32[/tex] lb, dry matter [tex]168[/tex] lb. After drying the grain weighs [tex]180[/tex] lb and contains [tex]32-20=12[/tex] lb of water:
[tex]\dfrac{12}{180}=\dfrac{1}{15}=6\frac{2}{3}\%[/tex]
Problem 8
A [tex]10[/tex]-gallon container is full of a [tex]70\%[/tex] acid solution. Twice, [tex]1[/tex] gallon of liquid was poured out and the container was refilled with water. What is the concentration of the acid after the second refill?
[tex]50\%[/tex]
[tex]56.7\%[/tex]
[tex]63\%[/tex]
[tex]49\%[/tex]
Solution:
At first there are [tex]7[/tex] gallons of acid. Each time [tex]\frac{1}{10}[/tex] of the liquid, and so [tex]\frac{1}{10}[/tex] of the acid, is poured out:
[tex]7\times 0.9=6.3, 6.3\times 0.9=5.67[/tex]
[tex]\dfrac{5.67}{10}=0.567=56.7\%[/tex]
Problem 9
An alloy of copper, zinc, lead and iron contains [tex]60\%[/tex] copper; the zinc is [tex]58\frac{1}{3}\%[/tex] of the copper, and the lead and the iron are in the ratio [tex]2:3[/tex]. How many pounds of copper, zinc, lead and iron (in this order) does the alloy contain if there is [tex]3.3[/tex] lb more zinc than lead?
[tex]6, 3.5, 0.3, 0.2[/tex]
[tex]6, 3.5, 0.2, 0.3[/tex]
[tex]5, 3, 0.8, 1.2[/tex]
[tex]6, 3.3, 0.3, 0.4[/tex]
Solution:
Let the alloy weigh [tex]m[/tex]. Copper: [tex]0.6m[/tex]; zinc: [tex]\frac{7}{12}\times 0.6m=0.35m[/tex] (because [tex]58\frac{1}{3}\%=\frac{7}{12}[/tex]). Lead and iron together are [tex]0.05m[/tex], split in the ratio [tex]2:3[/tex]: lead [tex]0.02m[/tex], iron [tex]0.03m[/tex].
[tex]0.35m-0.02m=3.3 \Rightarrow 0.33m=3.3 \Rightarrow m=10[/tex]
[tex]6, 3.5, 0.2, 0.3[/tex]
Problem 10
A jeweler mixed three gold–copper alloys containing [tex]50\%[/tex], [tex]40\%[/tex] and [tex]30\%[/tex] gold. He took [tex]100[/tex] oz more of the second alloy than of the first, and twice as much of the third as of the first. How many ounces of each alloy must be taken so that the new alloy contains [tex]340[/tex] oz of pure gold?
[tex]100, 200, 200[/tex]
[tex]300, 400, 600[/tex]
[tex]150, 250, 300[/tex]
[tex]200, 300, 400[/tex]
Solution:
Take [tex]x[/tex], [tex]x+100[/tex] and [tex]2x[/tex] of the three alloys:
[tex]0.5x+0.4(x+100)+0.3\times 2x=340 \Rightarrow 1.5x+40=340 \Rightarrow x=200[/tex]
[tex]200, 300, 400[/tex]
Check: [tex]100+120+120=340[/tex]
Problem 11
One alloy contains gold and copper in the ratio [tex]2:3[/tex], and another in the ratio [tex]3:7[/tex]. How many ounces of each alloy must be taken to get [tex]120[/tex] oz of a new alloy in which gold and copper are in the ratio [tex]3:5[/tex]?
[tex]30, 90[/tex]
[tex]60, 60[/tex]
[tex]90, 30[/tex]
[tex]80, 40[/tex]
Solution:
The gold makes up [tex]\frac{2}{5}=0.4[/tex] of the first alloy, [tex]\frac{3}{10}=0.3[/tex] of the second, and must be [tex]\frac{3}{8}\times 120=45[/tex] oz in the new one. Take [tex]x[/tex] oz of the first alloy:
[tex]0.4x+0.3(120-x)=45 \Rightarrow 0.1x=9 \Rightarrow x=90[/tex]
[tex]90[/tex] oz of the first alloy and [tex]30[/tex] oz of the second.
Problem 12
The octane number of gasoline equals the percentage of isooctane in it. A gas station owner topped up a partly filled tank of [tex]93[/tex]-octane gasoline with [tex]87[/tex]-octane gasoline. The full tank turned out to contain [tex]89[/tex]-octane gasoline, but he sold all of it at the price of [tex]93[/tex]-octane gasoline. The tank holds [tex]9000[/tex] gallons, and [tex]93[/tex]-octane gasoline costs [tex]30[/tex] cents per gallon more than [tex]87[/tex]-octane. How much extra profit did the dishonest dealer make?
[tex]900[/tex] dollars
[tex]1800[/tex] dollars
[tex]2700[/tex] dollars
[tex]540[/tex] dollars
Solution:
Let the tank have contained [tex]x[/tex] gallons of [tex]93[/tex]-octane gasoline:
[tex]93x+87(9000-x)=89\times 9000 \Rightarrow 6x=18\,000 \Rightarrow x=3000[/tex]
So [tex]6000[/tex] gallons of [tex]87[/tex]-octane gasoline were sold at the higher price: [tex]6000\times 0.30=1800[/tex] dollars.
Problem 13
One container holds a mixture of [tex]12[/tex] quarts of alcohol and [tex]20[/tex] quarts of water, and another holds [tex]9[/tex] quarts of alcohol and [tex]4[/tex] quarts of water. How many quarts must be poured from the first container into the second so that the second contains equal amounts of alcohol and water?
Solution:
The first mixture is [tex]\frac{12}{32}=\frac{3}{8}[/tex] alcohol and [tex]\frac{5}{8}[/tex] water. After pouring [tex]x[/tex] quarts into the second container:
[tex]9+\dfrac{3x}{8}=4+\dfrac{5x}{8} \Rightarrow 5=\dfrac{x}{4} \Rightarrow x=20[/tex]
Check: [tex]9+7.5=16.5, 4+12.5=16.5[/tex]
Problem 14
Melchior is an alloy of copper and nickel in the ratio [tex]4:1[/tex]. A workshop has [tex]6[/tex] lb of an alloy with copper and nickel in the ratio [tex]1:3[/tex] and [tex]8[/tex] lb of an alloy with copper and nickel in the ratio [tex]2:3[/tex]. How many pounds of pure copper must be added to the two alloys to get melchior?
[tex]37.2[/tex]
[tex]28[/tex]
[tex]32.5[/tex]
[tex]4.7[/tex]
Solution:
First alloy: copper [tex]\frac{1}{4}\times 6=1.5[/tex], nickel [tex]4.5[/tex]. Second alloy: copper [tex]\frac{2}{5}\times 8=3.2[/tex], nickel [tex]4.8[/tex]. Together: copper [tex]4.7[/tex], nickel [tex]9.3[/tex].
The nickel does not change, and the copper must be [tex]4[/tex] times the nickel: [tex]4\times 9.3=37.2[/tex]. So we add:
[tex]37.2-4.7=32.5[/tex]
Problem 15
Two containers hold acid solutions of unknown concentrations. Equal amounts from the two containers mixed together give a [tex]55\%[/tex] solution, and [tex]3[/tex] parts from the first container mixed with [tex]2[/tex] parts from the second give a [tex]51\%[/tex] solution. Find the concentration of the solution in each container.
[tex]75\%, 35\%[/tex]
[tex]45\%, 65\%[/tex]
[tex]35\%, 75\%[/tex]
[tex]40\%, 70\%[/tex]
Solution:
Let the concentrations be [tex]a\%[/tex] and [tex]b\%[/tex]:
[tex]\dfrac{a+b}{2}=55 \Rightarrow a+b=110[/tex]
[tex]\dfrac{3a+2b}{5}=51 \Rightarrow 3a+2b=255[/tex]
[tex]a=255-2\times 110=35, b=110-35=75[/tex]
Correct:
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