Algebra Quiz: Creating Solving One Variable Equations Inequalities

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Creating Solving One Variable Equations InequalitiesQuestion 1 of 20

Tickets to a school play cost $9 each, and there is a one-time online fee of $4 per order. Jordan has at most $40 to spend. Write an inequality representing the number of tickets Jordan can buy.

Let $t$ = the number of tickets.

9t + 4 \ge 40

9t - 4 \le 40

4t + 9 \le 40

9t + 4 \le 40

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Algebra Quiz

Algebra Quiz: Creating Solving One Variable Equations Inequalities

Practice Creating Solving One Variable Equations Inequalities in Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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.

All questions

Question 1

Tickets to a school play cost $9 each, and there is a one-time online fee of $4 per order. Jordan has at most $40 to spend. Write an inequality representing the number of tickets Jordan can buy.

Let $t$ = the number of tickets.

$9t + 4 \ge 40$
$9t - 4 \le 40$
$4t + 9 \le 40$
$9t + 4 \le 40$ (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

You have at most $50 to spend on notebooks that cost $4 each. Write an inequality representing this constraint and find the maximum number of notebooks you can buy. Let $n$ = number of notebooks.

Maximum $n=11$
Maximum $n=12$ (correct answer)
Maximum $n=13$
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 3

Pump A can fill a pool in 12 hours. Pump B can fill the same pool in $x$ hours. Working together, they fill the pool in 8 hours. Set up and solve an equation to find how long Pump B takes to fill the pool alone.

Let $x$ = Pump B's time (in hours).

$x = 20$
$x = 24$ (correct answer)
$x = 18$
$x = 16$

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 4

A ball is thrown upward from the ground. Its height (in feet) after $t$ seconds is given by $h = -16t^2 + 64t$ . Solve for the time(s) when the ball is on the ground.

Let $t$ = time in seconds.

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

A rectangular garden has a length that is 5 feet more than its width. The area is 84 square feet. Write and solve an equation to find the width of the garden.

Let $w$ = the width (in feet).

$w = 14$
$w = 12$
$w = 7$ (correct answer)
$w = 9$

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 6

A gym charges a one-time sign-up fee of $30 and then $18 per month. You have at most $150 to spend total. What is the maximum whole number of months you can afford?

Maximum $m$ is $7$
Maximum $m$ is $5$
Maximum $m$ is $6$ (correct answer)
Maximum $m$ is $8$

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 'A gym charges a one-time sign-up fee of $30 and then $18 per month. You have at most $150 to spend total' uses the phrase 'at most $150,' which signals ≤. Setting up: 30 + 18m ≤ 150. Solving: subtract 30 → 18m ≤ 120, divide by 18 → m ≤ 6.666. This means maximum of 6 whole months, since you can't have a fraction of a month here. Choice B is correct because it properly sets up the inequality from context, solves correctly, and interprets appropriately, giving maximum m=6 whole months. Choice A has the inequality symbol backwards or ignores the fee: perhaps doing 18m ≤ 150 → m ≤ 8.33 (to 8, but wrong), but 'at most $150' means ≤ after adding $30. These phrases can be tricky—memorizing which direction they go really helps! 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

An investment account grows according to the formula $A = 2500(1.06)^t$ , where $A$ is the account value and $t$ is time in years. At the same time, the investor makes annual withdrawals of $200. Which equation represents the net account balance when accounting for both growth and withdrawals over $ $t$ $ years?

$2500(1.06)^t - 200 = A$
$2500(1.06)^t - 200t = A$ (correct answer)
$2500(1.06 - 0.08)^t = A$
$2500(1.06)^t \cdot 0.92^t = A$

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

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 8

A store offers a membership where customers pay $30 annually and then receive a 15% discount on all purchases. Without membership, customers pay full price. For what annual spending amount $ $x$ $ (in dollars) would the membership cost exactly break even with non-membership shopping?

$0.85x = x - 30$
$30 + 0.15x = x$
$30 = 0.15x$
$30 + 0.85x = x$ (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 9

The height of a ball thrown upward can be modeled by $h(t) = -16t^2 + 48t + 6$ , where $h$ is the height in feet and $t$ is the time in seconds. Which equation would you solve to find when the ball reaches a height of 54 feet?

$-16t^2 + 48t + 6 = 54$ (correct answer)
$-16t^2 + 48t + 54 = 6$
$-16t^2 + 48t = 54$
$54 = -16t^2 + 48t - 6$

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 10

A rectangular garden has a length that is 4 feet more than twice its width. If the perimeter must be at least 32 feet but no more than 50 feet, which compound inequality represents the possible widths $w$ of the garden?

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

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 11

A bacteria population doubles every 3 hours. If the initial population is 500 bacteria, which equation can be used to find the time $t$ (in hours) when the population first exceeds 10,000?

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

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 12

A water tank is being drained at a constant rate. After 12 minutes, the tank contains 450 gallons. After 20 minutes, it contains 330 gallons. Which equation can be used to find the time $t$ (in minutes) when the tank will be completely empty?

$450 - 15t = 0$
$450 - 15(t - 12) = 0$ (correct answer)
$330 - 15(t - 20) = 0$
$450 + 15(t - 12) = 0$

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 13

The number of members in a chess club can be represented by $\frac{180}{n + 5}$ , where $n$ is the number of years since the club started. Which inequality represents when the club will have fewer than 12 members?

$180 < 12(n + 5)$
$\frac{180}{n + 5} > 12$
$\frac{12}{n + 5} < 180$
$\frac{180}{n + 5} < 12$ (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 14

The temperature $T$ (in degrees Celsius) of a cooling object after $t$ minutes follows the equation $T = 80 \cdot (0.95)^t + 20$ . Which equation would you solve to find when the temperature drops to exactly 35°C?

$80 \cdot (0.95)^t + 35 = 20$
$80 \cdot (0.95)^t = 35$
$80 \cdot (0.95)^t + 20 = 35$ (correct answer)
$35 = 80 \cdot (0.95)^t - 20$

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 15

A swimming pool is being filled at a rate of 8 gallons per minute. After 45 minutes, the pool contains 1,200 gallons. Which equation can be used to find the initial amount of water $w$ (in gallons) that was already in the pool?

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

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 16

A rental car company charges a flat fee of $25 plus $0.15 per mile driven. If Marcus wants to spend at most $85 on his rental, which inequality correctly represents the maximum number of miles $ $m$ $ he can drive?

$25 + 0.15m \leq 85$ (correct answer)
$25 + 0.15m \geq 85$
$0.15m - 25 \leq 85$
$25m + 0.15 \leq 85$

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 17

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). Write and solve an equation to find how many liters of water to add.

Let $x$ = liters of water added.

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

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!

Question 18

Two pumps fill a tank together in 4 hours. Pump A alone can fill the tank in 6 hours. Write and solve an equation to find how long Pump B would take to fill the tank alone.

Let $x$ = the number of hours for Pump B alone.

$x = 8$ hours
$x = 10$ hours
$x = 12$ hours (correct answer)
$x = 24$ hours

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 for working together, so (1/6) + (1/x) = 1/4. This gives us rational equation. Solving: 1/x = 1/4 - 1/6 = (3-2)/12 = 1/12, so x=12. Interpreting: Pump B takes 12 hours alone to fill the tank. Choice C is correct because it properly sets up the rational equation from the work rates context, solves correctly, and interprets appropriately, giving x=12 hours. Choice B makes an arithmetic error: perhaps subtracting rates incorrectly as 1/6 - 1/4 = -1/12, but it's together rate minus A gives B. 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 19

A student wants to buy notebooks that cost $4 each. They have a coupon for $6 off the total purchase. The student can spend at most $30. What is the maximum number of notebooks they can buy?

Let $n$ = the number of notebooks.

$n = 7$
$n = 8$
$n = 6$
$n = 9$ (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 'notebooks $4 each, $6 off total, at most $30' uses the phrase 'at most,' which signals ≤. Setting up: cost times number minus coupon ≤ budget, so 4n - 6 ≤ 30. Solving: add 6 → 4n ≤ 36, divide by 4 → n ≤ 9. This means maximum of 9 whole notebooks. Choice D is correct because it properly sets up the inequality from the context with the coupon subtracted and uses ≤ for 'at most,' giving n=9 as the maximum integer. Choice C solves the inequality correctly but doesn't check the context: n=8 is possible, but the question asks for maximum, so 9 fits since 4*9 -6=30 exactly. Always ask: does my answer make sense in the real world? This catches a lot of mistakes! 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 20

A ball is thrown upward from the ground. Its height (in feet) after $t$ seconds is given by $h = -16t^2 + 64t$ . Solve for the time(s) when the ball is on the ground.

Let $t$ = time in seconds, and set $h = 0$ .

$t = 0$ and $t = 2$
$t = 2$ only
$t = 4$ only
$t = 0$ and $t = 4$ (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 a motion problem, which means quadratic! Let t = time. The relationship 'height h = -16t² + 64t, find when on ground' translates to set h=0: -16t² + 64t = 0. Factoring: -16t(t - 4) = 0, so t=0 or t=4. Checking context: both make sense as start and landing times. So valid solutions t=0 and t=4 with interpretation: the ball is on the ground at launch and after 4 seconds. Choice B is correct because it properly sets up the quadratic from the context, solves correctly finding both roots, and interprets appropriately, giving the times when height is zero. Choice D finds one solution to the quadratic but misses the other (or keeps only one). When solving t² - 4t = 0, we get two values: 0 and 4. Checking context: both are valid for projectile motion! 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!