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📝 Mixture word problems with table (14 MCQs)

📖 From Digital SAT Algebra • 3. Mathematical Models in Algebra • 14 questions available

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What is Mixture word problems with table?

Definition:
Mixture word problems with tables involve combining two or more substances with different qualities (like concentrations, costs, or percentages) to achieve a desired mixture, and the table method organizes the data by listing each component with columns for the amount, the quality per unit (e.g., price, percentage), and the total quality, using the equation $\text{Amount}_1 \times \text{Quality}_1 + \text{Amount}_2 \times \text{Quality}_2 = \text{Total Amount} \times \text{Total Quality}$ .

Working:
To solve mixture problems with a table, first draw a table with rows for each component and the mixture, columns for quantity, quality (price, concentration, etc.), and total value, then fill in the known values, use variables for unknowns, and set up an equation for the total quantity and another for the total quality (e.g., total cost or acid amount), then solve the system of equations for the unknown quantities.

Example:
A coffee shop mixes 20 pounds of coffee worth $\8 \) per pound with some coffee worth $\12$ per pound to produce a mixture worth $\10 \) per pound, and the total mixture is 50 pounds. Let $x$ be the pounds of $\12$ coffee, then $20 + x = 50$ , so $x = 30$ pounds, and the total cost equation is $20(8) + 30(12) = 50(10)$ , giving $160 + 360 = 500$ , which holds true, so the mixture uses 20 pounds of $\8 \) coffee and 30 pounds of $\12$ coffee.

Reason:
Using a table for mixture problems simplifies complex data, reduces errors, and provides a clear visual representation of the relationships between quantities and qualities, making it an invaluable tool in algebra, chemistry, and business.

Easy

Medium

Hard

📝 All Mixture word problems with table MCQs

Q1. A café mixes 6 kg of coffee costing $\8 \) per kg with coffee costing $\12$ per kg. Which equation correctly represents the total cost of the mixture?

8x+12x=6

8(6-x)+12x

✅

8(6+x)+12x

8(6-x)+12(6-x)

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: If $x$ kilograms represent the amount of the $\$12$ coffee, then $6-x$ kilograms represent the cheaper coffee. Multiplying each quantity by its cost gives $8(6-x)+12x$ , which correctly models the total cost.

Q2. A student says that when two ingredients have different prices, the mixture price must always equal the average of their two prices. Which statement best evaluates this reasoning?

A.It is always correct because two prices are being combined

B.It is correct only when the quantities of the two ingredients are equal ✅

C.It is never correct because mixtures cannot have a single price

D.It is correct only when the more expensive ingredient is used in greater quantity

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: A simple average gives the correct mixture cost per unit only when equal quantities of the two ingredients are combined. If quantities differ, a weighted average is required because the larger quantity has greater influence on the final cost.

Q3. A fruit shop combines apples costing $\2 \) per kg and strawberries costing $\5$ per kg. The owner wants 10 kg of fruit costing $\$3.20$ per kg. How many kilograms of strawberries should be used?

A.2 kg

B.3 kg

C.4 kg ✅

D.6 kg

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Let $x$ be kilograms of strawberries, so $10-x$ kilograms are apples. The cost equation is $5x+2(10-x)=32$ . Simplifying gives $3x=12$ , so $x=4$ . Therefore, 4 kg of strawberries produces the desired average cost.

Q4. A trail mix contains peanuts costing $\4 \) per kg and premium nuts costing $\10$ per kg. A 15 kg batch must cost $\$7$ per kg. Which reasoning is most appropriate before solving?

A.Use 15 kg of each ingredient because the batch costs

\$7

B.Recognize that the target price lies between the two costs and determine quantities whose weighted average is

\$7

✅

C.Use only premium nuts because their price is closest to the target

D.Subtract 4 from 10 and use that difference as the amount of each ingredient

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The target cost of $\7 \) lies between $\4$ and $\$10$ , so both ingredients must contribute. The quantities cannot be chosen by ordinary averaging unless they are equal; instead, their amounts must produce the required weighted average.

Q5. A coffee shop has $x$ kg of coffee costing $\9$ per kg and $8-x$ kg costing $\15$ per kg. The desired mixture costs $\$12$ per kg. Which conclusion follows most efficiently?

A.More than 4 kg must come from the

\$15

coffee

B.Exactly 4 kg of each coffee is required ✅

C.Less than 4 kg must come from the

\$9

coffee

D.The quantities cannot be determined from the information

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The target cost $\12 \) is exactly halfway between $\9$ and $\$15$ . Therefore, equal quantities are required. Since the total is 8 kg, each type must contribute 4 kg to achieve the desired weighted average.

Q6. A market combines 3 kg of dried fruit costing $\6 \) per kg with some trail mix costing $\14$ per kg. The final mixture weighs 8 kg. A customer calculates the total cost as $8(10)=\80$ because $\10$ is halfway between the two prices. What is wrong?

A.The total weight should be multiplied by the cheaper price

B.The customer ignored that the quantities of the two products are unequal ✅

C.The customer should add the two prices before multiplying by 8

D.The customer must use the higher price for the entire mixture

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The halfway price of $\$10$ would apply only if the two ingredients were present in equal quantities. Here, 3 kg is known to be the cheaper ingredient and 5 kg is the other ingredient, so the costs must be calculated separately.

Q7. A cafeteria wants 20 kg of fruit salad using fruit costing $\3 \) per kg and premium fruit costing $\7$ per kg. The target cost is $\$5.50$ per kg. Which amount of premium fruit is required?

A.7.5 kg

B.10 kg

C.12.5 kg ✅

D.15 kg

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Let $x$ kilograms be premium fruit. Then $20-x$ kilograms cost $\$3$ per kg. The equation $7x+3(20-x)=110$ simplifies to $4x=50$ , giving $x=12.5$ . Thus, the correct amount is 12.5 kg, not 10 kg.

Q8. A bakery has 5 kg of nuts costing $\9 \) per kg and wants to add cheaper nuts costing $\5$ per kg to make a mixture costing $\$7$ per kg. How many kilograms of cheaper nuts are needed?

A.2.5 kg

B.5 kg ✅

C.7.5 kg

D.10 kg

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Let $x$ kilograms be the cheaper nuts. The total weight is $5+x$ , while the total cost is $45+5x$ . Setting the average cost to 7 gives $45+5x=7(5+x)$ , so $x=5$ .

Q9. A student models a 12 kg fruit mixture with $x$ kg of fruit costing $\4$ per kg and $12-x$ kg costing $\9$ per kg. The student writes $4x+9(12-x)=12$ . What does the equation fail to represent?

A.The total amount of fruit

B.The cost per kilogram of either fruit

C.The total cost of the mixture

D.The target average cost of the mixture ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: The expression $4x+9(12-x)$ represents total cost, not average cost. To model a target average of $\$12$ per kg, the total cost would need to equal 12 times the total weight, which is 144, not 12.

Q10. A graph of mixture cost versus kilograms of premium nuts is a straight line passing through $(0,40)$ and $(10,100)$ , where the vertical coordinate represents total cost in dollars. What does the slope represent?

A.The total amount of nuts

B.The starting cost of the cheaper nuts

C.The increase in total cost for each additional kilogram of premium nuts ✅

D.The average cost of the final mixture

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The slope is $(100-40)/(10-0)=6$ . Because the horizontal axis measures kilograms of premium nuts and the vertical axis measures total cost, the slope represents the additional cost contributed by each kilogram of premium nuts.

Q11. A chef needs 18 kg of salad costing $\6 \) per kg. She has vegetables costing $\4$ per kg and fruit costing $\$9$ per kg. Which system correctly models the situation if $x$ is kilograms of vegetables and $y$ is kilograms of fruit?

x+y=18,\ 4x+9y=108

✅

x-y=18,\ 4x+9y=108

x+y=108,\ 4x+9y=18

4x+9y=18,\ x-y=6

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The quantities must total 18 kg, giving $x+y=18$ . The total cost must equal $18(6)=108$ , giving $4x+9y=108$ . Together these equations model both the quantity and cost requirements.

Q12. A store has two coffees priced at $\6$ and $\10$ per kg. It creates a 25 kg blend. A graph of total blend cost against kilograms of $\10 \) coffee is linear, beginning at $\150$ when zero premium coffee is used and reaching $\250 \) at 25 kg. What quantity of premium coffee produces a blend costing $\200$ ?

A.6.25 kg

B.10 kg

C.12.5 kg ✅

D.18.75 kg

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The graph increases from $\150$ to $\250$ , a total increase of $\100 \) across 25 kg. A cost of $\200$ is halfway between the endpoints, so the corresponding quantity is halfway between 0 and 25 kg, giving 12.5 kg.

Q13. A supplier combines rice costing $\3 \) per kg with specialty rice costing $\11$ per kg. The supplier wants 30 kg at $\7 \) per kg. An employee claims that 15 kg of each is necessary because $\7$ is halfway between the prices. Is the claim correct?

A.Yes, because the target price is the arithmetic mean ✅

B.No, because 20 kg of cheap rice and 10 kg of specialty rice are required

C.No, because 10 kg of cheap rice and 20 kg of specialty rice are required

D.Yes, because any equal quantities always produce the target price

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Here the target price $\7 \) is exactly halfway between $\3$ and $\11 \). Equal quantities therefore produce the target weighted average. With 30 kg total, 15 kg of each ingredient gives $45+165=\210$ , which is $\$7$ per kg.

Q14. A fruit vendor can buy fruit at $\2$ , $\5$ , or $\8 \) per kg. He needs exactly 12 kg with an average cost of $\5$ per kg. He wants to use some of all three types. Which combination satisfies both requirements?

A.2 kg at

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' in math mode at position 3: 2 \̲)̲, 8 kg at

\" style="color:#cc0000">2

, 8 kg at

\</span>5

, 2 kg at

\$8

B.3 kg at

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' in math mode at position 3: 2 \̲)̲, 6 kg at

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, 6 kg at

\</span>5

, 3 kg at

\$8

C.4 kg at

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' in math mode at position 3: 2 \̲)̲, 4 kg at

\" style="color:#cc0000">2

, 4 kg at

\</span>5

, 4 kg at

\$8

✅

D.1 kg at

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' in math mode at position 3: 2 \̲)̲, 9 kg at

\" style="color:#cc0000">2

, 9 kg at

\</span>5

, 2 kg at

\$8

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: For option C, the quantities total $4+4+4=12$ kg. The cost is $4(2)+4(5)+4(8)=60$ , and $60/12=\$5$ per kg. The equal quantities also balance the lower and higher prices around the target.

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