Praxis Math Quiz: Model Real World Problems
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Model Real World ProblemsQuestion 1 of 20

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 mm miles long, which of the following expressions represents the total cost of the ride in dollars?

3.50+0.60m3.50 + 0.60m
3.50+2.40m3.50 + 2.40m
3.50m+0.603.50m + 0.60
3.50+0.15m3.50 + 0.15m
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Praxis Math Quiz

Praxis Math Quiz: Model Real World Problems

Practice Model Real World Problems in Praxis Math 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 Model Real World Problems, giving you a quick way to practice the rules, question types, and explanations that matter most for Praxis Math.

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.

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Question 1

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 mm miles long, which of the following expressions represents the total cost of the ride in dollars?

  1. 3.50+0.60m3.50 + 0.60m
  2. 3.50+2.40m3.50 + 2.40m (correct answer)
  3. 3.50m+0.603.50m + 0.60
  4. 3.50+0.15m3.50 + 0.15m
Explanation: 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 = $2.40.Forarideof. For a ride of mmiles,thevariablecostis(2.40m).Therefore,thetotalcostismiles, the variable cost is (2.40m). Therefore, the total cost is3.50 + 2.40m$.

Question 2

A rectangular garden has a length of LL feet and a width of WW feet. A brick walkway with a uniform width of xx feet is constructed around the entire perimeter of the garden. Which expression represents the area of the brick walkway in square feet?

  1. (L+x)(W+x)LW(L+x)(W+x) - LW
  2. 2Lx+2Wx2Lx + 2Wx
  3. (L+2x)(W+2x)LW(L+2x)(W+2x) - LW (correct answer)
  4. 2(L+x)+2(W+x)2(L+x) + 2(W+x)
Explanation: The area of the garden is LWLW. The walkway of width xx is added to all sides, so the new length is L+2xL+2x and the new width is W+2xW+2x. The total area of the garden and walkway combined is (L+2x)(W+2x)(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(L+2x)(W+2x) - LW.

Question 3

A baker is making a large batch of dough. The recipe requires ff cups of flour for every 13\frac{1}{3} cup of sugar. If the baker uses a total of SS cups of sugar, which expression represents the number of cups of flour needed?

  1. 3fS3fS (correct answer)
  2. fS3\frac{fS}{3}
  3. 3fS\frac{3f}{S}
  4. S3f\frac{S}{3f}
Explanation: The ratio of flour to sugar is ff to 13\frac{1}{3}. This means for every 1 cup of sugar, the baker needs 3f3f cups of flour (since f÷13=f×3=3ff \div \frac{1}{3} = f \times 3 = 3f). If the baker uses SS cups of sugar, the amount of flour needed is SS times the amount needed for one cup, which is 3fS3fS.

Question 4

A phone plan costs $40 per month, which includes 5 gigabytes (GB) of data. For each additional gigabyte of data used over the initial 5 GB, there is a charge of xx dollars. If a user consumes a total of dd gigabytes in a month, and d>5d > 5, which expression represents the total monthly bill?

  1. 40+xd40 + xd
  2. 40+x(d5)40 + x(d-5) (correct answer)
  3. 40+x(5d)40 + x(5-d)
  4. 40x+(d5)40x + (d-5)
Explanation: The base cost is $40. The number of additional gigabytes used is the total gigabytes, dd, minus the included 5 GB, which is (d5)(d-5). The cost for these additional gigabytes is xx dollars per gigabyte, so the additional charge is x(d5)x(d-5). The total monthly bill is the sum of the base cost and the additional charge: 40+x(d5)40 + x(d-5).

Question 5

A salesperson earns a base salary of SS dollars per month, plus a commission of cc% of their total sales, TT, 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?

  1. S+c100(T10000)S + \frac{c}{100}(T-10000) (correct answer)
  2. S+c100TS + \frac{c}{100}T
  3. S+c(T10000)S + c(T-10000)
  4. S+c100(T+10000)S + \frac{c}{100}(T+10000)
Explanation: The salesperson's earnings are their base salary, SS, plus the commission. The commission is calculated only on the portion of sales that exceeds $10,000. This amount is T10,000T - 10,000. The commission rate is cc%, which is c100\frac{c}{100} as a decimal. The commission amount is c100(T10,000)\frac{c}{100}(T - 10,000). Therefore, the total earnings are S+c100(T10000)S + \frac{c}{100}(T-10000).

Question 6

A tank is being filled with water at a constant rate of rr gallons per minute. The tank initially contained gg gallons of water. After mm minutes of filling, the tank is exactly half full. If the total capacity of the tank is CC gallons, which equation models this situation?

  1. g+rm=2Cg + rm = 2C
  2. rmg=C2rm - g = \frac{C}{2}
  3. g+rm=C2g + rm = \frac{C}{2} (correct answer)
  4. r(m+g)=C2r(m+g) = \frac{C}{2}
Explanation: The initial amount of water in the tank is gg gallons. Water is added at a rate of rr gallons per minute for mm minutes, so the amount of water added is rmrm gallons. The total amount of water in the tank after mm minutes is the initial amount plus the added amount: g+rmg + rm. The problem states this amount is equal to half the tank's capacity, C2\frac{C}{2}. Thus, the equation is g+rm=C2g + rm = \frac{C}{2}.

Question 7

An artist is creating a mixture of paint. She mixes xx ounces of a paint that is 20% blue pigment with yy ounces of a paint that is 50% blue pigment. Which expression represents the percentage of blue pigment in the final mixture?

  1. 100×0.2x+0.5yx+y100 \times \frac{0.2x + 0.5y}{x+y} (correct answer)
  2. 0.2x+0.5y2\frac{0.2x + 0.5y}{2}
  3. 100×(0.2x+0.5y)100 \times (0.2x + 0.5y)
  4. x+y0.2x+0.5y\frac{x+y}{0.2x+0.5y}
Explanation: The amount of blue pigment from the first paint is 0.2x0.2x ounces. The amount of blue pigment from the second paint is 0.5y0.5y ounces. The total amount of blue pigment is 0.2x+0.5y0.2x + 0.5y. The total volume of the mixture is x+yx+y ounces. The concentration (or proportion) of blue pigment is the total amount of pigment divided by the total volume: 0.2x+0.5yx+y\frac{0.2x + 0.5y}{x+y}. To express this as a percentage, we multiply by 100.

Question 8

At a conference, the registration fee is $75 per person. However, groups of 10 or more receive a discount of dd dollars off the total registration fee for the entire group. Which expression models the cost for a group of nn people, where n10n \ge 10?

  1. 75(nd)75(n-d)
  2. 75nd75n - d (correct answer)
  3. (75d)n(75-d)n
  4. 75n10d75n - 10d
Explanation: The standard cost for nn people would be 75n75n. The problem states that the discount of dd dollars is applied to the total registration fee, not per person. Therefore, the total cost for the group is the standard cost minus the single discount amount, which is 75nd75n - d.

Question 9

A factory produces widgets at a cost of cc dollars per widget, plus a fixed daily operational cost of FF dollars. The factory sells each widget for ss dollars. Which expression represents the factory's daily profit if it produces and sells xx widgets?

  1. sx(cx+F)sx - (cx + F) (correct answer)
  2. (sc)x+F(s-c)x + F
  3. scFs - c - F
  4. sxcxsx - cx
Explanation: Profit is calculated as Total Revenue - Total Cost. The total revenue from selling xx widgets at ss dollars each is sxsx. The total cost of producing xx widgets is the variable cost (cxcx) plus the fixed daily cost (FF), which is cx+Fcx + F. Therefore, the profit is RevenueCost=sx(cx+F)\text{Revenue} - \text{Cost} = sx - (cx + F).

Question 10

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 E1E_1, E2E_2, and E3E_3 on the exams and PP on the final paper, which expression represents their final grade?

  1. 0.2(E1+E2+E3)+0.4P0.2(E_1+E_2+E_3) + 0.4P (correct answer)
  2. 0.2(E1+E2+E3)+0.8P0.2(E_1+E_2+E_3) + 0.8P
  3. E1+E2+E33+P\frac{E_1+E_2+E_3}{3} + P
  4. 0.6(E1+E2+E3)+0.4P0.6(E_1+E_2+E_3) + 0.4P
Explanation: The three exams are each worth 20%, for a total of 3×20%=60%3 \times 20\% = 60\% of the grade. The final paper is worth the remainder, which is 100%60%=40%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 0.20E1+0.20E2+0.20E3+0.40P0.20E_1 + 0.20E_2 + 0.20E_3 + 0.40P, which can be factored as 0.2(E1+E2+E3)+0.4P0.2(E_1+E_2+E_3) + 0.4P.

Question 11

A farmer wants to fence a rectangular field. The length of the field is 10 meters more than its width, ww. The farmer needs to leave a 3-meter opening for a gate. Which expression represents the total length of fencing required in meters?

  1. 2w+2(w+10)32w + 2(w+10) - 3 (correct answer)
  2. w(w+10)3w(w+10) - 3
  3. 2w+2(w10)32w + 2(w-10) - 3
  4. 2w+2(w+10)+32w + 2(w+10) + 3
Explanation: The width of the field is ww. The length is 10 meters more than the width, so the length is w+10w+10. The perimeter of a rectangle is 2×(length+width)2 \times (\text{length} + \text{width}), so the total perimeter is 2((w+10)+w)=2(2w+10)=4w+202((w+10) + w) = 2(2w+10) = 4w+20. This can also be written as 2w+2(w+10)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)32w + 2(w+10) - 3.

Question 12

At a movie theater, adult tickets cost aa dollars and child tickets cost cc dollars. On Saturday, the theater sold 50 more adult tickets than child tickets. If kk is the number of child tickets sold, which expression represents the total revenue from ticket sales on Saturday?

  1. ak+c(k+50)ak + c(k+50)
  2. ac(2k+50)ac(2k+50)
  3. a(k+50)+cka(k+50) + ck (correct answer)
  4. (a+c)(2k+50)(a+c)(2k+50)
Explanation: Let kk be the number of child tickets sold. The revenue from child tickets is ckck. The number of adult tickets sold is 50 more than child tickets, which is k+50k+50. The revenue from adult tickets is a(k+50)a(k+50). The total revenue is the sum of the revenue from adult and child tickets, which is a(k+50)+cka(k+50) + ck.

Question 13

A city's population, PP, was 120,000 in the year 2010. The population has been decreasing by a constant rate of dd people per year since then. Which expression represents the city's population yy years after 2010?

  1. 120000+dy120000 + dy
  2. d(120000y)d(120000 - y)
  3. 120000dy120000 - dy (correct answer)
  4. 120000dy120000d - y
Explanation: The initial population is 120,000. The population decreases each year, so we will subtract from the initial value. The rate of decrease is dd people per year. Over a period of yy years, the total decrease is the rate multiplied by the number of years, which is dydy. Therefore, the population after yy years is the initial population minus the total decrease: 120000dy120000 - dy.

Question 14

A caterer charges a setup fee of FF dollars plus a cost of pp dollars per person served. A client has a budget that cannot exceed BB dollars. The client expects to have at least nn guests. Which pair of inequalities models this situation, where xx is the number of guests?

  1. F+pxB and xnF+px \leq B \text{ and } x \leq n
  2. F+pxB and xnF+px \geq B \text{ and } x \leq n
  3. F+pxB and xnF+px \leq B \text{ and } x \geq n (correct answer)
  4. F+p(x+n)BF+p(x+n) \leq B
Explanation: The total cost for xx guests is the setup fee FF plus the per-person cost pxpx, which gives F+pxF+px. This cost cannot exceed the budget BB, so F+pxBF+px \leq B. The client expects at least nn guests, which means the number of guests xx must be greater than or equal to nn, so xnx \geq n. Both conditions must be met.

Question 15

The sum of the ages of a parent and a child is 50. In yy years, the parent will be three times as old as the child. If the child's current age is cc, which equation models the relationship of their ages in yy years?

  1. (50c)+y=3(c+y)(50-c) + y = 3(c+y) (correct answer)
  2. (50c)+y=3c+y(50-c) + y = 3c+y
  3. 50+y=3(c+y)50+y = 3(c+y)
  4. 50c=3c+y50-c = 3c+y
Explanation: If the child's current age is cc, and the sum of their ages is 50, then the parent's current age is 50c50-c. In yy years, the child's age will be c+yc+y and the parent's age will be (50c)+y(50-c)+y. At that time, the parent will be three times as old as the child. This relationship is modeled by the equation: Parent's future age = 3 * Child's future age, or (50c)+y=3(c+y)(50-c) + y = 3(c+y).

Question 16

A square has a side length of ss. 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?

  1. (s+5)2=s2+100(s+5)^2 = s^2 + 100 (correct answer)
  2. s2+5=s2+100s^2 + 5 = s^2 + 100
  3. 4(s+5)=4s+1004(s+5) = 4s + 100
  4. (s+5)2=100(s+5)^2 = 100
Explanation: The area of the original square is s2s^2. The side length of the new square is s+5s+5. The area of the new square is (s+5)2(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=s2+100(s+5)^2 = s^2 + 100.

Question 17

A train leaves a station and travels north at a speed of rr miles per hour. Two hours later, a second train leaves the same station and travels north on a parallel track at a speed of ss miles per hour, where s>rs > r. If the second train catches up to the first train after tt hours, which equation represents the relationship between their distances traveled?

  1. r(t+2)=str(t+2) = st (correct answer)
  2. rt=s(t2)rt = s(t-2)
  3. r(t2)=str(t-2) = st
  4. rt=s(t+2)rt = s(t+2)
Explanation: Let tt be the travel time of the second train. Since the first train had a 2-hour head start, its total travel time is t+2t+2 hours. The distance traveled by the first train is its rate times its time, r(t+2)r(t+2). The distance traveled by the second train is its rate times its time, stst. When the second train catches up to the first, their distances from the station are equal. Therefore, the equation is r(t+2)=str(t+2) = st.

Question 18

A car travels at an average speed of ss miles per hour for tt 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?

  1. sts5\frac{st}{s-5} (correct answer)
  2. s5st\frac{s-5}{st}
  3. st5\frac{s}{t-5}
  4. t+5t+5
Explanation: The distance the car travels is given by d=rate×timed = \text{rate} \times \text{time}, so d=std = st. The truck travels this same distance. The truck's speed is 5 mph slower than the car's, so its speed is s5s-5. The time it takes for the truck to travel the distance dd is time=distancerate\text{time} = \frac{\text{distance}}{\text{rate}}. Substituting the expressions for the truck, we get timetruck=ds5\text{time}_{\text{truck}} = \frac{d}{s-5}. Since d=std = st, the expression becomes sts5\frac{st}{s-5}.

Question 19

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 xx be the number of type A widgets and yy be the number of type B widgets produced. Which inequality models the constraint on production time?

  1. 2x+3y1002x + 3y \leq 100 (correct answer)
  2. 2x+3y1002x + 3y \geq 100
  3. x+y100x+y \leq 100
  4. x2+y3100\frac{x}{2} + \frac{y}{3} \leq 100
Explanation: The total time spent producing type A widgets is the time per widget (2 hours) multiplied by the number of widgets (xx), which is 2x2x. The total time spent producing type B widgets is similarly 3y3y. 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+3y1002x + 3y \leq 100.

Question 20

The temperature in degrees Celsius, CC, can be converted to degrees Fahrenheit, FF, using the formula F=95C+32F = \frac{9}{5}C + 32. Which equation models the temperature in degrees Celsius, CC, in terms of a given temperature in degrees Fahrenheit, FF?

  1. C=59F32C = \frac{5}{9}F - 32
  2. C=95(F32)C = \frac{9}{5}(F-32)
  3. C=59(F32)C = \frac{5}{9}(F-32) (correct answer)
  4. C=5F329C = \frac{5F-32}{9}
Explanation: To solve the equation F=95C+32F = \frac{9}{5}C + 32 for CC, we first isolate the term with CC. Subtract 32 from both sides: F32=95CF - 32 = \frac{9}{5}C. Next, multiply both sides by the reciprocal of 95\frac{9}{5}, which is 59\frac{5}{9}, to solve for CC: 59(F32)=C\frac{5}{9}(F - 32) = C.