Unlock hundreds more features
Save your Quiz to the Dashboard
View and Export Results
Use AI to Create Quizzes and Analyse Results

OR
Sign inSign in with Facebook
Sign inSign in with Google
Quizzes > High School Quizzes > Mathematics

Mixtures Quick Check Practice Quiz

Boost your mixtures skills with practical test questions

Difficulty: Moderate
Grade: Grade 7
Study OutcomesCheat Sheet
Colorful paper art promoting a Quick Mix Check trivia quiz for 10th-grade math students.

A container holds 10 liters of a 20% acid solution. How many liters of acid does it contain?
20 liters
2 liters
8 liters
5 liters
20% of 10 liters is calculated by multiplying 10 by 0.20, which equals 2 liters. Thus, the correct answer is 2 liters.
When 2 liters of a 50% salt solution are mixed with 2 liters of water, what is the resulting concentration percentage?
30%
50%
75%
25%
The 2 liters of 50% solution provide 1 liter of salt. After mixing with 2 liters of water, the total volume is 4 liters, so the concentration becomes 1/4 which is 25%.
If 5 liters of a solution contains 2 liters of pure juice, what is the concentration percentage?
40%
60%
50%
20%
The concentration is determined by dividing the amount of pure juice by the total volume. Here, 2 divided by 5 gives 0.4, which is equivalent to 40%.
A farmer prepares a 10-liter fertilizer solution that is 10% fertilizer. How many liters of fertilizer are in the solution?
5 liters
10 liters
2 liters
1 liter
Calculating 10% of 10 liters gives 1 liter. Therefore, the solution contains 1 liter of fertilizer.
In a 100 mL solution that is 15% alcohol, what is the volume of pure alcohol?
85 mL
50 mL
15 mL
25 mL
15% of 100 mL is found by multiplying 100 by 0.15, which yields 15 mL of pure alcohol. Hence, the correct answer is 15 mL.
How much of a 40% solution must be mixed with 6 L of a 20% solution to obtain a 30% solution?
6 L
3 L
12 L
9 L
Let x be the volume of the 40% solution. The equation 0.4x + 0.2(6) = 0.3(x + 6) leads to x = 6 L. Therefore, 6 L is required.
A chemist needs a 25% acid solution. She has a 15% acid solution and a 35% acid solution. If she mixes them in equal volumes, what is the resulting concentration?
30%
35%
15%
25%
Mixing equal volumes of 15% and 35% solutions averages their concentrations: (15% + 35%) / 2 equals 25%. This is the resulting concentration of the mixture.
To obtain 8 L of a 60% mixture, how many liters of pure substance must be added to 5 L of a 40% mixture?
3 L
4 L
1.5 L
2.5 L
The 5 L of 40% mixture contains 2 L of pure substance. Let x be the liters of pure substance added; then (2 + x) / (5 + x) = 0.6. Solving this gives x = 2.5 L.
If 12 kg of a 30% alloy is mixed with 8 kg of a 50% alloy, what is the percentage of the resulting alloy?
35%
38%
42%
40%
The metal content is 0.30 × 12 + 0.50 × 8 = 3.6 + 4 = 7.6 kg in a total of 20 kg. The percentage is (7.6/20) × 100 = 38%.
How many mL of an 80% solution do you need to add to 200 mL of a 50% solution to obtain a 60% solution?
50 mL
100 mL
150 mL
200 mL
Let x be the mL of 80% solution added. The equation (0.8x + 0.5×200)/(x+200) = 0.6 leads to x = 100 mL. This is verified by solving the equation.
A solution is diluted by adding water. If 4 L of a 25% solution is diluted to reach a 20% concentration, what is the total volume of the diluted solution?
4.5 L
6 L
5 L
7 L
The original solution contains 1 L of solute (25% of 4 L). To have a 20% concentration, 1 L must be 20% of the final volume, which means the final volume is 5 L.
A bartender mixes two juices: one is 100% fruit juice and the other is 40% juice. If he mixes them in the ratio 1:2 respectively, what is the percentage of pure juice in the final mix?
50%
60%
40%
70%
Using the ratio 1:2, one part 100% and two parts 40% yields (1×1 + 2×0.4)/3 = (1 + 0.8)/3 = 1.8/3 = 0.6 or 60% concentration. Thus, the final mix is 60% pure juice.
Which formula correctly calculates the concentration (as a decimal) after mixing V1 liters of a solution with concentration C1 and V2 liters of a solution with concentration C2?
(C1 × V1 + C2 × V2) / (V1 + V2)
(V1 + V2) / (C1 + C2)
(C1 + C2) / (V1 + V2)
(C1 × V1 − C2 × V2) / (V1 − V2)
The correct method to find the concentration after mixing is to compute the weighted total of the solute divided by the total volume. This is given by the formula (C1 × V1 + C2 × V2) / (V1 + V2).
A 9 L mixture contains a 30% substance. How many liters of a 100% substance must be added to make the resulting mixture 50% substance?
4 L
3.6 L
3 L
5 L
The 9 L mixture contains 2.7 L of pure substance. Let x be the liters to add; then (2.7 + x)/(9 + x) = 0.5. Solving for x gives 3.6 L, which is the amount required.
In making a metal alloy, you mix 3 kg of a 20% alloy with 7 kg of a 50% alloy. What is the percentage of metal in the final 10 kg mixture?
45%
50%
41%
40%
The amount of metal is calculated as 0.20 × 3 + 0.50 × 7 = 0.6 + 3.5 = 4.1 kg. Dividing 4.1 kg by the total 10 kg mixture gives a metal percentage of 41%.
An inspector mixes two batches of metal alloys: one with 18% metal and the other with 42% metal. To produce 50 kg of an alloy with 30% metal, what is the required mass of the 42% alloy?
15 kg
30 kg
25 kg
20 kg
Let x represent the mass of the 42% alloy and (50 - x) the mass of the 18% alloy. Setting up the equation: 0.42x + 0.18(50 - x) = 0.30 × 50 leads to x = 25 kg. Thus, 25 kg of the 42% alloy is required.
An experiment requires a 3:1 ratio of solvent to solute. If you have 2.4 L of solvent, how much solute is needed to maintain the ratio?
0.8 L
3.0 L
2.4 L
1.2 L
A 3:1 ratio means for every 3 parts of solvent, 1 part of solute is required. Dividing 2.4 L of solvent by 3 gives 0.8 L of solute.
A solution contains 15% substance A and 85% substance B. After evaporating 5 L from the solution, the concentration of substance A increases to 20%. What was the initial volume of the solution?
20 L
25 L
30 L
15 L
Let V be the initial volume. Substance A remains constant at 0.15V, and after evaporation the new volume is V - 5. Setting up the equation 0.15V/(V - 5) = 0.20 and solving gives V = 20 L.
A 120 mL solution contains 36 mL of alcohol. If additional alcohol is added so that the final concentration becomes 50%, what is the volume of alcohol added?
48 mL
60 mL
36 mL
24 mL
Let x be the additional alcohol. The equation (36 + x)/(120 + x) = 0.50 leads to 36 + x = 60 + 0.50x, which simplifies to x = 48 mL. Therefore, 48 mL of alcohol must be added.
In a laboratory, 10 mL of a 70% solution and x mL of a 40% solution are mixed to obtain a 55% solution. Find x.
5 mL
15 mL
10 mL
20 mL
Setting up the equation: (0.70×10 + 0.40×x)/(10 + x) = 0.55 leads to 7 + 0.40x = 5.5 + 0.55x. Solving for x gives x = 10 mL, which is the correct amount of the 40% solution required.
0
{"name":"A container holds 10 liters of a 20% acid solution. How many liters of acid does it contain?", "url":"https://www.quiz-maker.com/QPREVIEW","txt":"A container holds 10 liters of a 20% acid solution. How many liters of acid does it contain?, When 2 liters of a 50% salt solution are mixed with 2 liters of water, what is the resulting concentration percentage?, If 5 liters of a solution contains 2 liters of pure juice, what is the concentration percentage?","img":"https://www.quiz-maker.com/3012/images/ogquiz.png"}

Study Outcomes

  1. Analyze mixture problems to determine the correct ratios and concentrations.
  2. Apply algebraic methods to solve multi-step mixture calculations.
  3. Evaluate solution strategies to choose the most effective approach for mixture problems.
  4. Demonstrate proficiency in calculating percentages and proportions within mixture contexts.
  5. Synthesize various mathematical concepts to address real-world mixture scenarios effectively.

Mixtures Quick Check Cheat Sheet

  1. Master the Law of Sines - A neat way to conquer oblique triangles by linking their angles and opposite sides; perfect for problems that don't have a right angle. Brush up on this formula and you'll slice through non-right triangle challenges like a pro. OpenStax: Law of Sines & Cosines
  2. Understand the Law of Cosines - This extends the Pythagorean theorem beyond right triangles, letting you calculate any side or angle in any triangle. It's your go-to tool when sides and angles play hide-and-seek in non-right triangles. OpenStax: Law of Cosines
  3. Familiarize yourself with the quadratic formula - Learning x = (-b ± √(b² - 4ac))❄(2a) gives you a magical key to solve any quadratic equation in a flash. It's a must-have in your algebra toolkit when equations get messy. Byju's: Quadratic Formula
  4. Learn properties of logarithms - Use log(ab) = log a + log b (and friends) to break down complex expressions into manageable bite-sized bits. Once you master these, you'll unlock the door to simplify and solve tricky log equations effortlessly. Topper's Bulletin: Logarithm Properties
  5. Practice the distance formula - d = √((x₂ - x₝)² + (y₂ - y₝)²) lets you find the straight-line gap between two points in the plane. Plot a couple of coordinates, and you'll be measuring distances faster than your calculator can say "radical." Topper's Bulletin: Distance Formula
  6. Review the binomial theorem - Expand (a + b)❿ like a boss to see exactly what each term looks like - no more guessing at coefficients! This theorem is your shortcut to break down big polynomial powers in a breeze. Topper's Bulletin: Binomial Theorem
  7. Understand surface area and volume formulas - Grab the equations for spheres (4πr², 4/3πr³), cylinders (2πr²+2πrh, πr²h), and cones (πr²+πrl, 1/3πr²h) to solve real-world geometry puzzles. From designing cans to wrapping balls, these formulas have got your back. Byju's: 3D Shape Formulas
  8. Learn area and circumference of a circle - Memorize Area = πr² for space inside, and Circumference = 2πr for the perimeter around; these are fundamental for everything circular. Whether it's a pizza's slice or gearing up for exam day, these formulas never let you down. Byju's: Circle Formulas
  9. Study arithmetic progression formulas - For sequences where each term steps by the same amount, use aₙ = a + (n - 1)d for the nth term and Sₙ = n/2 [2a + (n - 1)d] for sums. These handy formulas let you instantly jump to any term or total up a series in seconds. Byju's: Arithmetic Progression
  10. Master exponent rules - Rules like aᵝ × a❿ = aᵝ❺❿, (aᵝ)❿ = aᵝ❿, and a❰ = 1 turn complex power expressions into simple ones. Slash through exponential disasters by applying these properties to combine, separate, and simplify like a champ. OpenStax: Exponent Properties
Make a Quiz (FREE) Copy & Edit This Quiz Create a Quiz with AI

Mathematics Quizzes

Powered by: Quiz Maker