Solve Ratio Problems - Praxis Core Math

1. At a factory, Machine A produces widgets at a rate such that for every 4 widgets it produces, Machine B produces 3. If the two machines work together and produce a total of 1,050 widgets, how many widgets were produced by Machine A?
   Answer: 600 The ratio of production for Machine A to Machine B is 4:3. For every cycle, $4+3=7$ widgets are produced in total. To find out how many cycles occurred to produce 1,050 widgets, divide the total by the number of widgets per cycle: $1050 \div 7 = 150$ cycles. Machine A produces 4 widgets per cycle, so its total production is $150 \text{ cycles} \times 4 \frac{\text{widgets}}{\text{cycle}} = 600$ widgets.
2. If $3a = 4b$ and $5b = 6c$, what is the ratio of $a$ to $c$?
   Answer: 8:5 From $3a = 4b$, we can write the ratio $\frac{a}{b} = \frac{4}{3}$. From $5b = 6c$, we can write the ratio $\frac{b}{c} = \frac{6}{5}$. To find the ratio of $a$ to $c$, we can multiply these two ratios: $\frac{a}{c} = \frac{a}{b} \times \frac{b}{c}$. Substituting the values: $\frac{a}{c} = \frac{4}{3} \times \frac{6}{5} = \frac{24}{15}$. This fraction simplifies by dividing the numerator and denominator by their greatest common divisor, 3. $\frac{24 \div 3}{15 \div 3} = \frac{8}{5}$. Therefore, the ratio of $a$ to $c$ is 8:5.
3. A bag contains only red and blue marbles in a ratio of 3:7 (red to blue). If the bag contains a total of 80 marbles, how many red marbles must be added to the bag so that the ratio of red marbles to blue marbles becomes 1:2?
   Answer: 4 First, find the initial number of red and blue marbles. The total number of parts in the ratio is $3+7=10$. With 80 marbles total, each part represents $80 \div 10 = 8$ marbles. Initial red marbles: $3 \times 8 = 24$. Initial blue marbles: $7 \times 8 = 56$. Next, let $x$ be the number of red marbles to add. The number of blue marbles stays at 56. The new ratio is $\frac{24+x}{56} = \frac{1}{2}$. Cross-multiply to solve for $x$: $2(24+x) = 56$, which gives $48+2x=56$. Subtracting 48 from both sides gives $2x=8$, so $x=4$. 4 red marbles must be added.
4. A photocopier can print 150 pages in 4 minutes. At this constant rate, how many pages can it print in 10 minutes?
   Answer: 375 Set up a proportion to solve this problem. Let $p$ be the number of pages printed in 10 minutes. The ratio of pages to minutes is constant. $\frac{150 \text{ pages}}{4 \text{ minutes}} = \frac{p \text{ pages}}{10 \text{ minutes}}$. To solve for $p$, cross-multiply: $4p = 150 \times 10 = 1500$. Divide by 4: $p = \frac{1500}{4} = 375$. The photocopier can print 375 pages in 10 minutes.
5. A car travels 125 miles on 5 gallons of gasoline. At this rate, how much gasoline, in gallons, would be needed to travel 350 miles?
   Answer: 14 This is a proportion problem. Let $g$ be the number of gallons needed for 350 miles. We can set up the ratio of miles to gallons: $\frac{125 \text{ miles}}{5 \text{ gallons}} = \frac{350 \text{ miles}}{g \text{ gallons}}$. First, you can simplify the rate: $\frac{125}{5} = 25$ miles per gallon. Now, to find the gallons needed for 350 miles, divide the distance by the rate: $g = \frac{350 \text{ miles}}{25 \text{ mpg}} = 14$ gallons. Alternatively, using the original proportion: $125g = 350 \times 5$, so $125g = 1750$. Dividing both sides by 125 gives $g = 14$.
6. In a certain classroom, the ratio of students who prefer fiction to students who prefer nonfiction is 5 to 3. If there are 32 students in the classroom, how many more students prefer fiction than nonfiction?
   Answer: 8 The ratio of fiction to nonfiction is 5:3. This means there are $5+3=8$ total parts in the ratio. The 32 students represent these 8 parts. To find the number of students per part, divide the total students by the total parts: $32 \div 8 = 4$ students per part. The number of students who prefer fiction is $5 \text{ parts} \times 4 \frac{\text{students}}{\text{part}} = 20$ students. The number who prefer nonfiction is $3 \text{ parts} \times 4 \frac{\text{students}}{\text{part}} = 12$ students. The question asks for how many more students prefer fiction, which is the difference: $20 - 12 = 8$.
7. A city has a population of 80,000 and an area of 50 square miles. A neighboring city has a population of 120,000 and an area of 75 square miles. What is the ratio of the population density of the first city to the population density of the second city?
   Answer: 1:1 Population density is calculated as Population / Area. For the first city: Density₁ = $\frac{80,000}{50} = 1,600$ people per square mile. For the second city: Density₂ = $\frac{120,000}{75} = 1,600$ people per square mile. The ratio of the population density of the first city to the second city is $\frac{1600}{1600}$, which simplifies to 1:1.
8. If 5 machines can produce 300 widgets in 6 hours, how many hours would it take for 8 machines to produce 480 widgets, assuming all machines work at the same constant rate?
   Answer: 6 First, find the production rate of a single machine. If 5 machines produce 300 widgets in 6 hours, then one machine produces $\frac{300}{5} = 60$ widgets in 6 hours. The rate for one machine is $\frac{60 \text{ widgets}}{6 \text{ hours}} = 10$ widgets per hour. Now, find how long it would take 8 machines to produce 480 widgets. The combined rate of 8 machines is $8 \text{ machines} \times 10 \frac{\text{widgets/hr}}{\text{machine}} = 80$ widgets per hour. The time required is Total Widgets / Rate = $\frac{480 \text{ widgets}}{80 \text{ widgets/hr}} = 6$ hours.
9. To make a certain type of concrete, the ratio of cement to sand to gravel by weight is 2:3:5. A construction crew needs to make 1,500 pounds of this concrete. How many more pounds of gravel are needed than cement?
   Answer: 450 The ratio is 2:3:5. The total number of parts is $2+3+5=10$. The total weight of the concrete is 1,500 pounds. The weight of one part is $\frac{1,500 \text{ lbs}}{10 \text{ parts}} = 150$ pounds per part. The weight of cement is $2 \text{ parts} \times 150 \frac{\text{lbs}}{\text{part}} = 300$ pounds. The weight of gravel is $5 \text{ parts} \times 150 \frac{\text{lbs}}{\text{part}} = 750$ pounds. The difference between the weight of gravel and cement is $750 - 300 = 450$ pounds. Alternatively, the difference in parts is $5-2=3$ parts. The difference in weight is $3 \text{ parts} \times 150 \frac{\text{lbs}}{\text{part}} = 450$ pounds.
10. Two gears are connected. The first gear has 72 teeth and the second gear has 32 teeth. For every 8 complete revolutions of the first gear, how many complete revolutions does the second gear make?
    Answer: 18 The ratio of the number of revolutions of two connected gears is inversely proportional to the ratio of their number of teeth. Let $T_1$ and $R_1$ be the teeth and revolutions of the first gear, and $T_2$ and $R_2$ be for the second. The relationship is $T_1 \times R_1 = T_2 \times R_2$. We are given $T_1 = 72$, $T_2 = 32$, and $R_1 = 8$. We need to find $R_2$. Plugging in the values: $72 \times 8 = 32 \times R_2$. This gives $576 = 32 \times R_2$. Solving for $R_2$: $R_2 = \frac{576}{32} = 18$. The second gear makes 18 revolutions.
11. The ratio of width to length of a rectangular banner is 2:7. If the perimeter of the banner is 108 inches, what is the length of the banner in inches?
    Answer: 42 Let the width be $2x$ and the length be $7x$. The perimeter of a rectangle is given by the formula $P = 2(W+L)$. We are given $P=108$. So, $108 = 2(2x+7x)$. Simplify the expression in the parentheses: $108 = 2(9x)$, which is $108 = 18x$. Solve for $x$ by dividing by 18: $x = \frac{108}{18} = 6$. The question asks for the length of the banner, which is $7x$. So, the length is $7 \times 6 = 42$ inches.
12. A model airplane is built with a scale of 1:48. If the wingspan of the actual airplane is 96 feet, what is the wingspan of the model in inches?
    Answer: 24 The scale is 1:48, which means 1 unit on the model represents 48 units on the actual airplane. Let the model's wingspan be $m$. The proportion is $\frac{\text{model}}{\text{actual}} = \frac{1}{48}$. So, $\frac{m}{96 \text{ feet}} = \frac{1}{48}$. Solving for $m$: $48m = 96$, so $m = 2$ feet. The question asks for the wingspan in inches. Since there are 12 inches in a foot, convert the model's wingspan to inches: $2 \text{ feet} \times 12 \frac{\text{inches}}{\text{foot}} = 24$ inches.
13. A bronze alloy is made from copper and tin in the ratio 11:1 by mass. If a statue made of this alloy contains 24 kilograms of tin, what is the total mass of the statue?
    Answer: 288 kg The ratio of copper to tin is 11:1. This means there are 11 parts copper for every 1 part tin. The total number of parts in the alloy is $11+1=12$. We are told that the 1 part of tin corresponds to a mass of 24 kilograms. Therefore, the mass of one part is 24 kg. The total mass of the statue corresponds to 12 parts. So, the total mass is $12 \text{ parts} \times 24 \frac{\text{kg}}{\text{part}} = 288$ kg.
14. A recipe for 12 cookies requires $\frac{3}{4}$ cup of sugar. A baker needs to make 90 cookies for a catering event. If sugar is only sold in full 1-cup bags, what is the minimum number of bags of sugar the baker must purchase?
    Answer: 6 bags First, set up a proportion to find the total amount of sugar needed. Let $x$ be the cups of sugar for 90 cookies. $\frac{\frac{3}{4} \text{ cup}}{12 \text{ cookies}} = \frac{x \text{ cups}}{90 \text{ cookies}}$. Solving for $x$: $12x = 90 \times \frac{3}{4}$, which simplifies to $12x = \frac{270}{4} = 67.5$. Divide by 12: $x = \frac{67.5}{12} = 5.625$ cups. Since sugar is sold in full 1-cup bags, the baker cannot buy a fraction of a bag. The baker needs 5.625 cups, so they must purchase 6 full bags to have enough.
15. A map has a scale where 2 inches represents 75 miles. The distance on the map between two cities is 5.5 inches. A driver averages 60 miles per hour. How long will the drive between the two cities take?
    Answer: 3 hours, 26 minutes First, find the actual distance in miles. Set up a proportion: $\frac{2 \text{ inches}}{75 \text{ miles}} = \frac{5.5 \text{ inches}}{x \text{ miles}}$. Cross-multiply: $2x = 75 \times 5.5 = 412.5$. Solve for $x$: $x = 206.25$ miles. Next, use the formula Time = Distance / Speed. Time = $\frac{206.25 \text{ miles}}{60 \text{ mph}} = 3.4375$ hours. To convert this to hours and minutes, the whole number is the hours (3 hours). Convert the decimal part to minutes: $0.4375 \text{ hours} \times 60 \frac{\text{minutes}}{\text{hour}} = 26.25$ minutes. Rounded to the nearest minute, this is 26 minutes. The total time is 3 hours and 26 minutes.
16. A sum of money was shared between Liam and Mia in the ratio 5:8. Mia received $21 more than Liam. What was the total sum of money shared?
    Answer: $91 The ratio of Liam's share to Mia's share is 5:8. Let Liam's share be $5x$ and Mia's share be $8x$. The difference in their shares is $8x - 5x = 3x$. We are told this difference is $21. So, $3x = \$21$, which means $x = \$7$. The total sum of money is the sum of their shares, which is $5x + 8x = 13x$. Substitute the value of $x$: Total sum = $13 \times \$7 = \$91$.
17. A wildlife biologist catches, tags, and releases 80 deer in a forest. A month later, she catches a sample of 150 deer and finds that 12 of them have tags. What is the most reasonable estimate for the total deer population in the forest?
    Answer: 1000 This is a capture-recapture problem that can be solved with a proportion. The ratio of tagged deer to the total population should be approximately equal to the ratio of tagged deer in the sample to the sample size. Let $P$ be the total deer population. $\frac{\text{Tagged in population}}{\text{Total population}} = \frac{\text{Tagged in sample}}{\text{Sample size}}$. Plugging in the numbers: $\frac{80}{P} = \frac{12}{150}$. To solve for $P$, cross-multiply: $12P = 80 \times 150 = 12000$. Divide by 12: $P = \frac{12000}{12} = 1000$. The estimated total deer population is 1000.
18. A grocery store sells almonds in two sizes. A 12-ounce bag costs $5.40, and a 2-pound bag costs $12.80. What is the difference in the cost per ounce between the two bags? (Note: 1 pound = 16 ounces)
    Answer: $0.05 First, find the cost per ounce for the smaller bag. Cost per ounce = Total cost / Number of ounces = $\frac{\$5.40}{12 \text{ oz}} = \$0.45$ per ounce. Next, find the cost per ounce for the larger bag. First, convert the weight to ounces: $2 \text{ pounds} \times 16 \frac{\text{ounces}}{\text{pound}} = 32$ ounces. Then, calculate the cost per ounce: $\frac{\$12.80}{32 \text{ oz}} = \$0.40$ per ounce. Finally, find the difference in the cost per ounce: $\$0.45 - \$0.40 = \$0.05$.
19. A 6-foot-tall person casts a shadow that is 9 feet long. At the same time, a nearby flagpole casts a shadow that is 30 feet long. What is the height of the flagpole in yards?
    Answer: 6.7 The ratio of an object's height to its shadow's length is constant. Let $h$ be the height of the flagpole in feet. Set up the proportion: $\frac{\text{person's height}}{\text{person's shadow}} = \frac{\text{flagpole's height}}{\text{flagpole's shadow}}$. So, $\frac{6 \text{ ft}}{9 \text{ ft}} = \frac{h}{30 \text{ ft}}$. Simplify the ratio $\frac{6}{9}$ to $\frac{2}{3}$. Now we have $\frac{2}{3} = \frac{h}{30}$. Cross-multiply: $3h = 2 \times 30 = 60$. Solve for $h$: $h = 20$ feet. The question asks for the height in yards. Since there are 3 feet in 1 yard, divide the height in feet by 3: $20 \text{ feet} \div 3 \frac{\text{feet}}{\text{yard}} \approx 6.67$ yards. Rounded to one decimal place, this is 6.7 yards.
20. The ratio of the number of apples to the number of oranges in a basket is 3:5. The ratio of the number of oranges to the number of pears is 2:3. If there are 30 pears in the basket, how many apples are there?
    Answer: 12 Let A, O, and P be the number of apples, oranges, and pears. We are given $\frac{A}{O} = \frac{3}{5}$ and $\frac{O}{P} = \frac{2}{3}$. We are also given $P=30$. First, use the second ratio to find the number of oranges. $\frac{O}{30} = \frac{2}{3}$. Cross-multiply: $3O = 30 \times 2 = 60$, so $O=20$. Now use the first ratio to find the number of apples. $\frac{A}{20} = \frac{3}{5}$. Cross-multiply: $5A = 20 \times 3 = 60$, so $A=12$. There are 12 apples.

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