Proportion & Scale Problems - Accuplacer Qas

The time it takes to complete a construction project is inversely proportional to the number of workers. If it takes 12 workers 20 days to finish the project, how many fewer days would it take if 15 workers were assigned to the project?
Answer: 4 For an inverse proportion, the product of the two quantities is constant. The total 'worker-days' needed is $12 \text{ workers} \times 20 \text{ days} = 240 \text{ worker-days}$ . With 15 workers, the time needed is $\frac{240 \text{ worker-days}}{15 \text{ workers}} = 16 \text{ days}$ . The question asks for how many fewer days it would take. The difference is $20 \text{ days} - 16 \text{ days} = 4 \text{ fewer days}$ .
A recipe for 8 servings calls for 3 cups of flour and 2.25 cups of sugar. Maria wants to make enough food for 14 people, but she only has 5 cups of flour available. How many cups of sugar should she use to maintain the proper proportions with the flour she has?
Answer: 3.75 cups First, find the flour-to-sugar ratio: 3 cups flour to 2.25 cups sugar, which simplifies to 4:3. With 5 cups of flour available, use the proportion 3/2.25 = 5/x to find x = (5 × 2.25)/3 = 11.25/3 = 3.75 cups of sugar. Choice B incorrectly scales for 14 servings then tries to adjust. Choice C scales the sugar for 14 servings without considering flour limitation. Choice D assumes a 1:1 ratio minus the flour shortage.
A concrete mixture requires cement, sand, and gravel in the ratio 2:3:5. A contractor has 18 bags of cement, 24 bags of sand, and 50 bags of gravel. What is the maximum number of complete batches of concrete that can be made, and how many bags of sand will remain unused?
Answer: 8 batches, 0 bags of sand unused For ratio 2:3:5, each batch needs 2 bags cement, 3 bags sand, 5 bags gravel. Available batches limited by: cement: 18÷2=9 batches, sand: 24÷3=8 batches, gravel: 50÷5=10 batches. Sand is the limiting factor at 8 batches. For 8 batches: cement used = 8×2=16 (2 unused), sand used = 8×3=24 (0 unused), gravel used = 8×5=40 (10 unused). Choice A miscounts batches. Choice B assumes 9 batches possible. Choice D ignores the limiting factor constraint.
A 10-liter car radiator is filled with a solution that is 20% antifreeze. The owner needs to increase the concentration to 50% antifreeze. How many liters of the solution must be drained and replaced with pure antifreeze to achieve the desired concentration?
Answer: 3.75 Let $x$ be the number of liters to drain and replace. Initially, there are $10 \times 0.20 = 2$ liters of antifreeze. When $x$ liters of solution are drained, the amount of antifreeze removed is $0.20x$ . Then, $x$ liters of pure (100%) antifreeze are added. The new amount of antifreeze is $2 - 0.20x + x$ . The total volume is still 10 liters. The new concentration should be 50%, so $\frac{2 - 0.20x + x}{10} = 0.50$ . Simplify the numerator: $2 + 0.80x = 10 \times 0.50$ , which is $2 + 0.80x = 5$ . Subtract 2 from both sides: $0.80x = 3$ . Solve for $x$ : $x = \frac{3}{0.80} = 3.75$ liters.
A 6-foot tall person casts a 10-foot long shadow. At the same time of day, a nearby flagpole casts a 45-foot long shadow. What is the height of the flagpole?
Answer: 27 feet The ratio of an object's height to its shadow's length is constant. Set up a proportion: $\frac{\text{person's height}}{\text{person's shadow}} = \frac{\text{flagpole's height}}{\text{flagpole's shadow}}$ . This gives $\frac{6}{10} = \frac{x}{45}$ . Cross-multiply: $10x = 6 \times 45$ , which simplifies to $10x = 270$ . Divide by 10 to find $x = 27$ . The flagpole is 27 feet tall.
A cleaning solution is formed by mixing concentrate and water in a ratio of 3 parts concentrate to 8 parts water. If a total of 550 milliliters of the solution is prepared, how many milliliters of concentrate are used?
Answer: 150 The total number of parts in the solution is $3 + 8 = 11$ parts. To find the volume of one part, divide the total volume by the total number of parts: $\frac{550 \text{ mL}}{11 \text{ parts}} = 50 \text{ mL per part}$ . The amount of concentrate is 3 parts, so multiply the volume of one part by 3: $3 \text{ parts} \times 50 \frac{\text{mL}}{\text{part}} = 150 \text{ mL}$ .
On a map with a scale of 1 inch = 15 miles, two cities are 4.8 inches apart. A third city forms a triangle with the other two cities such that the actual distances form a 3:4:5 ratio triangle. If the longest side of the triangle is the distance between the first two cities, what is the map distance in inches for the shortest side?
Answer: 2.88 inches The actual distance between the first two cities is 4.8 × 15 = 72 miles. In a 3:4:5 triangle, if 72 miles is the longest side (ratio 5), then each ratio unit = 72/5 = 14.4 miles. The shortest side (ratio 3) = 3 × 14.4 = 43.2 miles. Converting to map distance: 43.2 ÷ 15 = 2.88 inches. Choice B uses ratio 4. Choice C assumes the given distance is the middle side. Choice D incorrectly applies the scale.
A recipe calls for ingredients in the ratio of 4 parts flour to 3 parts sugar to 2 parts butter by weight. If you want to make the recipe but only have 300 grams of butter available, and you need to reduce the sugar by 25% from the original proportion while keeping the flour-to-butter ratio constant, how much flour should you use?
Answer: 600 grams When you encounter ratio problems with modifications, break them down step by step rather than trying to solve everything at once. Start with the given information and work systematically through each constraint. The original recipe has a 4:3:2 ratio (flour:sugar:butter). With 300 grams of butter available, you can find the "unit size" by dividing: 300 ÷ 2 = 150 grams per ratio unit. This would normally give you 4 × 150 = 600 grams of flour and 3 × 150 = 450 grams of sugar. However, you're reducing sugar by 25% while keeping the flour-to-butter ratio constant. The flour-to-butter ratio is 4:2 or 2:1, so with 300 grams of butter, you need 600 grams of flour regardless of what happens to the sugar. Looking at the wrong answers: Choice A (450 grams) represents the original amount of sugar, not flour—this suggests confusing the ingredients. Choice B (525 grams) might come from incorrectly applying the 25% reduction to flour instead of sugar, then making calculation errors. Choice C (675 grams) could result from mistakenly increasing the flour amount when reducing sugar, perhaps thinking you need to compensate. The key insight is that reducing sugar by 25% doesn't affect the flour-to-butter ratio, which remains constant at 2:1. Therefore, 600 grams of flour is correct. Strategy tip: In modified ratio problems, identify which relationships stay constant and which change. Handle each constraint separately, then combine your results. Don't let modifications to one ingredient confuse your calculations for others.
A car travels 140 miles on 5 gallons of gasoline. At this same rate of fuel consumption, how many miles can the car travel on a full tank of 12 gallons?
Answer: 336 First, calculate the car's fuel efficiency in miles per gallon (mpg). $\text{Efficiency} = \frac{140 \text{ miles}}{5 \text{ gallons}} = 28 \text{ mpg}$ . To find the distance the car can travel on 12 gallons, multiply the efficiency by the amount of fuel: $28 \frac{\text{miles}}{\text{gallon}} \times 12 \text{ gallons} = 336 \text{ miles}$ .
Two gears are connected such that when one rotates, the other rotates in the opposite direction. The first gear has 60 teeth and the second has 24 teeth. For every 5 complete rotations of the second, smaller gear, how many complete rotations does the first, larger gear make?
Answer: 2 The number of rotations is inversely proportional to the number of teeth. Let $R_1$ and $T_1$ be the rotations and teeth of the first gear, and $R_2$ and $T_2$ be for the second. The relationship is $R_1 T_1 = R_2 T_2$ . We have $T_1 = 60$ , $T_2 = 24$ , and $R_2 = 5$ . Plugging these in gives $R_1 \times 60 = 5 \times 24$ , which simplifies to $60 R_1 = 120$ . Divide by 60: $R_1 = 2$ . The larger gear makes 2 rotations.
A scale model of a cube-shaped container is built with a length scale factor of 1:10. The volume of the actual container is 4,000 cubic meters. What is the volume of the scale model in cubic meters?
Answer: 4 The scale factor for length is 1:10. For volume, the scale factor must be cubed. The volume scale factor is $1^3 : 10^3$ , which is 1:1000. This means the model's volume is $\frac{1}{1000}$ of the actual volume. Set up the proportion: $\frac{\text{Volume}_{\text{model}}}{\text{Volume}_{\text{actual}}} = \frac{1}{1000}$ . So, $\frac{V_m}{4000} = \frac{1}{1000}$ . Solving for $V_m$ gives $V_m = \frac{4000}{1000} = 4$ cubic meters. Wait, the problem is flawed. A 1:10 model of something that is 4000 $m^3$ should be much smaller. 1:1000 would be 4 $m^3$ . Let's re-read. 1:10. Volume scale is $1^3$ : $10^3$ = 1:1000. So Volume_model = Volume_actual / 1000. 4000/1000 = 4. The math is correct. Let me re-write the explanation to be clearer. The scale factor for length is $\frac{1}{10}$ . The scale factor for volume is the cube of the length scale factor, which is $(\frac{1}{10})^3 = \frac{1}{1000}$ . The volume of the model is the actual volume multiplied by this scale factor: $4,000 \text{ m}^3 \times \frac{1}{1000} = 4 \text{ m}^3$ .
A blueprint for a rectangular room uses a scale of 1 inch to 4 feet. If the dimensions of the room on the blueprint are 3.5 inches by 5 inches, what is the actual area of the room in square feet?
Answer: 280 First, convert the blueprint dimensions to actual dimensions using the scale. The actual width is $3.5 \text{ inches} \times 4 \frac{\text{feet}}{\text{inch}} = 14 \text{ feet}$ . The actual length is $5 \text{ inches} \times 4 \frac{\text{feet}}{\text{inch}} = 20 \text{ feet}$ . The actual area is the product of the actual dimensions: $14 \text{ feet} \times 20 \text{ feet} = 280 \text{ square feet}$ .
A scale drawing of a triangular park has sides of length 6 cm, 8 cm, and 10 cm. The scale of the drawing is 1:1,500. What is the actual perimeter of the park in meters?
Answer: 360 First, find the perimeter of the park on the drawing: $6 + 8 + 10 = 24 \text{ cm}$ . Next, use the scale to find the actual perimeter in centimeters. The scale 1:1,500 means the actual perimeter is 1,500 times the drawing's perimeter: $24 \text{ cm} \times 1,500 = 36,000 \text{ cm}$ . Finally, convert the perimeter from centimeters to meters. Since there are 100 cm in 1 meter, divide by 100: $\frac{36,000 \text{ cm}}{100 \text{ cm/m}} = 360 \text{ meters}$ .
A rectangular photograph measuring 4 inches wide by 6 inches tall is enlarged proportionally. If the width of the enlarged photograph is 10 inches, what is its new height?
Answer: 15 inches The ratio of width to height must remain the same for the photograph to be enlarged proportionally. Set up the proportion: $\frac{\text{original width}}{\text{original height}} = \frac{\text{new width}}{\text{new height}}$ . This gives $\frac{4}{6} = \frac{10}{h}$ . Cross-multiply: $4h = 6 \times 10$ , which simplifies to $4h = 60$ . Divide by 4 to solve for $h$ : $h = 15$ . The new height is 15 inches.
A digital artist scales an image that is 800 pixels wide down by a factor of $\frac{3}{4}$ . She then takes the resulting image and scales it up by a factor of $\frac{5}{2}$ . What is the final width of the image in pixels?
Answer: 1,500 This problem involves applying two successive scale factors. First, scale the image down: $800 \text{ pixels} \times \frac{3}{4} = 600 \text{ pixels}$ . Next, take the new width and scale it up: $600 \text{ pixels} \times \frac{5}{2} = \frac{3000}{2} = 1500 \text{ pixels}$ . The final width is 1,500 pixels.
A particular metal has a density of 5.2 grams per cubic centimeter. A sculptor has a piece of this metal with a mass of 910 grams. What is the volume of the piece of metal?
Answer: 175 cm³ Density is the ratio of mass to volume ( $D = \frac{M}{V}$ ). This can be set up as a proportion. We are given $D = 5.2 \text{ g/cm³}$ and $M = 910 \text{ g}$ . So, $5.2 = \frac{910}{V}$ . To solve for $V$ , rearrange the equation: $V = \frac{910}{5.2}$ . Performing the division gives $V = 175 \text{ cm³}$ .
A standard medical dosage is 15 milligrams of medicine for every 10 kilograms of a patient's body weight. Following this standard, how many milligrams of medicine should be administered to a patient who weighs 84 kilograms?
Answer: 126 Set up a proportion to find the correct dosage. Let $x$ be the required amount of medicine. $\frac{15 \text{ mg}}{10 \text{ kg}} = \frac{x \text{ mg}}{84 \text{ kg}}$ . To solve for $x$ , cross-multiply: $10x = 15 \times 84$ , which gives $10x = 1260$ . Divide both sides by 10: $x = 126$ . The patient should receive 126 milligrams.
A water tank can be filled by pipe A alone in 12 hours or by pipe B alone in 18 hours. Both pipes are opened simultaneously to fill the empty tank, but after 4 hours, pipe A is closed and only pipe B continues. How many additional hours will pipe B need to finish filling the tank?
Answer: 8.0 hours When you encounter work rate problems, think in terms of "rate of work" - how much of the job each worker (or pipe) completes per unit time. This approach makes complex scenarios much more manageable. Pipe A fills the tank in 12 hours, so its rate is $\frac{1}{12}$ tank per hour. Pipe B takes 18 hours, so its rate is $\frac{1}{18}$ tank per hour. When both pipes work together, their combined rate is $\frac{1}{12} + \frac{1}{18} = \frac{3}{36} + \frac{2}{36} = \frac{5}{36}$ tank per hour. After 4 hours of working together, they fill $4 \times \frac{5}{36} = \frac{20}{36} = \frac{5}{9}$ of the tank. This leaves $1 - \frac{5}{9} = \frac{4}{9}$ of the tank remaining. Now only pipe B continues at its rate of $\frac{1}{18}$ tank per hour. To fill the remaining $\frac{4}{9}$ of the tank: $\frac{4/9}{1/18} = \frac{4}{9} \times 18 = 8$ hours. Choice A (6.0 hours) likely comes from incorrect calculation of the remaining work. Choice B (7.2 hours) might result from using the combined rate instead of just pipe B's rate for the final phase. Choice C (9.6 hours) could come from miscalculating the initial 4-hour period's work. The key strategy for work rate problems is to always convert to rates first, then multiply by time to find work completed. Remember that when workers combine, you add their individual rates.
If 5 identical machines can assemble 750 devices in 6 hours, how many hours would it take 3 of these machines to assemble 900 devices?
Answer: 12 First, find the rate of one machine. The rate of 5 machines is $\frac{750 \text{ devices}}{6 \text{ hours}} = 125 \text{ devices per hour}$ . The rate of one machine is $\frac{125}{5} = 25 \text{ devices per hour}$ . Now, find the rate of 3 machines: $3 \times 25 = 75 \text{ devices per hour}$ . Finally, find the time needed to assemble 900 devices at this rate: $\frac{900 \text{ devices}}{75 \text{ devices/hour}} = 12 \text{ hours}$ .
In a school, the ratio of students to teachers is 22 to 1. The ratio of teachers to administrators is 7 to 2. What is the ratio of students to administrators?
Answer: 77 to 1 Let S be students, T be teachers, and A be administrators. We are given $\frac{S}{T} = \frac{22}{1}$ and $\frac{T}{A} = \frac{7}{2}$ . To find $\frac{S}{A}$ , we can multiply the two ratios: $\frac{S}{T} \times \frac{T}{A} = \frac{22}{1} \times \frac{7}{2}$ . The T's cancel out, leaving $\frac{S}{A} = \frac{154}{2}$ . This ratio simplifies to $\frac{77}{1}$ , or 77 to 1.

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