College Algebra Quiz: Rate Problems Work Motion And Units

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.

Question 1

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 $6 \times 0.40 = 2.4$ liters of pure chemical. This amount won't change when you add water. Let $x$ = liters of water to add. After adding water, you'll have $(6 + x)$ total liters, but still only 2.4 liters of pure chemical. For a 25% concentration: $\frac{2.4}{6 + x} = 0.25$ Solving: $2.4 = 0.25(6 + x)$, so $2.4 = 1.5 + 0.25x$, which gives $0.9 = 0.25x$, therefore $x = 3.6$ liters. Choice A (3.0 liters) likely comes from incorrectly setting up the equation as $\frac{6}{6 + x} = 0.25$, 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 2

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 $\frac{5}{4.5} = \frac{10}{9}$ hours = $\frac{10}{9} \times 60 = \frac{200}{3} = 66\frac{2}{3}$ minutes. Second half time: $80 - 66\frac{2}{3} = 13\frac{1}{3}$ minutes = $\frac{1}{4.5}$ hours. Second half speed: $\frac{5 \text{ miles}}{\frac{1}{4.5} \text{ hours}} = 5 \times 4.5 \div 2 = 3.75$ mph. Choice A assumes equal time splits. Choice B uses incorrect calculations. Choice D assumes constant speed throughout.

Question 3

1. 15 mph wind speed
2. 20 mph wind speed
3. 30 mph wind speed (correct answer)
4. 25 mph wind speed

Explanation: Let $w$ = wind speed. With tailwind: speed = $180 + w$, against wind: speed = $180 - w$. Equal time means: $\frac{420}{180 + w} = \frac{300}{180 - w}$. Cross-multiplying: $420(180 - w) = 300(180 + w)$, so $75600 - 420w = 54000 + 300w$. Solving: $21600 = 720w$, thus $w = 30$ 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 4

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 $x$ = liters of 15% solution needed. Since the total is 20 liters, you'll need $(20-x)$ 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: $0.15x$ liters of pure saline From the 35% solution: $0.35(20-x)$ liters of pure saline In the final 28% solution: $0.28(20) = 5.6$ liters of pure saline Setting up the equation: $0.15x + 0.35(20-x) = 5.6$ Solving: $0.15x + 7 - 0.35x = 5.6$ $-0.20x = -1.4$ $x = 7$ 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

1. 4 mph current speed (correct answer)
2. 6 mph current speed
3. 5 mph current speed
4. 3 mph current speed

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