Mixture problems involve combining two or more quantities with different concentrations or values to form a single mixture with a desired property (such as concentration, strength, or cost).
General Principle:
The total amount of a substance from all parts is equal to the amount of that substance in the resulting mixture.
$$
v_1c_1 + v_2c_2 = (v_1 + v_2)c_f
$$
Where:
$v_1$, $v_2$ → volume or weight of each component
$c_1$, $c_2$ → concentration, strength, or unit value of each component
$c_f$ → final concentration or value of the resulting mixture
This principle applies whether you are working with salt in water, alcohol in liquid, or cost per kilogram in blended goods.
Problem:
The Jager-Bomb is a popular party-starting shot. Normally a 4.2-ounce mix contains 35% Jägermeister. Arataki Itto likes to act tough and wants to try a mix with 70% Jägermeister. How many ounces of pure Jägermeister must be added to a normal mix to obtain the desired mix? Ans. 4.9
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Problem:
Vessel A contains a mixture of 12 liters of wine and 18 liters of water. Vessel B contains 9 liters of wine and 3 liters of water. A certain amount must be drawn from vessel A and B to produce a mixture containing 7 liters of wine and 7 liters of water. How many liters must be drawn from Vessel A? Ans. 10 liters
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Problem:
A 100-kilogram salt solution is originally 4% by weight. Salt in water is boiled to reduce water content until the concentration is 5% by weight salt. How much water is evaporated? Ans. 20kg
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Problem:
A chemist has 300 grams of 20% 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 80% solution? Ans. 25 grams
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Exam Generator Problems
Additional board-style practice items for this topic.
Vessel A contains a mixture of 12 liters of wine and 18 liters of water. Vessel B contains 9 liters of wine and 3 liters of water. How many liters must be drawn from Vessel A to produce a mixture containing 7 liters of wine and 7 liters of water?
Liters from Vessel A
10
4
7
3
Liters from Vessel B
4
10
3
7
Part 1.
Vessel A is $12/30=0.40$ wine and Vessel B is $9/12=0.75$ wine. Let $a$ and $b$ be liters drawn from A and B. The final mixture has 14 L total and 7 L wine: $a+b=14$ $0.40a+0.75b=7$ Solving gives $a=10$. $\boxed{10}$
Two thousand kilograms of steel containing 8% nickel is to be made by mixing steel containing 14% nickel with another steel containing 6% nickel. How much of the steel containing 14% nickel is needed?
A 40-gram alloy containing 35% gold is to be melted with a 20-gram alloy containing 50% gold. How much percentage of gold is the resulting alloy?
Answer:
40%
30%
45%
35%
Gold in first alloy: $40 \times 0.35 = 14$ g Gold in second alloy: $20 \times 0.50 = 10$ g Total gold = 24 g in 60 g alloy $ \% \text{ gold} = \frac{24}{60} \times 100\%$ $\boxed{= 40\%}$
In what ratio must a peanut costing P240.00 per kg. be mixed with a peanut costing P340.00 per kg. so that a profit of 20% is made by selling the mixture at 360.00 per kg?
Answer:
2:3
1:2
3:2
3:5
Selling price = P360/kg with 20% profit → cost = $360/1.20 =$ P300/kg. Let $x$ = fraction of P240-peanut, $(1-x)$ = fraction of P340-peanut: $ 240x + 340(1-x) = 300 $ $ -100x = -40 \Rightarrow x = 0.40$ Ratio $x:(1-x) = 0.40:0.60$ $\boxed{= 2:3}$
A 100-kilogram salt solution is originally 4% by weight. Salt in water is boiled to reduce water content until the concentration is 5% by weight salt. How much water is evaporated?
Answer:
20
25
10
15
Salt content (unchanged): $100 \times 0.04 = 4$ kg For 5% concentration: $4 = 0.05 \times W_{\text{new}} \Rightarrow W_{\text{new}} = 80$ kg Water evaporated: $100 - 80$ $\boxed{= 20 \text{ kg}}$
A pound of alloy of lead and nickel weighs 14.4 ounces in water, where lead loses 1/11 of its weight and nickel loses 1/9 of its weight. How much of each metal is in alloy?
Answer:
Lead=8.8 ounces; Nickel=7.2 ounces;
Lead=7.2 ounces; Nickel=8.8 ounces;
Lead=6.5 ounces; Nickel=5.4 ounces;
Lead=5.4 ounces; Nickel=6.5 ounces;
Let $L$ = oz lead, $N$ = oz nickel. Total = 16 oz. $(1)\; L + N = 16$ Weight in water (lead loses $\frac{1}{11}$, nickel loses $\frac{1}{9}$): $(2)\; \frac{10L}{11} + \frac{8N}{9} = 14.4$ Multiply (2) by 99: $90L + 88N = 1425.6$ Substitute $L=16-N$: $90(16-N)+88N=1425.6 \Rightarrow -2N=-14.4 \Rightarrow N=7.2$ oz $L = 16 - 7.2 = 8.8$ oz $\boxed{\text{Lead} = 8.8 \text{ oz},\; \text{Nickel} = 7.2 \text{ oz}}$
A chemist has 300 grams of 20% 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 80% solution?
A radiator contains 8 quarts of a mixture of water and antifreeze. If 40% of the mixture is antifreeze, how much of the mixture should be drained and replaced by pure antifreeze so that the resultant mixture will contain 60% antifreeze?
Answer:
8/3 quarts
3/8 quarts
1/2 quarts
3/2 quarts
Let $x$ quarts be drained and replaced by pure antifreeze. Initial antifreeze is $0.40(8)=3.2$ qt. Draining removes $0.40x$ qt of antifreeze, and adding pure antifreeze adds $x$ qt: $3.2-0.40x+x=0.60(8)$ $3.2+0.60x=4.8$ $x=\frac{8}{3}$ $\boxed{8/3\text{ quarts}}$
The Jager-Bomb is a popular party-starting shot. Normally a 4.2-ounce mix contains 35% Jägermeister. Arataki Itto likes to act tough and wants to try a mix with 70% Jägermeister. How many ounces of pure Jägermeister must be added to a normal mix to obtain the desired mix?
Answer:
4.9
5.2
6.3
3.7
Initial J?germeister is $0.35(4.2)=1.47$ oz. Let $x$ ounces of pure J?germeister be added: $\frac{1.47+x}{4.2+x}=0.70$ $1.47+x=2.94+0.70x$ $0.30x=1.47$ $\boxed{x=4.9}$
How many grams of pure gold must be added to 36 grams of 60% pure gold to make an alloy that is 76% pure gold?
24
28
30
36
Initial pure gold is $0.60(36)=21.6$ g. Let $x$ grams of pure gold be added: $\frac{21.6+x}{36+x}=0.76$ $21.6+x=27.36+0.76x$ $0.24x=5.76$ $\boxed{x=24}$
A 10% fungicide solution is mixed with a 30% fungicide solution to produce 10 quarts of an 18% solution. How many quarts of the 10% and 30% solutions were used, respectively?
6 and 4
4 and 6
5 and 5
7 and 3
Let $a$ = quarts of 10%, $b$ = quarts of 30%, with $a + b = 10$. $0.10a + 0.30b = 0.18(10) = 1.8$ $0.10a + 0.30(10 - a) = 1.8 \Rightarrow 3 - 0.20a = 1.8 \Rightarrow a = 6,\ b = 4$ $\boxed{6 \text{ and } 4}$
A goldsmith needs 120 grams of gold alloy, of 75% pure gold. Only two alloys of gold are available, the first being 80% pure gold, and the other 60% pure gold. How many grams of 80% pure gold must he use to suit his requirements?
30
90
80
40
Let $x$ be the grams of 80% alloy. Then $120-x$ grams is 60% alloy. $0.80x+0.60(120-x)=0.75(120)$ $0.20x+72=90 \Rightarrow x=90$ $\boxed{90\text{ grams}}$
The acidity or alkalinity of any solution is determined by the concentration of hydrogen ions, [H⁺] in the substance... The acidity of rain over the northeastern United States... had a pH of 2.4. What was the concentration of hydrogen ions?
0.003
0.004
0.005
0.006
Solution pending in psadquestions/t1370.json.
Question Bank: t2083
MSTE - Algebra / Mixture Problems / Besavilla CE Pre-Board Math & Surveying
A block of monel alloy consists of 70% nickel and 30% copper. If it contains 88.2 g of nickel, determine the mass of copper in the block.
33.2 kg
30.6 kg
37.8 kg
45.8 kg
35.5 kg
Nickel is 70% of the alloy. If nickel mass is 88.2 g, total alloy mass is $T=\frac{88.2}{0.70}=126$ g Copper is 30% of the alloy: $m_c=0.30(126)=37.8$ g The printed choices use kg, but the given nickel mass is in grams. The keyed numeric answer is $\boxed{37.8}$.
