Middle School Math Quiz: Solve Real World System Problems
20 questions · exam conditions
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Solve Real World System ProblemsQuestion 1 of 20

A student claims that (x,y)=(4,7)(x,y)=(4,7) is the solution to the system that models a fundraiser where xx is the number of large posters and yy is the number of small posters. Large posters cost $5 each, small posters cost $2 each, 11 posters were sold, and $34 was collected. Is the student correct?​​

Yes, because 4+7=114+7=11 and 5(4)+2(7)=345(4)+2(7)=34.
No, because 4+7≠114+7\ne 11.
Yes, because 5+2=75+2=7 and 11+34=4511+34=45.
No, because 5(4)+2(7)=335(4)+2(7)=33, not 3434.
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Middle School Math Quiz

Middle School Math Quiz: Solve Real World System Problems

Practice Solve Real World System Problems in Middle School 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 Solve Real World System Problems, giving you a quick way to practice the rules, question types, and explanations that matter most for Middle School 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.

All questions

Question 1

A student claims that (x,y)=(4,7)(x,y)=(4,7) is the solution to the system that models a fundraiser where xx is the number of large posters and yy is the number of small posters. Large posters cost $5 each, small posters cost $2 each, 11 posters were sold, and $34 was collected. Is the student correct?​​

  1. Yes, because 4+7=114+7=11 and 5(4)+2(7)=345(4)+2(7)=34. (correct answer)
  2. No, because 4+7≠114+7\ne 11.
  3. Yes, because 5+2=75+2=7 and 11+34=4511+34=45.
  4. No, because 5(4)+2(7)=335(4)+2(7)=33, not 3434.
Explanation: This question tests setting up and solving systems of equations from real-world scenarios requiring two equations for two unknowns with context interpretation, then verifying a proposed solution. The process involves: (1) defining variables such as x for large posters and y for small, (2) writing x + y = 11 for total posters, (3) writing 5x + 2y = 34 for revenue, (4) checking if (4,7) satisfies both, (5) interpreting if it's the solution, and (6) verifying calculations. Plugging in: 4 + 7 = 11 (true) and 5(4) + 2(7) = 20 + 14 = 34 (true), so yes. Common errors include miscalculation like 20 + 14 = 33 or ignoring one equation. To avoid errors: set up system first, plug values into both equations carefully, and confirm both are satisfied before concluding.

Question 2

A store sells notebooks and pens. A notebook costs $3 and a pen costs $1. A student buys 18 items total and spends $38. Let $nbethenumberofnotebooksandbe the number of notebooks andp$ be the number of pens. How many notebooks did the student buy?

  1. 12
  2. 14
  3. 10 (correct answer)
  4. 8
Explanation: This question tests setting up and solving systems of equations from real-world scenarios requiring two equations for two unknowns with context interpretation. The process involves: (1) defining variables such as nn for notebooks and pp for pens, (2) writing n+p=18n + p = 18 for total items, (3) writing 3n+p=383n + p = 38 for total cost, (4) solving, (5) interpreting (number of notebooks), and (6) verifying. Solving correctly: subtract first from second: (3n+p)−(n+p)=38−18(3n + p) - (n + p) = 38 - 18, 2n=202n = 20, n=10n = 10, then p=8p = 8. These satisfy: 10+8=1810 + 8 = 18 and 3(10)+8=30+8=383(10) + 8 = 30 + 8 = 38. Common errors include incorrect cost equation or not solving the system fully. To avoid errors: define clearly, translate costs and counts, use elimination or substitution, interpret the asked quantity, and verify.

Question 3

A juice mix is made using apple juice and grape juice. The total amount of mix is 10 liters. Apple juice costs $3 per liter and grape juice costs $5 per liter. The total cost is $38. Let $abelitersofapplejuiceandbe liters of apple juice andg$ be liters of grape juice. How many liters of each were used?

  1. a=6, g=4a=6,\ g=4 (correct answer)
  2. a=5, g=5a=5,\ g=5
  3. a=4, g=6a=4,\ g=6
  4. a=7, g=3a=7,\ g=3
Explanation: This problem tests setting up and solving systems from real-world scenarios requiring two equations for two unknowns with context interpretation. The process involves: (1) defining variables with units (let a = liters of apple juice, g = liters of grape juice), (2) writing total volume equation (a + g = 10), (3) writing total cost equation (3a + 5g = 38), (4) solving (a = 10 - g, 3(10 - g) + 5g = 30 - 3g + 5g = 30 + 2g = 38, 2g = 8, g = 4, a = 6), (5) interpreting (6 liters apple, 4 liters grape), (6) verifying (6 + 4 = 10, 18 + 20 = 38). This matches choice A. Equations correctly represent totals. Errors: swapping costs or wrong solving like B. Setup: identify quantities, define, translate, solve, interpret, verify. Ensure non-negative sensible values.

Question 4

The sum of two siblings' ages is 45 years, and the older sibling is 5 years older than the younger sibling. Let oo be the older sibling's age and yy be the younger sibling's age. What are their ages?​​

  1. o=20, y=25o=20,\ y=25
  2. o=25, y=20o=25,\ y=20 (correct answer)
  3. o=30, y=15o=30,\ y=15
  4. o=22, y=23o=22,\ y=23
Explanation: This question tests setting up and solving systems of equations from real-world scenarios requiring two equations for two unknowns with context interpretation. The process involves: (1) defining variables such as o for older sibling's age and y for younger's age in years, (2) writing the equation for sum: o + y = 45, (3) writing the equation for difference: o = y + 5, (4) solving by substitution, (5) interpreting the results, and (6) verifying. Solving correctly: substitute o = y + 5 into the first to get (y + 5) + y = 45, 2y + 5 = 45, 2y = 40, y = 20, o = 25. These satisfy both: 25 + 20 = 45 and 25 - 20 = 5. Common errors include reversing older and younger or misstating the difference, leading to pairs like o=20, y=25. To avoid errors: identify relationships clearly, define variables precisely, solve systematically, check for sensibility (ages positive), and verify both conditions.

Question 5

The sum of two siblings' ages is 45 years, and the older sibling is 5 years older than the younger sibling. Let oo be the older sibling's age and yy be the younger sibling's age. What are their ages?

  1. o=22, y=23o=22,\ y=23
  2. o=30, y=15o=30,\ y=15
  3. o=25, y=20o=25,\ y=20 (correct answer)
  4. o=20, y=25o=20,\ y=25
Explanation: This question tests setting up and solving systems of equations from real-world scenarios requiring two equations for two unknowns with context interpretation. The process involves: (1) defining variables such as o for older sibling's age and y for younger's age in years, (2) writing the equation for sum: o + y = 45, (3) writing the equation for difference: o = y + 5, (4) solving by substitution, (5) interpreting the results, and (6) verifying. Solving correctly: substitute o = y + 5 into the first to get (y + 5) + y = 45, 2y + 5 = 45, 2y = 40, y = 20, o = 25. These satisfy both: 25 + 20 = 45 and 25 - 20 = 5. Common errors include reversing older and younger or misstating the difference, leading to pairs like o=20, y=25. To avoid errors: identify relationships clearly, define variables precisely, solve systematically, check for sensibility (ages positive), and verify both conditions.

Question 6

Two friends start biking toward each other from towns that are 54 miles apart. One bikes at 12 mph and the other bikes at 6 mph. Let tt be the time in hours until they meet, and let dd be the distance (in miles) the 12 mph biker travels. Which system correctly models the situation, and what is tt?

  1. System: d=12td=12t and d+6t=54d+6t=54; t=4.5t=4.5 (correct answer)
  2. System: d=12td=12t and d−6t=54d-6t=54; t=3t=3
  3. System: d=12td=12t and d+6t=54d+6t=54; t=3t=3
  4. System: d=6td=6t and d+12t=54d+12t=54; t=3t=3
Explanation: This problem tests setting up and solving systems from real-world scenarios requiring two equations for two unknowns with context interpretation. The process involves: (1) defining variables with units (t = time in hours, d = distance in miles by 12 mph biker), (2) writing distance equation (d = 12t), (3) writing meeting equation (d + 6t = 54), (4) solving (substitute: 12t + 6t = 54, 18t = 54, t = 3), (5) interpreting (they meet after 3 hours), (6) verifying (d=36, 36+18=54). This matches choice A. System correct as it models distances summing to 54. Errors: wrong speeds or signs like C, wrong t like D. Setup: understand motion, define, equations from relations, solve, check. Consider relative speed for intuition.

Question 7

A rectangle has perimeter 40 cm. Its length is 3 cm more than its width. Let ll be the length and ww be the width. Solve the system to find the dimensions of the rectangle.

  1. w=10w=10 cm and l=13l=13 cm
  2. w=8.5w=8.5 cm and l=11.5l=11.5 cm (correct answer)
  3. w=7w=7 cm and l=13l=13 cm
  4. w=9.5w=9.5 cm and l=12.5l=12.5 cm
Explanation: This problem tests setting up and solving systems of equations from real-world scenarios requiring two equations for two unknowns, along with interpreting the solution in context. The process involves: (1) defining variables with units (let l = length in cm, w = width in cm), (2) writing the perimeter equation (2l + 2w = 40 or l + w = 20), (3) writing the relation (l = w + 3), (4) solving by substitution (w + 3 + w = 20, 2w = 17, w = 8.5, l = 11.5), (5) interpreting (width 8.5 cm, length 11.5 cm), and (6) verifying perimeter and relation. The correct system is l + w = 20 and l = w + 3, yielding w = 8.5 cm and l = 11.5 cm. These equations are correct as they simplify the perimeter and capture the dimension difference. Common errors include forgetting to halve the perimeter or reversing the relation, leading to integer but incorrect dimensions like w=7, l=13. To set up correctly: express perimeter properly, define relations clearly, solve accurately, ensure dimensions make sense, and verify. Avoid errors like using incorrect coefficients or not checking units.

Question 8

A store sells notebooks and pens. A notebook costs $3 and a pen costs $1. A student buys 18 items total and spends $38. Let $nbethenumberofnotebooksandbe the number of notebooks andp$ be the number of pens. How many notebooks did the student buy?​​

  1. 10 (correct answer)
  2. 12
  3. 8
  4. 14
Explanation: This question tests setting up and solving systems of equations from real-world scenarios requiring two equations for two unknowns with context interpretation. The process involves: (1) defining variables such as n for notebooks and p for pens, (2) writing n + p = 18 for total items, (3) writing 3n + p = 38 for total cost, (4) solving, (5) interpreting (number of notebooks), and (6) verifying. Solving correctly: subtract first from second: (3n + p) - (n + p) = 38 - 18, 2n = 20, n = 10, then p = 8. These satisfy: 10 + 8 = 18 and 3(10) + 8 = 30 + 8 = 38. Common errors include incorrect cost equation or not solving the system fully. To avoid errors: define clearly, translate costs and counts, use elimination or substitution, interpret the asked quantity, and verify.

Question 9

Two friends start biking toward each other from towns that are 300 miles apart. One bikes at 60 mph and the other bikes at 40 mph. Let tt be the time in hours until they meet, and let dd be the distance (in miles) the 60 mph biker travels. Which system correctly models the situation, and what is tt?​​

  1. System: d=40td=40t and d+60t=300d+60t=300; t=5t=5
  2. System: d=60td=60t and d+40t=300d+40t=300; t=3t=3 (correct answer)
  3. System: 60t+40t=30060t+40t=300; t=6t=6
  4. System: 60t−40t=30060t-40t=300; t=15t=15
Explanation: This question tests setting up and solving systems of equations from real-world scenarios requiring two equations for two unknowns with context interpretation. The process involves: (1) defining variables such as t for time in hours and d for distance traveled by the 60 mph biker in miles, (2) writing d = 60t for the faster biker's distance, (3) writing d + 40t = 300 for total distance, (4) solving, (5) interpreting, and (6) verifying. Solving correctly: substitute d = 60t into the second to 60t + 40t = 300, 100t = 300, t = 3. This fits: d = 180, and 180 + 120 = 300. Common errors include switching speeds or using subtraction instead of addition for approaching. To avoid errors: understand relative motion (combined speed when toward each other), set up equations based on definitions, solve, check units, and verify.

Question 10

A gym sells day passes and monthly passes. A day pass costs $6 and a monthly pass costs $30. On one day, the gym sold 40 total passes and collected $720. Let $dbethenumberofdaypassesandbe the number of day passes andmbethenumberofmonthlypasses.Whichsystemrepresentsthesituationandwhatarebe the number of monthly passes. Which system represents the situation and what aredandandm$?​​

  1. System: d+m=720d+m=720 and 6d+30m=406d+30m=40; d=20, m=20d=20,\ m=20
  2. System: d+m=40d+m=40 and 6d+30m=7206d+30m=720; d=20, m=20d=20,\ m=20 (correct answer)
  3. System: d+m=40d+m=40 and 6d+30m=7206d+30m=720; d=10, m=30d=10,\ m=30
  4. System: d+m=40d+m=40 and 30d+6m=72030d+6m=720; d=20, m=20d=20,\ m=20
Explanation: This question tests setting up and solving systems of equations from real-world scenarios requiring two equations for two unknowns with context interpretation. The process involves: (1) defining variables such as d for day passes and m for monthly passes, (2) writing d + m = 40 for total passes, (3) writing 6d + 30m = 720 for revenue, (4) solving, (5) interpreting, and (6) verifying. Solving correctly: from d = 40 - m, 6(40 - m) + 30m = 720, 240 - 6m + 30m = 720, 240 + 24m = 720, 24m = 480, m = 20, d = 20. These fit both equations. Common errors include swapping costs or totals, like using d + m = 720. To avoid errors: match constraints to equations carefully, solve step-by-step, ensure non-negative integers, and verify.

Question 11

A movie theater sells small popcorn for $4 and large popcorn for $6. On one night, they sold 50 popcorns total and collected $240. Let $sbethenumberofsmallpopcornsandbe the number of small popcorns andl$ be the number of large popcorns. Which solution matches the situation?

  1. s=30, l=20s=30,\ l=20 (correct answer)
  2. s=20, l=30s=20,\ l=30
  3. s=10, l=40s=10,\ l=40
  4. s=40, l=10s=40,\ l=10
Explanation: This problem tests setting up and solving systems from real-world scenarios requiring two equations for two unknowns with context interpretation. The process involves: (1) defining variables (s = number of small popcorns, l = number of large popcorns), (2) writing total sold (s + l = 50), (3) writing revenue (4s + 6l = 240), (4) solving (s = 50 - l, 4(50 - l) + 6l = 200 + 2l = 240, 2l = 40, l = 20, s = 30), (5) interpreting (30 small, 20 large), (6) verifying (30 + 20 = 50, 120 + 120 = 240). This matches choice A. Equations correct for counts and earnings. Errors: reversing like B or extremes like C. Setup: identify, define, equations, solve, check positives.

Question 12

A science club needs 20 liters of a 30% salt solution. They have a 20% solution and a 50% solution. Let xx be liters of the 20% solution and yy be liters of the 50% solution. Which system correctly represents the situation?

  1. x+y=30x+y=30 and 0.2x+0.5y=200.2x+0.5y=20 (correct answer)
  2. x+y=20x+y=20 and 0.2x+0.5y=60.2x+0.5y=6
  3. x+y=20x+y=20 and 0.2x+0.5y=300.2x+0.5y=30
  4. x+y=20x+y=20 and 0.3x+0.3y=60.3x+0.3y=6
Explanation: This problem tests setting up and solving systems from real-world scenarios requiring two equations for two unknowns with context interpretation. The process involves: (1) defining variables with units (x = liters of 20% solution, y = liters of 50% solution), (2) writing total volume (x + y = 20), (3) writing salt amount (0.2x + 0.5y = 6, since 30% of 20 is 6), (4) solving if needed, but question asks for system, (5) interpreting mixture, (6) verifying consistency. This matches choice A. System correct for volume and concentration. Errors: wrong total salt like B or swapped totals like C. Setup: understand mixing, define, balance quantities and amounts, check units.

Question 13

A gym sells day passes and monthly passes. A day pass costs $6 and a monthly pass costs $30. On one day, the gym sold 40 total passes and collected $720. Let $dbethenumberofdaypassesandbe the number of day passes andmbethenumberofmonthlypasses.Whichsystemrepresentsthesituationandwhatarebe the number of monthly passes. Which system represents the situation and what aredandandm$?

  1. System: d+m=40d+m=40 and 30d+6m=72030d+6m=720; d=20, m=20d=20,\ m=20
  2. System: d+m=720d+m=720 and 6d+30m=406d+30m=40; d=20, m=20d=20,\ m=20
  3. System: d+m=40d+m=40 and 6d+30m=7206d+30m=720; d=10, m=30d=10,\ m=30
  4. System: d+m=40d+m=40 and 6d+30m=7206d+30m=720; d=20, m=20d=20,\ m=20 (correct answer)
Explanation: This question tests setting up and solving systems of equations from real-world scenarios requiring two equations for two unknowns with context interpretation. The process involves: (1) defining variables such as d for day passes and m for monthly passes, (2) writing d + m = 40 for total passes, (3) writing 6d + 30m = 720 for revenue, (4) solving, (5) interpreting, and (6) verifying. Solving correctly: from d = 40 - m, 6(40 - m) + 30m = 720, 240 - 6m + 30m = 720, 240 + 24m = 720, 24m = 480, m = 20, d = 20. These fit both equations. Common errors include swapping costs or totals, like using d + m = 720. To avoid errors: match constraints to equations carefully, solve step-by-step, ensure non-negative integers, and verify.

Question 14

Two friends start biking toward each other from towns that are 300 miles apart. One bikes at 60 mph and the other bikes at 40 mph. Let tt be the time in hours until they meet, and let dd be the distance (in miles) the 60 mph biker travels. Which system correctly models the situation, and what is tt?

  1. System: d=60td=60t and d−40t=300d-40t=300; t=15t=15
  2. System: d=60td=60t and d+40t=300d+40t=300; t=3t=3 (correct answer)
  3. System: 50t=30050t=300; t=6t=6
  4. System: d=60td=60t and d=300d=300; t=5t=5
Explanation: Since the bikers move toward each other, the 60 mph biker's distance is d=60td=60t, and the 40 mph biker covers the rest of the 300 miles, so d+40t=300d+40t=300. Substituting gives 60t+40t=30060t+40t=300, so 100t=300100t=300 and t=3t=3 hours. Choice A is wrong because it subtracts the two bikers' distances, as if one were chasing down the other, instead of adding them, which is what happens when two people move toward each other. Choice C is wrong because it averages the two speeds together instead of adding the distances each biker actually covers. Choice D is wrong because it ignores the 40 mph biker's contribution entirely, treating the whole 300 miles as covered by the 60 mph biker alone.

Question 15

A swimming pool is being filled by two pipes. Pipe A can fill the pool in 8 hours working alone, while Pipe B can fill it in 12 hours working alone. If both pipes work together for 3 hours, then Pipe A is shut off and only Pipe B continues, how many additional hours will Pipe B need to finish filling the pool?

  1. Pipe B will need 3.5 additional hours to complete filling
  2. Pipe B will need 4.0 additional hours to complete filling
  3. Pipe B will need 4.5 additional hours to complete filling (correct answer)
  4. Pipe B will need 5.0 additional hours to complete filling
Explanation: Pipe A fills 1/8 of the pool per hour, Pipe B fills 1/12 per hour. Together they fill 1/8 + 1/12 = 3/24 + 2/24 = 5/24 of the pool per hour. In 3 hours working together, they fill 3 × 5/24 = 15/24 = 5/8 of the pool. Remaining to fill = 1 - 5/8 = 3/8 of the pool. Pipe B alone fills 1/12 per hour, so time needed = (3/8) ÷ (1/12) = 3/8 × 12 = 36/8 = 4.5 hours. Choice A uses incorrect combined rate. Choice B forgets to account for work already done. Choice D uses Pipe A's rate instead of Pipe B's.

Question 16

Two cars start 300 miles apart and drive toward each other. Car A drives 60 mph and Car B drives 40 mph. Let tt be the time in hours until they meet, and let dd be the distance Car A travels. Which system correctly models this situation?

  1. d=60td=60t and d+40t=300d+40t=300 (correct answer)
  2. d=60td=60t and d+40t=500d+40t=500
  3. d=40td=40t and d+60t=300d+60t=300
  4. d=60td=60t and d−40t=300d-40t=300
Explanation: Since Car A travels at 60 mph, d=60td=60t. The cars start 300 miles apart and drive toward each other, so their distances together must add up to the full 300 miles: d+40t=300d+40t=300. Choice B is wrong because the total distance between the cars is 300 miles, not 500. Choice C is wrong because dd represents Car A's distance, so its coefficient must be 60, not 40. Choice D is wrong because the cars are driving toward each other, so their distances add; subtraction would model the cars moving apart instead.

Question 17

A student solves a system for a fundraiser and gets (x,y)=(18,12)(x,y)=(18,12), where xx is the number of large water bottles sold and yy is the number of small water bottles sold. The student also knows 30 bottles were sold in total. What does y=12y=12 mean in this context?

  1. 12 large bottles were sold.
  2. Each bottle cost $12.
  3. $12 was collected in total.
  4. 12 small bottles were sold. (correct answer)
Explanation: The variables were defined so that xx represents large bottles and yy represents small bottles. Since y=12y=12, this means 12 small bottles were sold. Checking, x+y=18+12=30x+y=18+12=30, which matches the total number of bottles sold. Choice A incorrectly swaps which variable represents which type of bottle. Choices B and C misinterpret yy as a price or a total amount of money rather than a count of bottles.

Question 18

At a school fundraiser, hamburgers cost $5 each and hot dogs cost $3 each. The goal was to sell 200 items and raise $800. Due to supply issues, they were short 20 hamburgers but had extra hot dogs. If they still sold exactly 200 items, how much less money did they raise compared to their goal?

  1. They raised $30 less than their original goal amount
  2. They raised $40 less than their original goal amount (correct answer)
  3. They raised $50 less than their original goal amount
  4. They raised $60 less than their original goal amount
Explanation: Let h = hamburgers and d = hot dogs in the original plan. We have h + d = 200 and 5h + 3d = 800. From the first equation: d = 200 - h. Substituting: 5h + 3(200 - h) = 800, so 5h + 600 - 3h = 800, giving 2h = 200 and h = 100. Originally planned: 100 hamburgers, 100 hot dogs. Actually sold: 80 hamburgers (20 fewer), 120 hot dogs (20 more to reach 200 total). Actual revenue: 5(80) + 3(120) = 400 + 360 = $760. Difference: $800 - $760 = $40. Choice A miscalculates the price difference. Choice C uses wrong quantities. Choice D doubles the correct answer.

Question 19

The sum of two siblings' ages is 45 years, and the older sibling is 5 years older than the younger sibling. Let xx be the older sibling's age and yy be the younger sibling's age. Set up a system of equations and solve for xx and yy.

  1. x=25, y=20x=25,\ y=20 (correct answer)
  2. x=22.5, y=22.5x=22.5,\ y=22.5
  3. x=20, y=25x=20,\ y=25
  4. x=30, y=15x=30,\ y=15
Explanation: This problem tests setting up and solving systems from real-world scenarios requiring two equations for two unknowns with context interpretation. The process involves: (1) defining variables with units (let x = older sibling's age in years, y = younger sibling's age in years), (2) writing an equation from the sum constraint (x + y = 45), (3) writing an equation from the age difference (x = y + 5 or x - y = 5), (4) solving the system (substitute x = y + 5 into first: y + 5 + y = 45, 2y = 40, y = 20, x = 25), (5) interpreting (older is 25 years, younger is 20 years), and (6) verifying (25 + 20 = 45, 25 - 20 = 5). The correct solution is x=25, y=20, which matches choice C. The equations are correct as they capture the total age and the difference stated. Common errors include reversing the difference (y = x + 5) or arithmetic slips leading to other pairs like in A or D. To set up: identify constraints carefully, define variables with context, form equations accurately, solve and check for sensibility (ages positive and logical).

Question 20

Two cars start from the same point and travel in opposite directions. Car A travels 15 mph faster than Car B. After 2.5 hours, they are 275 miles apart. If Car A reduces its speed by 5 mph for the next hour while Car B maintains its speed, how far apart will they be after the additional hour?

  1. The cars will be 350 miles apart after the additional hour
  2. The cars will be 365 miles apart after the additional hour
  3. The cars will be 380 miles apart after the additional hour (correct answer)
  4. The cars will be 395 miles apart after the additional hour
Explanation: Let b = Car B's speed and a = Car A's speed = b + 15. After 2.5 hours: 2.5b + 2.5(b + 15) = 275. This gives 2.5b + 2.5b + 37.5 = 275, so 5b = 237.5 and b = 47.5 mph. Car A's speed = 62.5 mph. In the next hour, Car A travels at 57.5 mph and Car B at 47.5 mph. Additional distance = 57.5 + 47.5 = 105 miles. Total distance apart = 275 + 105 = 380 miles. Choice A uses original speeds incorrectly. Choice B forgets Car A's speed reduction. Choice D adds the speed reduction instead of subtracting.