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📝 Mixture problems finding quantities with total cost or interest (12 MCQs)

📖 From Digital SAT Algebra • 3. Mathematical Models in Algebra • 12 questions available

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What is Mixture problems finding quantities with total cost or interest?

Definition:
Mixture problems finding quantities with total cost or interest involve determining the amounts of two or more components that, when combined, yield a specified total cost or interest, and these problems are solved by setting up equations that balance the total quantity and the total monetary value or interest, using the principles of weighted averages and linear equations.

Working:
To find quantities, define variables for the unknown amounts, use the equation for total quantity (sum of components equals total) and the equation for total cost or interest (sum of each component's cost or interest equals the given total), then solve the system of equations, with the table method being highly effective for organizing data, especially when dealing with percentages, prices, or interest rates.

Example:
A mixture of 50 pounds of peanuts and almonds costs $\6 \) per pound for the mixture, with peanuts costing $\4$ per pound and almonds costing $\$10$ per pound. Let $p$ be pounds of peanuts and $a$ be pounds of almonds, then $p + a = 50$ and $4p + 10a = 50 \times 6 = 300$ , solving the system: $a = 50 - p$ , substitute into $4p + 10(50 - p) = 300$ gives $4p + 500 - 10p = 300$ , so $-6p = -200$ , hence $p = \frac{100}{3} \approx 33.33$ pounds of peanuts and $a = 50 - 33.33 = 16.67$ pounds of almonds.

Reason:
These problems are fundamental in business, economics, and everyday decisions, such as blending products, investing funds, or optimizing resources, and mastering them helps develop analytical skills for managing costs and maximizing efficiency.

Easy

Medium

Hard

📝 All Mixture problems finding quantities with total cost or interest MCQs

Q1. A shop mixes $x$ kg of rice costing $' in math mode at position 11: 4/kg with \̲(̲20-x$ kg costi…" style="color:#cc0000">4/kg with $20-x$ kg costing7/kg. The final mixture costs $5.20/kg. How many kilograms of the cheaper rice are used?

A.8 kg

B.10 kg

C.12 kg ✅

D.14 kg

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The total cost is $4x+7(20-x)$ , while the mixture costs $20(5.20)=104$ . Solving $4x+140-7x=104$ gives $x=12$ . Therefore, 12 kg of the $4/kg rice is required.

Q2. An investor places $\8,000 \) into two accounts. One earns $4\%$ simple interest and the other earns $7\%$ . If the total annual interest is $\440$ , how much was invested at $4\%$ ?

\$2,000

\$3,000

\$4,000

✅

\$5,000

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Let $x$ be the amount at $4\%$ , so $8000-x$ earns $7\%$ . The interest equation is $0.04x+0.07(8000-x)=440$ . Simplifying gives $560-0.03x=440$ , so $x=4000$ .

Q3. A 30-liter mixture contains liquid A costing $6 per liter and liquid B costing$ 10 per liter. The mixture costs $7.60 per liter. Which statement best describes the amounts?

A.There must be more B than A.

B.There must be equal amounts of A and B.

C.There must be more A than B. ✅

D.The amounts cannot be determined from the average cost.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Because $7.60 is closer to$ 6 than to $10, the mixture must contain more of the cheaper liquid A. Equal quantities would produce an average of$ 8, while the given average is lower, confirming that A is the larger component.

Q4. A mixture of two investments totals $\15,000 \). One investment earns $5\%$ simple interest and the other earns $9\%$ . Which equation correctly models an annual interest total of $\1,050$ if $x$ represents the amount invested at $5\%$ ?

0.05x+0.09x=1050

0.05x+0.09(15000-x)=1050

✅

0.05(15000-x)+0.09(15000-x)=1050

0.05(15000)+0.09x=1050

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: If $x$ is invested at $5\%$ , the remaining $15000-x$ must be invested at $9\%$ . Their annual interests add to $1,050, so $0.05x+0.09(15000-x)=1050$ correctly represents both the total principal and interest conditions.

Q5. A laboratory needs 50 liters of a solution worth $18 per liter by mixing a$ 12/L solution with a $24/L solution. How many liters of the$ 24/L solution are needed?

A.20 L

B.25 L ✅

C.30 L

D.35 L

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Let $x$ be liters of the $' in math mode at position 19: …solution. Then \̲(̲50-x$ liters c…" style="color:#cc0000">24 solution. Then $50-x$ liters cost12 each. The total-value equation is $24x+12(50-x)=18(50)$ . This simplifies to $12x=300$ , giving $x=25$ liters.

Q6. A student invests $\10,000 \) between two certificates. Certificate A earns $6\%$ , while Certificate B earns $9\%$ . After one year, the total simple interest is $\780$ . What amount must be placed in Certificate B?

\$2,000

\$3,000

\$4,000

\$6,000

✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: Let $x$ be invested at $9\%$ , leaving $10000-x$ at $6\%$ . The equation $0.09x+0.06(10000-x)=780$ becomes $0.03x=180$ , so $x=6000$ . Therefore $\$6,000$ must be invested at $9\%$ .

Q7. A café wants 40 kg of a coffee blend costing $15/kg. It combines coffee costing$ 11/kg with coffee costing $19/kg. The owner accidentally prepares 25 kg of the$ 11 coffee first. How many kilograms of the $19 coffee should be added to obtain the target blend?

A.15 kg ✅

B.20 kg

C.25 kg

D.30 kg

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: With 25 kg of the $11 coffee, the total target quantity requires 15 kg of the$ 19 coffee. Checking cost gives $25(11)+15(19)=275+285=560$ , and $560/40=\14\), not15. Thus none of the listed quantities reaches the target; the scenario is inconsistent because the preselected amount cannot produce a $15/kg blend.

Q8. A student models a $\9,000 \) investment split between accounts earning $4\%$ and $6\%$ , with $\480$ total annual interest, by writing $0.04x+0.06(9000+x)=480$ . What is the student's main error?

A.The interest rates should be added.

B.The second investment should be

9000-x

, not

9000+x

. ✅

C.The total principal should be

\$4,800

D.Simple interest cannot be used for this problem.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Since the entire $\9,000 \) is divided between two accounts, if $x$ is placed at $4\%$ , only $9000-x$ remains for the $6\%$ account. Using $9000+x$ makes the two amounts exceed the available $\9,000$ and violates the mixture condition.

Q9. A student claims that mixing equal amounts of $5/kg and$ 13/kg materials produces a mixture costing $9/kg, so a$ 10/kg mixture must contain equal amounts plus an extra $1/kg charge. What is wrong?

A.Equal amounts produce

10/kg, not

9/kg. ✅

B.The mixture cost must always equal $5/kg.

C.Equal amounts cannot be mixed.

D.The average price is found by multiplying the two prices.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: For equal quantities, the average cost is $(5+13)/2=\9\) per kilogram, so the first claim is correct. However, a10/kg target must be created by using more of the $13 material, not by adding an unexplained$ 1 charge.

Q10. A graph plots the total annual interest $I$ against the amount $x$ invested at $4\%$ , with the remaining money invested at $8\%$ . The line decreases from $\800 \) when $x=0$ to $\400$ when $x=10,000$ . If the target interest is $\$560$ , what value of $x$ should the graph indicate?

\$4,000

\$5,000

\$6,000

✅

\$7,000

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The line falls $400 over$ 10,000, so its slope is $-0.04$ . Using $I=800-0.04x$ , set $560=800-0.04x$ . This gives $0.04x=240$ , so $x=\$6,000$ . Therefore option C is correct.

Q11. An investor has $\20,000 \) and wants an average simple-interest rate of $6.5\%$ by combining accounts paying $5\%$ and $8\%$ . The investor puts $\2,000$ more into the $8\%$ account than into the $5\%$ account. Which total annual interest results?

\$1,100

\$1,200

\$1,300

✅

\$1,400

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Let the $5\%$ amount be $x$ , so the $8\%$ amount is $x+2000$ . Since their sum is $' in math mode at position 9: 20,000, \̲(̲2x+2000=20000$ …" style="color:#cc0000">20,000, $2x+2000=20000$ , giving $x=9000$ and11,000 at $8\%$ . Interest is $450+880=\1,330\), so none of the listed choices is correct; the data imply1,330.

Q12. A $\16,000 \) investment is split between accounts earning $4\%$ , $7\%$ , and $10\%$ . The amounts in the first and third accounts are equal, while the total annual interest is $\1,120$ . How much is invested at $7\%$ ?

\$4,000

\$5,000

\$6,000

\$8,000

✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: Let $x$ be invested at both $4\%$ and $10\%$ , and $y$ at $7\%$ . Then $2x+y=16000$ . Interest gives $0.04x+0.07y+0.10x=1120$ . Substituting $y=16000-2x$ gives $1120=1120-0.10x$ , so $x=0$ , making $y=16000$ . Thus none of the listed answers is correct; the constraints force the entire investment into the $7\%$ account.

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