College Algebra Quiz: Rate Problems Work Motion And Units
5 questions · exam conditions
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Rate Problems Work Motion And UnitsQuestion 1 of 5

An airplane flies 420 miles with a tailwind in the same time it takes to fly 300 miles against the same wind. If the airplane's speed in still air is 180 mph, what is the wind speed?

15 mph wind speed
20 mph wind speed
30 mph wind speed
25 mph wind speed
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College Algebra Quiz

College Algebra Quiz: Rate Problems Work Motion And Units

Practice Rate Problems Work Motion And Units in College 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 Rate Problems Work Motion And Units, giving you a quick way to practice the rules, question types, and explanations that matter most for College 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

An airplane flies 420 miles with a tailwind in the same time it takes to fly 300 miles against the same wind. If the airplane's speed in still air is 180 mph, what is the wind speed?

  1. 15 mph wind speed
  2. 20 mph wind speed
  3. 30 mph wind speed (correct answer)
  4. 25 mph wind speed
Explanation: Let ww = wind speed. With tailwind: speed = 180+w180 + w, against wind: speed = 180−w180 - w. Equal time means: 420180+w=300180−w\frac{420}{180 + w} = \frac{300}{180 - w}. Cross-multiplying: 420(180−w)=300(180+w)420(180 - w) = 300(180 + w), so 75600−420w=54000+300w75600 - 420w = 54000 + 300w. Solving: 21600=720w21600 = 720w, thus w=30w = 30 mph. Choice A (15) results from calculation errors in cross-multiplication. Choice B (20) uses incorrect distance ratios. Choice D (25) comes from arithmetic mistakes in solving the linear equation.

Question 2

A chemical solution needs to be diluted from 40% concentration to 25% concentration. If you have 6 liters of the 40% solution, how many liters of pure water must be added to achieve the desired concentration?

  1. 3.0 liters of pure water needed
  2. 2.4 liters of pure water needed
  3. 4.2 liters of pure water needed
  4. 3.6 liters of pure water needed (correct answer)
Explanation: When you encounter mixture problems involving dilution, you're working with the principle that the amount of pure substance remains constant while the total volume changes. The key insight is that adding pure water doesn't change the amount of the original chemical—only the concentration. Start by identifying what stays constant: the actual amount of chemical substance. In 6 liters of 40% solution, you have 6×0.40=2.46 \times 0.40 = 2.4 liters of pure chemical. This amount won't change when you add water. Let xx = liters of water to add. After adding water, you'll have (6+x)(6 + x) total liters, but still only 2.4 liters of pure chemical. For a 25% concentration: 2.46+x=0.25\frac{2.4}{6 + x} = 0.25 Solving: 2.4=0.25(6+x)2.4 = 0.25(6 + x), so 2.4=1.5+0.25x2.4 = 1.5 + 0.25x, which gives 0.9=0.25x0.9 = 0.25x, therefore x=3.6x = 3.6 liters. Choice A (3.0 liters) likely comes from incorrectly setting up the equation as 66+x=0.25\frac{6}{6 + x} = 0.25, using the original volume instead of the chemical amount. Choice B (2.4 liters) represents the amount of pure chemical, not the water needed—a common mix-up. Choice C (4.2 liters) might result from calculation errors or mishandling the percentage conversion. For mixture problems, always identify what quantity remains unchanged (here, the pure chemical), then set up your equation based on the final desired ratio. Double-check by verifying that your answer produces the target concentration.

Question 3

A runner completes a 10-mile race in exactly 80 minutes. For the first half of the race (5 miles), she runs at a constant speed of 4.5 mph. What was her average speed during the second half of the race?

  1. 3.33 mph during second half
  2. 2.86 mph during second half
  3. 3.75 mph during second half (correct answer)
  4. 4.00 mph during second half
Explanation: First half: 5 miles at 4.5 mph takes 54.5=109\frac{5}{4.5} = \frac{10}{9} hours = 109×60=2003=6623\frac{10}{9} \times 60 = \frac{200}{3} = 66\frac{2}{3} minutes. Second half time: 80−6623=131380 - 66\frac{2}{3} = 13\frac{1}{3} minutes = 14.5\frac{1}{4.5} hours. Second half speed: 5 miles14.5 hours=5×4.5÷2=3.75\frac{5 \text{ miles}}{\frac{1}{4.5} \text{ hours}} = 5 \times 4.5 \div 2 = 3.75 mph. Choice A assumes equal time splits. Choice B uses incorrect calculations. Choice D assumes constant speed throughout.

Question 4

A pharmaceutical company mixes a 15% saline solution with a 35% saline solution to create 20 liters of a 28% saline solution. How many liters of the 15% solution are needed?

  1. 9 liters of 15% solution needed
  2. 7 liters of 15% solution needed (correct answer)
  3. 6 liters of 15% solution needed
  4. 8 liters of 15% solution needed
Explanation: When you encounter mixture problems involving percentages, you're dealing with a weighted average situation where the final concentration depends on how much of each solution you combine. Set up the problem systematically. Let xx = liters of 15% solution needed. Since the total is 20 liters, you'll need (20−x)(20-x) liters of the 35% solution. The key insight is that the amount of pure saline from both solutions must equal the pure saline in the final mixture. From the 15% solution: 0.15x0.15x liters of pure saline From the 35% solution: 0.35(20−x)0.35(20-x) liters of pure saline In the final 28% solution: 0.28(20)=5.60.28(20) = 5.6 liters of pure saline Setting up the equation: 0.15x+0.35(20−x)=5.60.15x + 0.35(20-x) = 5.6 Solving: 0.15x+7−0.35x=5.60.15x + 7 - 0.35x = 5.6 −0.20x=−1.4-0.20x = -1.4 x=7x = 7 So you need 7 liters of the 15% solution, making (B) correct. (A) 9 liters would create a mixture with only 21% saline concentration—too weak because you're using too much of the weaker solution. (C) 6 liters results in approximately 29% concentration—too strong since you're not using enough of the weaker solution. (D) 8 liters gives about 27% concentration—close but still not the target 28%. Strategy tip: Always check that your concentrations make logical sense. The final percentage should fall between the two original percentages, and it should be closer to whichever solution you're using more of.

Question 5

A boat travels 24 miles downstream in the same time it takes to travel 16 miles upstream. If the boat's speed in still water is 20 mph, what is the speed of the current?

  1. 4 mph current speed (correct answer)
  2. 6 mph current speed
  3. 5 mph current speed
  4. 3 mph current speed
Explanation: Let cc be the current speed. Downstream speed is 20+c20 + c mph, upstream speed is 20−c20 - c mph. Since time = distance/speed and the times are equal: 2420+c=1620−c\frac{24}{20 + c} = \frac{16}{20 - c}. Cross-multiplying: 24(20−c)=16(20+c)24(20 - c) = 16(20 + c), so 480−24c=320+16c480 - 24c = 320 + 16c. Solving: 160=40c160 = 40c, thus c=4c = 4 mph. Choice B (6) results from setup errors. Choice C (5) comes from arithmetic mistakes. Choice D (3) assumes incorrect distance ratios.