Model Real World Problems - Praxis Core Math

A salesperson earns a base salary of $S$ dollars per month, plus a commission of $c$ % of their total sales, $T$ , for any sales exceeding $10,000. If the salesperson's total sales for a month are greater than $10,000, which expression represents their total monthly earnings?
Answer: $S + \frac{c}{100}(T-10000)$ The salesperson's earnings are their base salary, $S$ , plus the commission. The commission is calculated only on the portion of sales that exceeds $10,000. This amount is $T - 10,000$ . The commission rate is $c$ %, which is $\frac{c}{100}$ as a decimal. The commission amount is $\frac{c}{100}(T - 10,000)$ . Therefore, the total earnings are $S + \frac{c}{100}(T-10000)$ .
A company has a budget of $B$ dollars for an event. It spends a fixed amount of $500 on venue rental. The remainder of the budget will be spent on meals for attendees, with each meal costing $c$ dollars. Which inequality correctly models the maximum number of attendees, $n$ , that can be served a meal?
Answer: $cn + 500 \leq B$ The total cost of the meals is the cost per meal, $c$ , multiplied by the number of attendees, $n$ , which is $cn$ . The total cost for the event is the sum of the meal costs and the fixed venue rental cost, which is $cn + 500$ . This total cost must be less than or equal to the total budget, $B$ . Therefore, the correct inequality is $cn + 500 \leq B$ .
A company produces two types of widgets, A and B. It takes 2 hours to produce one widget of type A and 3 hours to produce one widget of type B. The company has a maximum of 100 hours of production time available per week. Let $x$ be the number of type A widgets and $y$ be the number of type B widgets produced. Which inequality models the constraint on production time?
Answer: $2x + 3y \leq 100$ The total time spent producing type A widgets is the time per widget (2 hours) multiplied by the number of widgets ( $x$ ), which is $2x$ . The total time spent producing type B widgets is similarly $3y$ . The sum of the time spent on both types of widgets must be less than or equal to the maximum available time, which is 100 hours. This gives the inequality $2x + 3y \leq 100$ .
A train leaves a station and travels north at a speed of $r$ miles per hour. Two hours later, a second train leaves the same station and travels north on a parallel track at a speed of $s$ miles per hour, where $s > r$ . If the second train catches up to the first train after $t$ hours, which equation represents the relationship between their distances traveled?
Answer: $r(t+2) = st$ Let $t$ be the travel time of the second train. Since the first train had a 2-hour head start, its total travel time is $t+2$ hours. The distance traveled by the first train is its rate times its time, $r(t+2)$ . The distance traveled by the second train is its rate times its time, $st$ . When the second train catches up to the first, their distances from the station are equal. Therefore, the equation is $r(t+2) = st$ .
A machine can produce $p$ parts in 8 hours. A new, faster machine can produce the same number of parts in 5 hours. If both machines work together, which equation can be used to find the time, $t$ , in hours, it would take them to produce $p$ parts?
Answer: $\frac{t}{8} + \frac{t}{5} = 1$ This is a work-rate problem. The rate of the first machine is $\frac{1 \text{ job}}{8 \text{ hours}}$ , where one job is producing $p$ parts. The rate of the second machine is $\frac{1 \text{ job}}{5 \text{ hours}}$ . Their combined rate is $\frac{1}{8} + \frac{1}{5}$ . The work done is rate multiplied by time. If they work together for time $t$ to complete 1 job, the equation is $(\frac{1}{8} + \frac{1}{5})t = 1$ . This can be rewritten as $\frac{t}{8} + \frac{t}{5} = 1$ . The variable $p$ is extraneous information used to define one 'job'.
The value of a rare coin, initially $V_0$ , appreciates by $r$ % each year. Which expression represents the value of the coin after $t$ years?
Answer: $V_0(1 + \frac{r}{100})^t$ This is a model of compound growth. The percentage rate $r$ % must be converted to a decimal, $\frac{r}{100}$ . Each year, the value is multiplied by $(1 + \frac{r}{100})$ . After $t$ years, this multiplication has been applied $t$ times. This is represented by the exponential expression $V_0(1 + \frac{r}{100})^t$ .
A farmer wants to fence a rectangular field. The length of the field is 10 meters more than its width, $w$ . The farmer needs to leave a 3-meter opening for a gate. Which expression represents the total length of fencing required in meters?
Answer: $2w + 2(w+10) - 3$ The width of the field is $w$ . The length is 10 meters more than the width, so the length is $w+10$ . The perimeter of a rectangle is $2 \times (\text{length} + \text{width})$ , so the total perimeter is $2((w+10) + w) = 2(2w+10) = 4w+20$ . This can also be written as $2w + 2(w+10)$ . Since there is a 3-meter opening for a gate, we must subtract this from the total perimeter. The required length of fencing is $2w + 2(w+10) - 3$ .
A car travels at an average speed of $s$ miles per hour for $t$ hours. A truck travels the same distance, but its average speed is 5 miles per hour slower than the car's speed. Which expression represents the time, in hours, that the truck takes to travel the distance?
Answer: $\frac{st}{s-5}$ The distance the car travels is given by $d = \text{rate} \times \text{time}$ , so $d = st$ . The truck travels this same distance. The truck's speed is 5 mph slower than the car's, so its speed is $s-5$ . The time it takes for the truck to travel the distance $d$ is $\text{time} = \frac{\text{distance}}{\text{rate}}$ . Substituting the expressions for the truck, we get $\text{time}_{\text{truck}} = \frac{d}{s-5}$ . Since $d = st$ , the expression becomes $\frac{st}{s-5}$ .
An investor deposits $P$ dollars into an account that earns 3% annual interest, compounded annually. At the end of each year, the investor withdraws $100. Which expression represents the account balance after two years?
Answer: $P(1.03)^2 - 100(1.03) - 100$ After the first year, the balance is the principal plus interest, $P(1.03)$ . Then $100 is withdrawn, leaving $P(1.03) - 100$ . For the second year, this new balance earns interest, becoming $(P(1.03) - 100)(1.03)$ . Then another $100 is withdrawn. The final balance is $(P(1.03) - 100)(1.03) - 100$ . Distributing the $1.03$ gives $P(1.03)^2 - 100(1.03) - 100$ .
The sum of three consecutive even integers is $S$ . If the smallest of these integers is $x$ , which equation correctly relates $x$ and $S$ ?
Answer: $x + (x+2) + (x+4) = S$ Consecutive even integers differ by 2. If the smallest integer is $x$ , the next even integer is $x+2$ , and the one after that is $x+4$ . The sum of these three integers is $x + (x+2) + (x+4)$ . Setting this sum equal to $S$ gives the equation $x + (x+2) + (x+4) = S$ .
A square has a side length of $s$ . If the side length is increased by 5 units, the area of the new square is 100 square units greater than the area of the original square. Which equation models this relationship?
Answer: $(s+5)^2 = s^2 + 100$ The area of the original square is $s^2$ . The side length of the new square is $s+5$ . The area of the new square is $(s+5)^2$ . The problem states that the new area is 100 square units greater than the original area. This translates to the equation: New Area = Original Area + 100, or $(s+5)^2 = s^2 + 100$ .
A baker is making a large batch of dough. The recipe requires $f$ cups of flour for every $\frac{1}{3}$ cup of sugar. If the baker uses a total of $S$ cups of sugar, which expression represents the number of cups of flour needed?
Answer: $3fS$ The ratio of flour to sugar is $f$ to $\frac{1}{3}$ . This means for every 1 cup of sugar, the baker needs $3f$ cups of flour (since $f \div \frac{1}{3} = f \times 3 = 3f$ ). If the baker uses $S$ cups of sugar, the amount of flour needed is $S$ times the amount needed for one cup, which is $3fS$ .
The temperature in degrees Celsius, $C$ , can be converted to degrees Fahrenheit, $F$ , using the formula $F = \frac{9}{5}C + 32$ . Which equation models the temperature in degrees Celsius, $C$ , in terms of a given temperature in degrees Fahrenheit, $F$ ?
Answer: $C = \frac{5}{9}(F-32)$ To solve the equation $F = \frac{9}{5}C + 32$ for $C$ , we first isolate the term with $C$ . Subtract 32 from both sides: $F - 32 = \frac{9}{5}C$ . Next, multiply both sides by the reciprocal of $\frac{9}{5}$ , which is $\frac{5}{9}$ , to solve for $C$ : $\frac{5}{9}(F - 32) = C$ .
A small business owner's profit, $P$ , is the difference between revenue, $R$ , and cost, $C$ . The revenue from selling $x$ items is $25x$ . The cost of producing $x$ items is $500 + 15x$ . The owner wants to make a profit of at least $1000. Which inequality models this situation?
Answer: $25x - (500+15x) \geq 1000$ Profit is Revenue minus Cost, so $P = R - C$ . Here, $R = 25x$ and $C = 500 + 15x$ . So the profit is $P = 25x - (500 + 15x)$ . The owner wants the profit to be 'at least $1000', which means greater than or equal to $1000. Therefore, the correct inequality is $25x - (500+15x) \geq 1000$ .
A library has $b$ hardcover books. The number of paperback books is 15 fewer than twice the number of hardcover books. Which expression represents the total number of books in the library?
Answer: $b + (2b - 15)$ Let $b$ be the number of hardcover books. The number of paperback books is '15 fewer than twice the number of hardcover books,' which translates to $2b - 15$ . The total number of books is the sum of hardcover and paperback books: $b + (2b - 15)$ .
A rectangular garden has a length of $L$ feet and a width of $W$ feet. A brick walkway with a uniform width of $x$ feet is constructed around the entire perimeter of the garden. Which expression represents the area of the brick walkway in square feet?
Answer: $(L+2x)(W+2x) - LW$ The area of the garden is $LW$ . The walkway of width $x$ is added to all sides, so the new length is $L+2x$ and the new width is $W+2x$ . The total area of the garden and walkway combined is $(L+2x)(W+2x)$ . To find the area of only the walkway, subtract the garden's area from the total area: $(L+2x)(W+2x) - LW$ .
An artist is creating a mixture of paint. She mixes $x$ ounces of a paint that is 20% blue pigment with $y$ ounces of a paint that is 50% blue pigment. Which expression represents the percentage of blue pigment in the final mixture?
Answer: $100 \times \frac{0.2x + 0.5y}{x+y}$ The amount of blue pigment from the first paint is $0.2x$ ounces. The amount of blue pigment from the second paint is $0.5y$ ounces. The total amount of blue pigment is $0.2x + 0.5y$ . The total volume of the mixture is $x+y$ ounces. The concentration (or proportion) of blue pigment is the total amount of pigment divided by the total volume: $\frac{0.2x + 0.5y}{x+y}$ . To express this as a percentage, we multiply by 100.
A water delivery service charges a monthly fee of $20, which includes the first $n$ bottles of water. For each additional bottle delivered beyond the first $n$ bottles, the service charges $1.50. If a customer has $b$ bottles delivered in a month, and $b > n$ , which expression represents the customer's total monthly charge?
Answer: $20 + 1.50(b-n)$ The monthly fee is $20. The number of additional bottles is the total number of bottles, $b$ , minus the number included in the plan, $n$ . This amount is $b-n$ . The cost for these additional bottles is $1.50 each, so the additional charge is $1.50(b-n)$ . The total monthly charge is the base fee plus the additional charge: $20 + 1.50(b-n)$ .
A taxi service charges a flat fee of $3.50 for every ride, plus an additional $0.60 per quarter-mile. If a ride is $m$ miles long, which of the following expressions represents the total cost of the ride in dollars?
Answer: $3.50 + 2.40m$ The total cost is the flat fee plus the variable cost. The flat fee is $3.50. The variable cost is $0.60 per quarter-mile. Since there are 4 quarter-miles in a mile, the cost per mile is $4 \times \$0.60 = \$2.40$ . For a ride of $m$ miles, the variable cost is (2.40m). Therefore, the total cost is $3.50 + 2.40m$ .
A student's grade in a class is determined by three exams, each worth 20% of the final grade, and a final paper, worth the remainder. If the student scores $E_1$ , $E_2$ , and $E_3$ on the exams and $P$ on the final paper, which expression represents their final grade?
Answer: $0.2(E_1+E_2+E_3) + 0.4P$ The three exams are each worth 20%, for a total of $3 \times 20\% = 60\%$ of the grade. The final paper is worth the remainder, which is $100\% - 60\% = 40\%$ . The weighted score is calculated by multiplying each score by its weight (in decimal form) and summing the results. The expression for the final grade is $0.20E_1 + 0.20E_2 + 0.20E_3 + 0.40P$ , which can be factored as $0.2(E_1+E_2+E_3) + 0.4P$ .

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