Solve Ratio Problems - HSPT Math

1. A machine produces widgets at rates proportional to $3:4:5$ for its three production lines. In an $8$-hour shift, the total production is $2400$ widgets. Due to maintenance, the second production line operates at only $75\%$ efficiency for the next $6$-hour shift. How many widgets will be produced in this $6$-hour shift?
   Answer: $1650$ widgets When you encounter problems involving proportional rates and changing conditions, break them down into clear steps: find the individual rates, then apply any efficiency changes. First, let's determine each production line's rate during the normal 8-hour shift. The lines operate in a $3:4:5$ ratio, totaling $3+4+5=12$ parts. With 2400 total widgets, each part represents $2400÷12=200$ widgets. So the hourly rates are: - Line 1: $3×200÷8=75$ widgets/hour - Line 2: $4×200÷8=100$ widgets/hour - Line 3: $5×200÷8=125$ widgets/hour For the 6-hour shift with Line 2 at 75% efficiency: - Line 1: $75×6=450$ widgets - Line 2: $100×0.75×6=450$ widgets - Line 3: $125×6=750$ widgets - Total: $450+450+750=1650$ widgets This confirms answer A is correct. Answer B (1575) likely results from incorrectly applying the 75% efficiency to all three lines instead of just Line 2. Answer C (1725) might come from miscalculating the reduced efficiency as 85% instead of 75%. Answer D (1800) represents production if all lines operated at full capacity for 6 hours, ignoring the maintenance issue entirely. Study tip: In multi-step proportion problems, always verify your ratios add up correctly and double-check which specific conditions (like efficiency changes) apply to which parts of the system.
2. Two pumps working together can fill a pool in $4$ hours. The pumps have flow rates in the ratio $3:5$. If only the faster pump is used, how long would it take to fill $\frac{3}{4}$ of the pool?
   Answer: $4.8$ hours Work rate problems involve finding how much work each component can do per unit time. When you see pumps, pipes, or workers operating together, think about their individual rates and how they combine. Since the pumps work together to fill the pool in 4 hours, their combined rate is $\frac{1}{4}$ pool per hour. With flow rates in the ratio $3:5$, the slower pump works at rate $3x$ and the faster pump at rate $5x$ for some value $x$. Together: $3x + 5x = 8x = \frac{1}{4}$, so $x = \frac{1}{32}$. This means the faster pump alone works at rate $5x = 5 \cdot \frac{1}{32} = \frac{5}{32}$ pool per hour. To fill $\frac{3}{4}$ of the pool at this rate: time = $\frac{\frac{3}{4}}{\frac{5}{32}} = \frac{3}{4} \cdot \frac{32}{5} = \frac{96}{20} = 4.8$ hours. Answer A ($4.8$ hours) is correct. Answer B ($6$ hours) incorrectly assumes the faster pump takes $\frac{3}{4}$ as long as both pumps working together ($\frac{3}{4} \times 4 + 4 = 6$), but this ignores the $3:5$ ratio. Answer C ($3.6$ hours) mistakenly uses the time for the faster pump to fill the entire pool ($\frac{32}{5} = 6.4$ hours) and multiplies by $\frac{3}{4}$, getting $4.8$, but then makes an arithmetic error. Answer D ($5.4$ hours) appears to use incorrect ratio calculations. Remember: in combined work problems, always find individual rates first using the given ratio, then apply the specific rate to the required task. Don't shortcut with proportional thinking—it often leads to traps.
3. A paint mixture requires red, blue, and yellow paint in the ratio $7:4:3$. If the total volume of paint needed is $2.8$ liters, but the painter only has $0.6$ liters of blue paint available, what is the maximum volume of paint mixture that can be made using the correct ratio?
   Answer: $2.1$ liters When you encounter ratio problems with constraints, you need to identify which component limits your total mixture. The ratio $7:4:3$ tells you that for every 7 parts red, you need 4 parts blue and 3 parts yellow, totaling 14 parts. First, determine how much mixture you could make if blue paint is your limiting factor. With 0.6 liters of blue paint representing 4 parts of the ratio, each part equals $0.6 ÷ 4 = 0.15$ liters. Therefore, the maximum total mixture is $14 × 0.15 = 2.1$ liters, which is answer A. Let's see why the other answers miss the mark. Answer B (1.8 liters) incorrectly assumes each part is $1.8 ÷ 14 = 0.129$ liters, which would only require $4 × 0.129 = 0.514$ liters of blue paint—but you have 0.6 liters available, so this underuses your constraint. Answer C (2.4 liters) makes the error of assuming you can use the full 2.8 liters mentioned in the problem, ignoring that the blue paint limitation prevents this. Answer D (1.5 liters) seems to assume each part equals $1.5 ÷ 14 = 0.107$ liters, which would need only 0.429 liters of blue paint, again underutilizing what's available. Remember this key strategy: in constrained ratio problems, always identify your limiting component first, then work backward to find the maximum total. The component with the smallest available amount relative to its ratio requirement determines your ceiling.
4. The cost to host a banquet is a fixed rental fee plus a per-guest charge. For 40 guests, the total cost is $2100. For 60 guests, the total cost is $2900. A budget of $4100 is set for a future banquet. If the per-guest charge is increased by 25% while the rental fee remains the same, what is the maximum number of guests that can be invited?
   Answer: 72 This problem requires finding a linear relationship, then applying a change. Let F be the fixed fee and C be the per-guest charge. We have a system of two equations: 1) $F + 40C = 2100$ and 2) $F + 60C = 2900$. Subtracting equation (1) from (2) gives: $(F + 60C) - (F + 40C) = 2900 - 2100$, which simplifies to $20C = 800$, so $C = \$40$. Substitute C back into equation (1): $F + 40(40) = 2100$, so $F + 1600 = 2100$, which gives $F = \$500$. Now, the per-guest charge is increased by 25%. The new charge, C', is $C' = 40 \times (1 + 0.25) = 40 \times 1.25 = \$50$. The fixed fee F remains $500. The budget is $4100. Let G be the number of guests. The new cost equation is $500 + 50G = 4100$. Solving for G: $50G = 4100 - 500$, so $50G = 3600$. $G = \frac{3600}{50} = 72$. The maximum number of guests is 72.
5. A contractor mixes concrete using cement, sand, and gravel in the ratio $1:3:4$. For a large job, he needs to make $120$ cubic yards of concrete, but his supplier can only deliver sand in $10$-cubic-yard increments. What is the minimum amount of sand he should order, and how much total concrete can he make with that amount?
   Answer: Order $50$ cubic yards of sand; can make $133.33$ cubic yards of concrete When you encounter ratio problems with real-world constraints, you need to work backwards from the limiting factor rather than the total amount needed. The concrete ratio is $1:3:4$ (cement:sand:gravel), meaning for every 8 total parts, 3 parts are sand. To make exactly 120 cubic yards, you'd need $120 \times \frac{3}{8} = 45$ cubic yards of sand. However, sand only comes in 10-cubic-yard increments, so you must round up to 50 cubic yards to have enough material. With 50 cubic yards of sand, you can determine the maximum concrete possible. Since sand represents 3 parts out of 8 total parts, you can make $50 \times \frac{8}{3} = 133.33$ cubic yards of concrete. This confirms answer A is correct. Answer B incorrectly calculates the total concrete as 130 cubic yards, likely from a rounding error in the ratio calculation. Answer C suggests ordering only 40 cubic yards of sand, but this falls short of the 45 cubic yards needed for 120 cubic yards of concrete. Answer D makes the impossible claim that 40 cubic yards of sand can produce 120 cubic yards of concrete, which violates the given ratio entirely. Strategy tip: In ratio problems with quantity constraints, always identify which component has the restriction, then work backwards to find the maximum possible output. Don't assume you can make exactly what's "needed" when real-world limitations apply.
6. A recipe calls for milk and water in the ratio 2 : 7. If 1.5 cups of milk are used, how many cups of water are needed?
   Answer: 5.25 When you see a ratio problem, you're working with proportional relationships where the quantities maintain a constant relationship to each other. Here, milk and water must always be in the ratio 2:7, meaning for every 2 parts milk, you need 7 parts water. To solve this, set up a proportion. If the ratio is 2:7 (milk to water), then $\frac{2}{7} = \frac{1.5}{x}$, where x is the cups of water needed. Cross multiply: $2x = 7 \times 1.5 = 10.5$. Therefore, $x = 5.25$ cups of water. You can verify this: $\frac{1.5}{5.25} = \frac{2}{7}$ when you simplify both fractions to $\frac{2}{7}$. Looking at the wrong answers: Choice A (3.75) comes from incorrectly setting up the proportion as $\frac{2}{5} = \frac{1.5}{x}$ - this treats the ratio as 2:5 instead of 2:7. Choice B (4.5) results from multiplying 1.5 by 3, which would be correct if the ratio were 1:3 instead of 2:7. Choice D (6.0) comes from simply multiplying 1.5 by 4, showing a misunderstanding of how ratios work. Remember: in ratio problems, always set up your proportion carefully with the correct corresponding parts. Double-check by verifying that your answer maintains the original ratio when you divide the quantities you found.
7. An inheritance is divided between a son and a daughter in the ratio 7:5. The son invests his entire share for two years at a 10% annual simple interest rate. The daughter invests her share for the same period at a 12% annual simple interest rate. After two years, what is the ratio of the total amount the son has to the total amount the daughter has?
   Answer: 42:31 Let the son's share be 7k and the daughter's share be 5k. Simple interest for 2 years at 10% adds 2 × 10% = 20%, so the son ends with 7k × 1.20 = 8.4k. Simple interest for 2 years at 12% adds 2 × 12% = 24%, so the daughter ends with 5k × 1.24 = 6.2k. The ratio of their totals is 8.4k : 6.2k. Multiply both terms by 10 to get 84 : 62, then divide by 2 to get 42 : 31. The ratio 7:5 ignores the interest, but the daughter earns a higher rate, so the ratio changes. The ratio 6:7 compares only the interest rates' effects, 1.2 to 1.4, and leaves out the different starting amounts. The ratio 35:31 multiplies the son's share by 5 instead of applying his own growth factor. The ratio of the son's total to the daughter's total is 42:31.
8. In a chemistry lab, three solutions must be mixed in the ratio $3:7:5$ by volume. However, the available containers can only measure in increments of $25$ mL. What is the smallest total volume of mixture that can be prepared using this ratio with the measuring constraint?
   Answer: $375$ mL When you encounter ratio problems with measurement constraints, you need to find the smallest actual volumes that maintain the given ratio while meeting the physical limitations. The ratio $3:7:5$ means for every 3 parts of solution A, you need 7 parts of solution B and 5 parts of solution C. To find actual volumes, multiply each ratio part by the same factor $k$: $3k : 7k : 5k$. Since containers only measure in 25 mL increments, each volume ($3k$, $7k$, and $5k$) must be divisible by 25. This means $k$ itself must make each product divisible by 25. For $3k$ to be divisible by 25: since $\gcd(3,25) = 1$, we need $k$ divisible by 25. For $7k$ to be divisible by 25: since $\gcd(7,25) = 1$, we need $k$ divisible by 25. For $5k$ to be divisible by 25: since $5 \times 5 = 25$, we need $k$ divisible by 5. The smallest $k$ satisfying all conditions is $k = 25$. This gives volumes: $3(25) = 75$ mL, $7(25) = 175$ mL, and $5(25) = 125$ mL. Total volume: $75 + 175 + 125 = 375$ mL. Answer A (375 mL) is correct. Answer B (750 mL) uses $k = 50$, which works but isn't minimal. Answer C (300 mL) doesn't maintain the ratio with 25 mL increments. Answer D (450 mL) also fails to satisfy both the ratio and measurement constraints. Remember: when ratios meet measurement constraints, find the least common multiple that makes each ratio component measurable.
9. In a basketball league, the ratio of wins to losses for Team A is $7:3$, and for Team B it's $5:2$. If Team A has played $40$ games and Team B has played $35$ games, which team has the better winning percentage and by how much?
   Answer: Team B has better percentage: $71.4\%$ vs Team A's $70\%$, difference of $1.4\%$ When you encounter ratio problems involving percentages, you need to convert the ratios into actual numbers of wins and losses, then calculate the winning percentage for each team. For Team A with a $7:3$ win-to-loss ratio over $40$ games: The ratio parts total $7+3=10$. So Team A wins $\frac{7}{10} \times 40 = 28$ games. Their winning percentage is $\frac{28}{40} = 0.70 = 70\%$. For Team B with a $5:2$ win-to-loss ratio over $35$ games: The ratio parts total $5+2=7$. So Team B wins $\frac{5}{7} \times 35 = 25$ games. Their winning percentage is $\frac{25}{35} = \frac{5}{7} ≈ 0.714 = 71.4\%$. Team B has the better winning percentage at $71.4\%$ versus Team A's $70\%$, with a difference of $71.4\% - 70\% = 1.4\%$. Choice A correctly identifies this relationship. Choice B makes a statement error—it claims Team A has the better percentage but then shows Team B's percentage ($71.4\%$) is actually higher than Team A's ($70\%$). Choice C has the right percentages and comparison but calculates the wrong difference ($2.4\%$ instead of $1.4\%$). Choice D incorrectly calculates Team B's percentage as $68.6\%$, likely from a ratio misinterpretation. Study tip: Always convert ratios to actual numbers first, then calculate percentages. Double-check that your final comparison statement matches your calculated values—many wrong answers contain contradictory statements.
10. Two candles of the same height are lit at the same time. The first is consumed in 6 hours and the second in 4 hours. Assuming that each candle burns at a constant rate, in how many hours after being lit will the ratio of the height of the first candle to the second candle be 2:1?
    Answer: 3 hours Let the initial height of both candles be H. The rate of burning for the first candle (C1) is H/6 per hour. The rate of burning for the second candle (C2) is H/4 per hour. Let 't' be the time in hours after they are lit. The height of C1 at time 't' is $H_1(t) = H - \frac{H}{6}t = H(1 - \frac{t}{6})$. The height of C2 at time 't' is $H_2(t) = H - \frac{H}{4}t = H(1 - \frac{t}{4})$. We want to find the time 't' when the ratio of their heights is 2:1, i.e., $\frac{H_1(t)}{H_2(t)} = \frac{2}{1}$. Substituting the expressions for the heights: $\frac{H(1 - t/6)}{H(1 - t/4)} = 2$. The H terms cancel out: $1 - \frac{t}{6} = 2(1 - \frac{t}{4})$. Distribute the 2 on the right side: $1 - \frac{t}{6} = 2 - \frac{2t}{4} = 2 - \frac{t}{2}$. Now, solve for t. It's helpful to clear the denominators by multiplying the entire equation by the least common multiple of 6 and 2, which is 6. $6(1) - 6(\frac{t}{6}) = 6(2) - 6(\frac{t}{2})$. This gives $6 - t = 12 - 3t$. Rearranging the terms to solve for t: $3t - t = 12 - 6$, which simplifies to $2t = 6$. Therefore, $t = 3$ hours.
11. Two similar polygons have perimeters in the ratio 3:5. The sum of their areas is 544 square centimeters. What is the area of the larger polygon?
    Answer: 400 sq cm The ratio of the perimeters of two similar polygons is equal to the ratio of their corresponding sides. Let this ratio be $s_1:s_2 = 3:5$. The ratio of the areas of two similar polygons is the square of the ratio of their corresponding sides. So, the ratio of the areas is $A_1:A_2 = (s_1)^2:(s_2)^2 = 3^2:5^2 = 9:25$. Let the areas be 9x and 25x. The sum of the areas is given as 544 sq cm. So, $9x + 25x = 544$. This simplifies to $34x = 544$. Solving for x, we get $x = \frac{544}{34} = 16$. The area of the larger polygon is 25x. Therefore, the area is $25 \times 16 = 400$ square centimeters.
12. In a bag of marbles, the ratio of red marbles to blue marbles is 5 : 3. If the bag contains 24 blue marbles, how many red marbles are in the bag?
    Answer: 40 When you encounter ratio problems, you're working with proportional relationships between quantities. The key is understanding that ratios tell you the relative amounts, and you can use them to find actual quantities when given one piece of information. The ratio 5 : 3 means that for every 5 red marbles, there are 3 blue marbles. Since you know there are 24 blue marbles, you can set up a proportion. If 3 parts equal 24 marbles, then each part equals $24 ÷ 3 = 8$ marbles. Since red marbles make up 5 parts, there are $5 × 8 = 40$ red marbles. You can also solve this using cross-multiplication: $\frac{\text{red}}{\text{blue}} = \frac{5}{3} = \frac{x}{24}$, which gives you $3x = 5 × 24 = 120$, so $x = 40$. Looking at the wrong answers: Choice B (35) might result from incorrectly adding the ratio parts (5 + 3 = 8) and somehow manipulating that with 24. Choice C (30) could come from mistakenly thinking the ratio means red marbles are $\frac{5}{4}$ times the blue marbles (since $24 × \frac{5}{4} = 30$). Choice D (15) likely comes from confusing the ratio direction and thinking red marbles are $\frac{3}{5}$ times the blue marbles. Remember: in ratio problems, always identify what each "part" represents in real quantities first. This prevents confusion about which number in the ratio corresponds to which quantity in the problem.
13. On a map, 1 inch represents 12 miles. If two cities are 3.5 inches apart on the map, what is their actual distance in miles?
    Answer: 42 Map scale problems test your ability to use proportional reasoning to convert between map measurements and real-world distances. When you see a scale like "1 inch represents 12 miles," you're looking at a ratio that stays constant across the entire map. To find the actual distance, set up a proportion using the given scale. Since 1 inch on the map equals 12 miles in reality, you can write this as $\frac{1 \text{ inch}}{12 \text{ miles}} = \frac{3.5 \text{ inches}}{x \text{ miles}}$. Cross-multiplying gives you $1 \times x = 12 \times 3.5$, so $x = 42$ miles. Alternatively, you can think of this as a simple multiplication: if each inch represents 12 miles, then 3.5 inches represents $3.5 \times 12 = 42$ miles. Looking at the wrong answers: Choice A (30) might result from incorrectly calculating $3.5 \times 8$ or making an arithmetic error. Choice B (36) could come from multiplying $3.5 \times 10$ instead of 12, possibly misremembering the scale. Choice D (48) might occur if you mistakenly thought the scale was 1 inch = 13.7 miles, or made an error in your multiplication. Study tip: For map scale problems, always identify the scale ratio first, then multiply the map distance by the scale factor. Double-check your arithmetic since these problems often include answer choices that result from common calculation mistakes. Practice recognizing that "1 inch represents X miles" means you multiply map inches by X to get real miles.
14. A paint mixture requires red, blue, and yellow paint in the ratio $4:3:2$. If a painter has $36$ gallons of blue paint available and wants to use it all, how many gallons of red and yellow paint combined will be needed?
    Answer: $72$ gallons of red and yellow paint combined When you encounter ratio problems, your key strategy is to find the scaling factor that connects the given information to the ratio. Here, the paint mixture uses red, blue, and yellow in the ratio $4:3:2$, and you have $36$ gallons of blue paint to use completely. Since blue paint corresponds to the "$3$" in the ratio $4:3:2$, you can set up the equation: $3x = 36$, where $x$ is your scaling factor. Solving gives you $x = 12$. Now you can find the actual amounts needed: - Red paint: $4x = 4(12) = 48$ gallons - Yellow paint: $2x = 2(12) = 24$ gallons - Combined red and yellow: $48 + 24 = 72$ gallons This confirms answer A is correct. Let's examine why the other answers are wrong: Answer B ($60$ gallons) would result from incorrectly thinking the ratio parts for red and yellow are $3 + 2 = 5$, then calculating $5 \times 12 = 60$. This misses that red requires $4$ parts, not $3$. Answer C ($54$ gallons) might come from adding all three paint amounts incorrectly or using the wrong scaling factor. Answer D ($48$ gallons) gives only the red paint amount, forgetting to include the yellow paint in the total. Study tip: In ratio problems, always identify which part of the ratio corresponds to your given information, find the scaling factor by dividing, then multiply each ratio part by that factor. Double-check by ensuring your given quantity matches what you calculate.
15. In a gear system, three gears A, B, and C have teeth in the ratio $12:16:20$. If gear A makes $80$ complete rotations, and the gears are connected such that adjacent gears rotate in opposite directions, how many rotations does gear C make, and in which direction relative to gear A?
    Answer: $48$ rotations in the same direction as A When you encounter gear system problems, remember two key principles: gears with more teeth rotate slower, and adjacent gears always rotate in opposite directions. First, let's find how many rotations gear C makes. In any gear system, the product of teeth and rotations remains constant. So if gear A (12 teeth) makes 80 rotations, we can write: $12 \times 80 = 20 \times R_C$, where $R_C$ is gear C's rotations. Solving: $R_C = \frac{12 \times 80}{20} = \frac{960}{20} = 48$ rotations. Next, we need to determine the direction. Since adjacent gears rotate in opposite directions, we must trace through the connections. Gear A connects to gear B (opposite direction), then gear B connects to gear C (opposite direction again). Two direction changes means gear C rotates in the same direction as gear A. Looking at the wrong answers: Choice B gives the correct number of rotations (48) but incorrectly states the direction as opposite to A. This ignores that there are two gear connections, not one. Choices C and D both incorrectly calculate 64 rotations, likely from using gear B's tooth count (16) in the calculation instead of gear C's count (20). Choice D compounds this error with the wrong direction as well. For gear problems, always set up the relationship that teeth times rotations equals a constant, and carefully count the number of gear connections to determine final direction. Each connection reverses the direction once.
16. If $a, b, c$ are in continued proportion, where $a = x+4$, $b = 2x+2$, and $c = 3x+3$, which of the following is the ratio of $a$ to $c$? Assume $x>0$.
    Answer: 4:9 If a, b, and c are in continued proportion, it means that $a:b = b:c$, or $b^2 = ac$. We are given the expressions for a, b, and c in terms of x. Substitute these into the equation: $(2x+2)^2 = (x+4)(3x+3)$. Expand both sides of the equation. Left side: $(2x+2)^2 = 4x^2 + 8x + 4$. Right side: $(x+4)(3x+3) = 3x^2 + 3x + 12x + 12 = 3x^2 + 15x + 12$. Now, set the expanded forms equal to each other: $4x^2 + 8x + 4 = 3x^2 + 15x + 12$. Rearrange this into a standard quadratic equation form ($ax^2+bx+c=0$): $x^2 - 7x - 8 = 0$. Factor the quadratic: $(x-8)(x+1) = 0$. The possible solutions for x are $x=8$ and $x=-1$. Since the problem states that $x>0$, we must have $x=8$. Now find the values of a and c using x=8. $a = x+4 = 8+4 = 12$. $c = 3x+3 = 3(8)+3 = 24+3 = 27$. The ratio of a to c is 12:27. Simplify this ratio by dividing both numbers by their greatest common divisor, which is 3. The simplified ratio is 4:9.
17. A photo is enlarged so that the ratio of the new area to the original area is $25:4$. If the original photo was $6$ inches wide, what is the width of the enlarged photo, and what is the ratio of the new perimeter to the original perimeter?
    Answer: New width is $15$ inches; perimeter ratio is $5:2$ When you see questions about enlarging or scaling geometric figures, remember that area scales with the square of the linear scale factor, while perimeter scales directly with the linear scale factor. Since the area ratio is $25:4$, you need to find the linear scale factor first. Because area scales as the square of linear dimensions, if the linear scale factor is $k$, then $k^2 = \frac{25}{4}$. Taking the square root: $k = \frac{5}{2} = 2.5$. For the new width: multiply the original width by the scale factor. The original width was 6 inches, so the new width is $6 \times 2.5 = 15$ inches. For the perimeter ratio: since perimeter is a linear measurement, it scales directly with the linear scale factor. Therefore, the perimeter ratio is $5:2$. Looking at the wrong answers: Choice B gives the correct width but incorrectly uses the area ratio ($25:4$) for the perimeter ratio. This is a common mistake—forgetting that perimeter scales linearly, not quadratically. Choice C calculates the new width as $6 \times \frac{25}{4} = 37.5$ inches, incorrectly applying the area ratio directly to a linear dimension. Choice D makes both errors: using the area ratio for both the width calculation and the perimeter ratio. Study tip: Always identify whether you're working with linear measurements (length, width, perimeter) or area measurements. Linear measurements scale with the scale factor $k$, while areas scale with $k^2$. When given an area ratio, take its square root to find the linear scale factor.
18. The ratio of the areas of two similar triangles is $9:16$. If the perimeter of the smaller triangle is $24$ cm, what is the perimeter of the larger triangle?
    Answer: $32$ cm, because the ratio of perimeters is $3:4$ When working with similar triangles, you need to understand the relationship between different types of measurements. Similar triangles have the same shape but different sizes, which creates predictable ratios between corresponding parts. The key insight is that while areas have a squared relationship to linear dimensions, perimeters have a direct linear relationship. If two similar triangles have corresponding sides in the ratio $a:b$, then their perimeters are also in the ratio $a:b$, but their areas are in the ratio $a^2:b^2$. Since the area ratio is $9:16$, you need to find the ratio of corresponding linear dimensions. Taking the square root of both parts: $\sqrt{9}:\sqrt{16} = 3:4$. This $3:4$ ratio applies to all linear measurements, including perimeters. If the smaller triangle's perimeter is $24$ cm and the perimeter ratio is $3:4$, then: $\frac{24}{x} = \frac{3}{4}$, where $x$ is the larger triangle's perimeter. Cross-multiplying: $3x = 96$, so $x = 32$ cm. Answer A gives the correct numerical result but incorrectly states that perimeter ratios equal area ratios. Answer C incorrectly uses the area ratio $9:16$ directly for perimeters, calculating $24 \times \frac{16}{9} = 42.67$ cm, and compounds this by claiming perimeters equal area ratios. Answer D uses the same incorrect calculation as C but correctly identifies the $3:4$ ratio relationship. Remember: for similar figures, linear measurements (like perimeter) relate by taking the square root of the area ratio. Always convert area ratios to linear ratios first.
19. At a school, the ratio of freshmen to sophomores to juniors is 6 : 5 : 4. If there are 150 sophomores, how many total students are in these three classes combined?
    Answer: 450 When you encounter ratio problems, you're working with proportional relationships between different groups. The key insight is that ratios tell you the relative sizes of groups, and you can use any known quantity to find the others. Given the ratio 6:5:4 for freshmen:sophomores:juniors, this means that for every 5 sophomores, there are 6 freshmen and 4 juniors. Since you know there are 150 sophomores, you can find the ratio multiplier. If the ratio calls for 5 sophomores but you actually have 150, then $150 ÷ 5 = 30$. This means each part of the ratio represents 30 students. Therefore: freshmen = $6 × 30 = 180$, sophomores = $5 × 30 = 150$ (which confirms our calculation), and juniors = $4 × 30 = 120$. The total is $180 + 150 + 120 = 450$ students. Looking at the wrong answers: (A) 390 likely comes from miscalculating the ratio multiplier or making an arithmetic error in the final addition. (C) 480 might result from incorrectly treating 150 as corresponding to a different part of the ratio, perhaps thinking it represents the "6" portion instead of the "5" portion. (D) 525 could come from adding 150 to each ratio number (6+5+4=15, then 15×150÷something) or another fundamental misunderstanding of how ratios work. Remember: in ratio problems, always identify which quantity corresponds to which part of the ratio, then find the multiplier to scale up the entire ratio proportionally.
20. Two gears are connected such that their rotation speeds are inversely proportional to their radii. If gear A has radius $6$ cm and rotates at $150$ rpm, while gear B has radius $9$ cm, and gear C is added such that all three gears maintain the inverse relationship, what must be the radius of gear C if it rotates at $75$ rpm?
    Answer: $12$ cm When you encounter gear problems involving inverse proportionality, remember that as one quantity increases, the other decreases at a predictable rate. Here, rotation speed and radius have an inverse relationship, meaning $\text{speed} \times \text{radius} = \text{constant}$ for all connected gears. First, find the constant using gear A's information: $150 \text{ rpm} \times 6 \text{ cm} = 900$. You can verify this works for gear B: if its radius is 9 cm, then its speed should be $900 ÷ 9 = 100$ rpm, which makes sense since larger gears rotate slower. For gear C rotating at 75 rpm, use the same constant: $75 \times r_C = 900$, so $r_C = 900 ÷ 75 = 12$ cm. Looking at the wrong answers: Choice A (18 cm) would give a speed of $900 ÷ 18 = 50$ rpm, not 75 rpm. Choice C (15 cm) would result in $900 ÷ 15 = 60$ rpm. Choice D (20 cm) would produce $900 ÷ 20 = 45$ rpm. These represent common calculation errors or misunderstanding the inverse relationship. The answer is B) 12 cm. Study tip: For inverse proportion problems, always establish the constant first by multiplying the two given quantities. Then use this constant to find missing values. Watch for the key phrase "inversely proportional" – it signals that one value increases as the other decreases, and their product remains constant.

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