Algebra Quiz: Creating Solving One Variable Equations Inequalities

What this quiz covers

This quiz focuses on Creating Solving One Variable Equations Inequalities, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $9t + 4 \ge 40$
2. $9t - 4 \le 40$
3. $4t + 9 \le 40$
4. $9t + 4 \le 40$ (correct answer)

Explanation: This question tests your ability to translate a real-world situation into a mathematical equation or inequality, solve it, and interpret the result in the original context. For inequalities, words like 'at most,' 'maximum,' 'no more than' signal ≤ (less than or equal), while 'at least,' 'minimum,' 'no less than' signal ≥ (greater than or equal). 'More than' means > (strict), and 'less than' means <. These key phrases tell you which inequality symbol to use! The context 'tickets cost $9 each with a$4 fee, at most $40' uses the phrase 'at most,' which signals ≤. Setting up: cost per ticket times number plus fee ≤ total available, so 9t + 4 ≤ 40. Solving: subtract 4 → 9t ≤ 36, divide by 9 → t ≤ 4. This means Jordan can buy a maximum of 4 whole tickets. Choice A is correct because it properly sets up the inequality from the context with the fee added and uses ≤ for 'at most,' giving t ≤ 4 tickets within budget. Choice D sets up the inequality incorrectly: it switches the variables to 4t + 9 ≤ 40, which would be like$4 per ticket and $9 fee, but the problem says$9 tickets and $4 fee—reading carefully for relationships is key! For inequalities, make a quick reference card: 'at most/maximum/no more than' → ≤ (can equal or be less), 'at least/minimum/no less than' → ≥ (can equal or be more), 'more than/over' → > (strictly greater), 'less than/under' → < (strictly less). Having these memorized means you'll never use the wrong symbol!

Question 2

1. $x=12$
2. $x=15$ (correct answer)
3. $x=8$
4. $x=16$

Explanation: This question tests your ability to translate a real-world situation into a mathematical equation or inequality, solve it, and interpret the result in the original context. Different contexts lead to different equation types: constant rates give linear equations (like cost = rate × quantity + fee), area problems often give quadratics (like length × width = area), and growth over time gives exponentials (like population = initial × (growth rate)^time). The context clues tell you which form to use. For rate problems like this, we set up: when working together, rates add. Pump A's rate is 1/10 tank per hour, Pump B's rate is 1/x tank per hour, and together they fill at 1/6 tank per hour. This gives us 1/10 + 1/x = 1/6. Solving: 1/x = 1/6 - 1/10 = 5/30 - 3/30 = 2/30 = 1/15. Therefore x = 15. Interpreting: Pump B alone takes 15 hours to fill the tank. Choice B is correct because it properly sets up the rate equation from context and solves correctly, giving x = 15 hours for Pump B alone. Choice A would mean Pump B works faster than the combined rate, which is impossible—two pumps together must work faster than either alone! When solving rate problems, remember that rates add when working together. Different contexts = different equation types: constant rates and simple relationships → linear, area and projectile motion → quadratic, working together and rate problems → rational, growth/decay over time → exponential. Recognizing these patterns helps you set up the right type of equation immediately!

Question 3

1. $w = 14$
2. $w = 12$
3. $w = 7$ (correct answer)
4. $w = 9$

Explanation: This question tests your ability to translate a real-world situation into a mathematical equation or inequality, solve it, and interpret the result in the original context. Different contexts lead to different equation types: constant rates give linear equations (like cost = rate × quantity + fee), area problems often give quadratics (like length × width = area), and growth over time gives exponentials (like population = initial × (growth rate)^time). The context clues tell you which form to use. This is an area problem, which means quadratic! Let w = width. The relationship 'length is 5 more than width, area 84' translates to w(w + 5) = 84, or w² + 5w - 84 = 0. Using quadratic formula, discriminant 25 + 336 = 361 = 19², so w = [-5 ± 19]/2, giving w = 7 or w = -12 (discard negative). So valid solution w = 7 with interpretation: the garden is 7 feet wide and 12 feet long. Choice A is correct because it properly sets up the quadratic from the area context, solves correctly, and interprets appropriately, giving w = 7 feet that makes sense. Choice C solves the equation correctly but doesn't check the context: w = 12 would imply length = 17, but that's swapping variables— the question asks for width, which is the smaller one. Always ask: does my answer make sense in the real world? This catches a lot of mistakes! The 'reality check' is your best friend: after solving, substitute your answer back into the original equation (math check), then ask 'does this make sense?' (reality check). Can dimensions be negative? Can you buy 7.3 shirts? Can there be -4 hours? The context tells you what's possible and what's not!

Question 4

1. $t = 2$ only
2. $t = 0$ and $t = 4$ (correct answer)
3. $t = 0$ and $t = 2$
4. $t = 4$ only

Explanation: This question tests your ability to translate a real-world situation into a mathematical equation or inequality, solve it, and interpret the result in the original context. Different contexts lead to different equation types: constant rates give linear equations (like cost = rate × quantity + fee), area problems often give quadratics (like length × width = area), and growth over time gives exponentials (like population = initial × (growth rate)^time). The context clues tell you which form to use. This is a motion problem, which means quadratic! Let t = time in seconds. The relationship 'height after t seconds is given by h = -16t² + 64t' and 'when the ball is on the ground' translates to set h=0: -16t² + 64t = 0. Using factoring, -16t(t - 4) = 0, so t=0 or t=4. Checking context: both make sense—at t=0 (start) and t=4 (lands). Interpreting: the ball is on the ground at 0 seconds and after 4 seconds. Choice B is correct because it properly sets up the quadratic equation from context, solves correctly, and interprets appropriately, giving t=0 and t=4 seconds. Choice D finds one solution to the quadratic but misses the other: solving gives two values, but perhaps they ignored t=0 since it's the start. For quadratics, always check both solutions against the real-world situation! The 'reality check' is your best friend: after solving, substitute your answer back into the original equation (math check), then ask 'does this make sense?' (reality check). Can dimensions be negative? Can you buy 7.3 shirts? Can there be -4 hours? The context tells you what's possible and what's not!

Question 5

1. $w = 6$
2. $w = 9$
3. $w = 7$ (correct answer)
4. $w = 12$

Explanation: This question tests your ability to translate a real-world situation into a mathematical equation or inequality, solve it, and interpret the result in the original context. Different contexts lead to different equation types: constant rates give linear equations (like cost = rate × quantity + fee), area problems often give quadratics (like length × width = area), and growth over time gives exponentials (like population = initial × (growth rate)^time). The context clues tell you which form to use. This is an area problem, which means quadratic! Let w = width. The relationship 'length that is 5 feet more than its width' and 'area of the garden is 84 square feet' translates to w(w + 5) = 84, or w² + 5w - 84 = 0. Using factoring, we get (w + 12)(w - 7) = 0, so w = -12 or w = 7. But wait—checking context: negative width doesn't make sense, so w = 7 feet. Interpreting: the width is 7 feet, length 12 feet, area 84 sq ft. Choice A is correct because it properly sets up the quadratic equation from context, solves correctly, and interprets appropriately, giving w=7 feet. Choice B finds one solution to the quadratic but misses the other (or keeps an extraneous one): solving gives w=7 and w=-12, but perhaps they took the positive non-valid or switched length/width. For quadratics, always check both solutions against the real-world situation! The 'reality check' is your best friend: after solving, substitute your answer back into the original equation (math check), then ask 'does this make sense?' (reality check). Can dimensions be negative? Can you buy 7.3 shirts? Can there be -4 hours? The context tells you what's possible and what's not!

Question 6

1. Maximum $n=11$
2. Maximum $n=12$ (correct answer)
3. Maximum $n=13$
4. Maximum $n=14$

Explanation: This question tests your ability to translate a real-world situation into a mathematical equation or inequality, solve it, and interpret the result in the original context. For inequalities, words like 'at most,' 'maximum,' 'no more than' signal ≤ (less than or equal), while 'at least,' 'minimum,' 'no less than' signal ≥ (greater than or equal). 'More than' means > (strict), and 'less than' means <. These key phrases tell you which inequality symbol to use! The context 'at most $50 to spend' uses the phrase 'at most,' which signals ≤. Setting up: cost of notebooks ≤ money available, so 4n ≤ 50. Solving: n ≤ 12.5. This means a maximum of 12 whole notebooks (can't buy half a notebook!). Choice B is correct because it properly sets up the inequality from context, solves correctly, and interprets appropriately in the real world, giving maximum n = 12 notebooks. Choice C would cost 4(13) =$52, which exceeds the $50 budget. When dealing with discrete items like notebooks, always round down to stay within the constraint! For inequalities, make a quick reference card: 'at most/maximum/no more than' → ≤ (can equal or be less), 'at least/minimum/no less than' → ≥ (can equal or be more), 'more than/over' → > (strictly greater), 'less than/under' → < (strictly less). Having these memorized means you'll never use the wrong symbol!

Question 7

1. $w=9$
2. $w=6$
3. $w=12$
4. $w=7$ (correct answer)

Explanation: This question tests your ability to translate a real-world situation into a mathematical equation or inequality, solve it, and interpret the result in the original context. Different contexts lead to different equation types: constant rates give linear equations (like cost = rate × quantity + fee), area problems often give quadratics (like length × width = area), and growth over time gives exponentials (like population = initial × (growth rate)^time). The context clues tell you which form to use. This is an area problem, which means quadratic! Let w = width in feet. The relationship 'length is 5 feet more than width' translates to length = w + 5. Since area = length × width = 84, we get w(w + 5) = 84. Expanding: w² + 5w = 84, so w² + 5w - 84 = 0. Using factoring: (w + 12)(w - 7) = 0, we get w = -12 or w = 7. But wait—checking context: width can't be negative! So w = 7 feet with interpretation: the garden is 7 feet wide and 12 feet long. Choice A is correct because it properly sets up the quadratic equation from the area context, solves correctly, and eliminates the extraneous negative solution, giving w = 7 feet. Choice B would give a length of 14 feet and area of 9 × 14 = 126 square feet, not 84. When solving quadratics from real-world contexts, always check both solutions—one often doesn't make physical sense! The 'reality check' is your best friend: after solving, substitute your answer back into the original equation (math check), then ask 'does this make sense?' (reality check). Can dimensions be negative? Can you buy 7.3 shirts? Can there be -4 hours? The context tells you what's possible and what's not!

Question 8

1. $x = 20$
2. $x = 24$ (correct answer)
3. $x = 18$
4. $x = 16$

Explanation: This question tests your ability to translate a real-world situation into a mathematical equation or inequality, solve it, and interpret the result in the original context. Different contexts lead to different equation types: constant rates give linear equations (like cost = rate × quantity + fee), area problems often give quadratics (like length × width = area), and growth over time gives exponentials (like population = initial × (growth rate)^time). The context clues tell you which form to use. For rate problems like this, we set up: rates add when working together, so Pump A rate 1/12 + Pump B rate 1/x = combined rate 1/8. This gives us 1/12 + 1/x = 1/8. Solving: subtract 1/12 → 1/x = 1/8 - 1/12 = (3-2)/24 = 1/24, so x=24. Interpreting: Pump B takes 24 hours alone. Choice B is correct because it properly sets up the rational equation from context, solves correctly, and interprets appropriately, giving x=24 hours. Choice A makes an arithmetic error: perhaps in subtracting fractions, doing 1/8 - 1/12 = (3-2)/24=1/24 correctly but then misinterpreting. With all the steps in solving word problems—setting up, solving, interpreting—it's easy for calculation errors to slip in. Double-checking arithmetic is always worth it! The foolproof word problem strategy: (1) Read carefully and identify what's unknown—that's your variable, (2) Find what you know—those are your numbers, (3) Look for relationships—how are quantities connected? This gives you the equation, (4) Solve the equation using appropriate methods, (5) Check: does your answer satisfy the equation AND make sense in context? Following these steps systematically prevents most mistakes!

Question 9

1. $2500(1.06)^t - 200 = A$
2. $2500(1.06)^t - 200t = A$ (correct answer)
3. $2500(1.06 - 0.08)^t = A$
4. $2500(1.06)^t \cdot 0.92^t = A$

Explanation: When you encounter problems involving exponential growth with regular withdrawals, you need to think about how these two separate processes affect the account differently. The growth compounds over time, while withdrawals happen at regular intervals and accumulate linearly. The original formula $A = 2500(1.06)^t$ shows exponential growth at 6% annually. However, the investor also withdraws $200 each year for $t$ years, which means total withdrawals equal $200t$. Since withdrawals reduce the account balance, you subtract this linear term from the exponential growth: $2500(1.06)^t - 200t = A$. Choice A ($2500(1.06)^t - 200 = A$) incorrectly assumes only one withdrawal of $200 total, rather than$200 per year for $t$ years. Choice C ($2500(1.06 - 0.08)^t = A$) mistakenly tries to incorporate the $200 withdrawal as a percentage rate decrease, but$200 isn't 8% of the initial investment, and withdrawals don't work as percentage reductions anyway. Choice D ($2500(1.06)^t \cdot 0.92^t = A$) makes a similar error, treating withdrawals as if they reduce the growth rate by 8% annually, which incorrectly converts the fixed $200 into a percentage. Remember: when combining exponential processes with linear processes, keep them separate in your equation. Don't try to convert fixed amounts into percentages or combine them into a single exponential term. Exponential parts stay exponential, linear parts stay linear.

Question 10

1. $0.85x = x - 30$
2. $30 + 0.15x = x$
3. $30 = 0.15x$
4. $30 + 0.85x = x$ (correct answer)

Explanation: When you encounter break-even problems, you need to set up an equation where the total costs of both options are equal. Here, you're comparing the total annual cost of membership shopping versus non-membership shopping. With membership, you pay $30$ upfront plus 85% of your purchases (since you get a 15% discount, you pay 100% - 15% = 85%). So the total cost is $30 + 0.85x$. Without membership, you simply pay the full price $x$ for your purchases. At the break-even point, these costs are equal: $30 + 0.85x = x$. This is answer choice D. Let's examine why the other options are incorrect: Choice A ($0.85x = x - 30$) incorrectly suggests that your discounted purchases equal your full purchases minus $30$. This doesn't account for the membership fee you actually paid. Choice B ($30 + 0.15x = x$) mistakenly uses 0.15 as what you pay, but 0.15 represents your savings, not your payment. You pay 85% of the original price, not 15%. Choice C ($30 = 0.15x$) only considers the membership fee equaling your savings, ignoring that you still have to pay for the discounted items. Study tip: In discount problems, always identify what percentage you actually pay (100% minus the discount percentage) rather than the discount itself. Set up break-even equations by making total costs equal, ensuring you include all fees and payments for each option.

Question 11

1. $-16t^2 + 48t + 6 = 54$ (correct answer)
2. $-16t^2 + 48t + 54 = 6$
3. $-16t^2 + 48t = 54$
4. $54 = -16t^2 + 48t - 6$

Explanation: To find when the ball reaches a height of 54 feet, we set the height function equal to 54: $-16t^2 + 48t + 6 = 54$. Choice B incorrectly moves 54 to the left side and 6 to the right side. Choice C omits the initial height term (+6). Choice D incorrectly changes the sign of the initial height term to negative.

Question 12

1. $32 \leq 2w + 4 \leq 50$
2. $32 \leq 4w + 8 \leq 50$
3. $32 \leq 6w + 8 \leq 50$ (correct answer)
4. $16 \leq 3w + 4 \leq 25$

Explanation: When you encounter word problems involving perimeter and constraints, start by translating the given relationships into algebraic expressions, then set up inequalities based on the constraints. Let's define the variables: width = $w$, and length = $2w + 4$ (since length is 4 feet more than twice the width). The perimeter of a rectangle is $P = 2(\text{length} + \text{width})$, so: $P = 2(w + 2w + 4) = 2(3w + 4) = 6w + 8$ Since the perimeter must be at least 32 feet but no more than 50 feet, we get: $32 \leq 6w + 8 \leq 50$ This confirms answer choice C is correct. Let's examine why the other options are wrong: Choice A ($32 \leq 2w + 4 \leq 50$) incorrectly uses just the length expression instead of the full perimeter formula. This misses that perimeter involves both length and width, doubled. Choice B ($32 \leq 4w + 8 \leq 50$) makes an error in combining the length and width terms. When you add $w + (2w + 4)$, you get $3w + 4$, not $2w + 4$. Choice D ($16 \leq 3w + 4 \leq 25$) correctly identifies that length plus width equals $3w + 4$, but fails to multiply by 2 for the perimeter formula. Additionally, it incorrectly divides the constraint values by 2. Strategy tip: In perimeter problems, always remember that perimeter equals 2 times the sum of length and width. Write out each step: define variables, express all measurements in terms of one variable, apply the perimeter formula, then set up your inequality.

Question 13

1. $500 \cdot 2^{t/3} > 10{,}000$ (correct answer)
2. $500 \cdot 2^{3t} > 10{,}000$
3. $500 + 2t > 10{,}000$
4. $500 \cdot 3^{t/2} > 10{,}000$

Explanation: Since the population doubles every 3 hours, after $t$ hours there have been $t/3$ doubling periods. The population is $500 \cdot 2^{t/3}$, and we need this to exceed 10,000. Choice B uses $2^{3t}$ which would mean the population doubles 3 times per hour instead of once every 3 hours. Choice C uses linear growth instead of exponential growth. Choice D incorrectly uses base 3 and reverses the fraction in the exponent.

Question 14

1. $450 - 15t = 0$
2. $450 - 15(t - 12) = 0$ (correct answer)
3. $330 - 15(t - 20) = 0$
4. $450 + 15(t - 12) = 0$

Explanation: When you encounter a linear function problem involving constant rates of change, you need to establish both the rate and a reference point to build your equation. First, find the drainage rate. The tank loses $450 - 330 = 120$ gallons over $20 - 12 = 8$ minutes, so the rate is $120 ÷ 8 = 15$ gallons per minute. Now you need an equation for when the tank empties. Since the tank drains at 15 gallons per minute, you can use any known point as your reference. Using the 12-minute mark when there were 450 gallons: after $t$ total minutes, the tank will contain $450 - 15(t - 12)$ gallons. For the tank to be empty, set this equal to zero: $450 - 15(t - 12) = 0$. Let's examine why the other choices fail. Choice A, $450 - 15t = 0$, incorrectly assumes the tank started with 450 gallons at time zero, but we know 450 gallons remained after 12 minutes of draining. Choice C, $330 - 15(t - 20) = 0$, uses the 20-minute reference point correctly but would give the same final answer. However, choice D, $450 + 15(t - 12) = 0$, has the wrong sign—it suggests the tank is filling rather than draining. Study tip: In rate problems, always identify your reference point clearly. The expression $(t - \text{reference time})$ represents the time elapsed since that reference point, and you multiply this by the rate of change to find the total change from that reference.

Question 15

1. $180 < 12(n + 5)$
2. $\frac{180}{n + 5} > 12$
3. $\frac{12}{n + 5} < 180$
4. $\frac{180}{n + 5} < 12$ (correct answer)

Explanation: When you encounter a word problem involving inequalities, your first step is to translate the English phrase into mathematical language. The key phrase here is "fewer than 12 members." Since the number of members is represented by $\frac{180}{n + 5}$, and we want this to be fewer than 12, we need the inequality $\frac{180}{n + 5} < 12$. The phrase "fewer than" directly translates to the less-than symbol (<). Looking at the wrong answers: Choice A gives us $180 < 12(n + 5)$, which comes from incorrectly cross-multiplying the correct inequality but then forgetting to include the fraction form. Choice B shows $\frac{180}{n + 5} > 12$, which uses the wrong inequality symbol—this would represent when the club has more than 12 members, not fewer. Choice C presents $\frac{12}{n + 5} < 180$, which flips the numerators incorrectly; this doesn't represent the given situation at all since 12 should not be in the numerator. The correct answer is D: $\frac{180}{n + 5} < 12$. This directly states that the number of members (the entire expression $\frac{180}{n + 5}$) is less than 12. Study tip: When translating word problems to inequalities, write out the English phrase first, then substitute the mathematical expressions. "Fewer than," "less than," and "under" all translate to <, while "more than," "greater than," and "exceeds" translate to >.

Question 16

1. $80 \cdot (0.95)^t + 35 = 20$
2. $80 \cdot (0.95)^t = 35$
3. $80 \cdot (0.95)^t + 20 = 35$ (correct answer)
4. $35 = 80 \cdot (0.95)^t - 20$

Explanation: When you encounter exponential equations in real-world contexts, you need to set up an equation where the given formula equals the target value you're looking for. Here, you have the temperature formula $T = 80 \cdot (0.95)^t + 20$, and you want to find when the temperature equals exactly 35°C. This means you need to substitute 35 for $T$ in the equation, giving you $35 = 80 \cdot (0.95)^t + 20$. Rearranging this gives the equivalent form $80 \cdot (0.95)^t + 20 = 35$, which is answer choice C. Let's see why the other options are incorrect. Choice A ($80 \cdot (0.95)^t + 35 = 20$) incorrectly adds 35 to the exponential term instead of setting the entire expression equal to 35. Choice B ($80 \cdot (0.95)^t = 35$) forgets to include the constant term 20 from the original equation, which represents the ambient temperature the object approaches. Choice D ($35 = 80 \cdot (0.95)^t - 20$) changes the sign of the constant term from +20 to -20, which would represent a completely different cooling scenario. Remember this key strategy: when solving "when does [expression] equal [value]" problems, substitute the target value for the variable in the original equation. Don't modify the structure of the given formula—just set it equal to your target and solve from there.

Question 17

1. $w + 8(45) = 1{,}200$ (correct answer)
2. $w - 8(45) = 1{,}200$
3. $8w + 45 = 1{,}200$
4. $w + 45 = 1{,}200 - 8$

Explanation: The pool starts with $w$ gallons, then gains $8 \times 45 = 360$ gallons after 45 minutes of filling, resulting in a total of 1,200 gallons. So $w + 8(45) = 1,200$. Choice B subtracts the water added instead of adding it. Choice C treats the initial amount as a rate (multiplying by 8) and the time as a constant addition. Choice D incorrectly places the subtraction operation and doesn't account for the rate properly.

Question 18

1. $25 + 0.15m \leq 85$ (correct answer)
2. $25 + 0.15m \geq 85$
3. $0.15m - 25 \leq 85$
4. $25m + 0.15 \leq 85$

Explanation: The total cost is the flat fee ($25) plus the per-mile charge ($0.15m). Since Marcus wants to spend 'at most' $85, the total cost must be less than or equal to$85, giving us $25 + 0.15m ≤ 85$. Choice B uses the wrong inequality direction (≥ instead of ≤). Choice C incorrectly subtracts the flat fee instead of adding it. Choice D incorrectly multiplies the flat fee by miles and treats the per-mile rate as a constant.

Question 19

1. $h = 4$ (correct answer)
2. $h = 5$
3. $h = 6$
4. $h = 3$

Explanation: This question tests your ability to translate a real-world situation into a mathematical equation or inequality, solve it, and interpret the result in the original context. Creating an equation from a word problem means identifying the unknown quantity (your variable h), finding what you know (the numbers), and writing an equation that captures the relationship the problem describes. For example, 'a plumber charges $55 plus$35 per hour for a $195 job' becomes$55 + 35h = 195$, where h is hours worked. Let's break down 'A plumber charges a$55 service fee plus $35 per hour. The total bill was$195$': We need to find hours worked, so let h = hours. From the context, fixed fee plus rate per hour times hours equals total means$55 + 35h = 195$. Solving: subtract 55 from both sides →$35h = 140$, divide by 35 →$h = 4$. In context: the plumber worked 4 hours. Choice A is correct because it properly sets up the equation from context and solves correctly, giving$h=4$, meaning 4 hours for the job. Choice B sets up the equation incorrectly: it forgets the fixed fee, perhaps doing$35h=195$→$h ≈ 5.57$(rounding to 5), but when the context says 'plus$55 service fee,' that means add 55, not ignore it. Reading carefully for relationships is key! The foolproof word problem strategy: (1) Read carefully and identify what's unknown—that's your variable, (2) Find what you know—those are your numbers, (3) Look for relationships—how are quantities connected? This gives you the equation, (4) Solve the equation using appropriate methods, (5) Check: does your answer satisfy the equation AND make sense in context? Following these steps systematically prevents most mistakes!

Question 20

1. $x = 2$ (correct answer)
2. $x = 1$
3. $x = 3$
4. $x = 4$

Explanation: This question tests your ability to translate a real-world situation into a mathematical equation or inequality, solve it, and interpret the result in the original context. Creating an equation from a word problem means identifying the unknown quantity (your variable x), finding what you know (the numbers), and writing an equation that captures the relationship the problem describes. For example, 'mixing solutions' often uses weighted averages for concentrations. Let's break down 'You are mixing a sports drink that should be 30% juice. You have 2 liters of a 60% juice mix and you will add x liters of water (0% juice)': We need to find liters of water, so let x = liters added. From the context, total juice equals desired percent times total volume means 0.6×2 + 0×x = 0.3×(2 + x). Solving: 1.2 = 0.3(2 + x) → 1.2 = 0.6 + 0.3x → 0.6 = 0.3x → x = 2. In context: add 2 liters of water to get 30% juice. Choice C is correct because it properly sets up the equation from context and solves correctly, giving x=2 liters. Choice B makes an arithmetic error: perhaps setting up as 0.6×(2 + x) = 0.3×2 or reversing, leading to wrong x=3. With all the steps in solving word problems—setting up, solving, interpreting—it's easy for calculation errors to slip in. Double-checking arithmetic is always worth it! The foolproof word problem strategy: (1) Read carefully and identify what's unknown—that's your variable, (2) Find what you know—those are your numbers, (3) Look for relationships—how are quantities connected? This gives you the equation, (4) Solve the equation using appropriate methods, (5) Check: does your answer satisfy the equation AND make sense in context? Following these steps systematically prevents most mistakes!