Ratios, Rates, and Proportions

46 soru

Soru 1Soru

A beverage mixture is prepared by combining apple juice, orange juice, and grape juice in a ratio of 3:4:53 : 4 : 5 by volume. If a pitcher contains a total of 4848 fluid ounces of this mixture, how many fluid ounces of orange juice are in the pitcher?

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Cevap: 1616 fluid ounces

Cevap

1616 fluid ounces
The ratio 3:4:53 : 4 : 5 indicates that the total mixture is divided into 3+4+5=123 + 4 + 5 = 12 equal parts. Orange juice accounts for 44 of these 1212 parts, representing 412=13\frac{4}{12} = \frac{1}{3} of the total volume. Taking 13\frac{1}{3} of 4848 fluid ounces gives 1616 fluid ounces.

Adım Adım Çözüm

1
Calculate the total number of ratio parts.
Total parts = 3+4+5=123 + 4 + 5 = 12.
To find the fraction of each ingredient in the mixture, we must sum all parts of the ratio.
2
Determine the fraction of the total volume that represents orange juice.
Orange juice fraction = 412=13\frac{4}{12} = \frac{1}{3}.
Orange juice corresponds to 44 parts out of the 1212 total parts.
3
Multiply the orange juice fraction by the total volume.
13×48=16\frac{1}{3} \times 48 = 16 fluid ounces.
Multiplying the part-to-whole fraction by the total volume yields the exact volume of orange juice.

Anahtar Kavram

Part-to-Whole Ratio Relationships
Tahmini Süre:45s
Soru 2Soru

A renewable energy facility distributes stored electricity among three battery banks: Bank 1, Bank 2, and Bank 3. Initially, the ratio of the energy stored in Bank 1 to Bank 2 to Bank 3 is 3:4:53 : 4 : 5.

To balance the system, two sequential transfers of energy are performed without any energy loss:
1. Energy is transferred from Bank 3 to Bank 1 such that the ratio of the energy in Bank 1 to Bank 2 becomes 5:45 : 4.
2. Energy is then transferred from Bank 2 to Bank 3 such that the ratio of the energy in Bank 2 to Bank 3 becomes 1:41 : 4.

If Bank 3 contains 60 MWh60\text{ MWh} more energy after the second transfer than it did initially, what was the total amount of energy, in MWh, stored across all three battery banks?

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Cevap: 1,200 MWh1,200\text{ MWh}

Cevap

The total energy stored across all three battery banks is 1,200 MWh1,200\text{ MWh}.
The correct answer is 1,200 MWh1,200\text{ MWh}. By setting up algebraic expressions for the energy in each bank relative to total energy TT, we track how each transfer modifies individual bank totals while conserving total energy. The intermediate transfer reduces Bank 3 from 2560T\frac{25}{60}T to 1560T\frac{15}{60}T, and the subsequent transfer increases it to 2860T\frac{28}{60}T. The resulting difference of 360T=120T\frac{3}{60}T = \frac{1}{20}T equals 60 MWh60\text{ MWh}, solving directly to T=1,200 MWhT = 1,200\text{ MWh}.

Adım Adım Çözüm

1
Express initial energy quantities in terms of total energy TT
Bank 1 = 312T\frac{3}{12}T, Bank 2 = 412T\frac{4}{12}T, Bank 3 = 512T=2560T\frac{5}{12}T = \frac{25}{60}T
The ratio 3:4:53 : 4 : 5 sums to 1212 total parts.
2
Calculate energy amounts after the first transfer (Bank 3 to Bank 1)
Bank 2 remains 412T\frac{4}{12}T; Bank 1 becomes 512T\frac{5}{12}T; Bank 3 becomes T512T412T=312TT - \frac{5}{12}T - \frac{4}{12}T = \frac{3}{12}T
The new ratio of Bank 1 to Bank 2 is 5:45 : 4, and Bank 2's energy did not change.
3
Calculate energy amounts after the second transfer (Bank 2 to Bank 3)
Combined energy in Banks 2 and 3 = 412T+312T=712T\frac{4}{12}T + \frac{3}{12}T = \frac{7}{12}T. Final Bank 3 = 45×712T=2860T\frac{4}{5} \times \frac{7}{12}T = \frac{28}{60}T
Bank 1 remains unchanged at 512T\frac{5}{12}T. The remaining energy is divided between Bank 2 and Bank 3 in a 1:41 : 4 ratio.
4
Determine the net change in Bank 3 and solve for total energy TT
Net change = 2860T2560T=360T=120T\frac{28}{60}T - \frac{25}{60}T = \frac{3}{60}T = \frac{1}{20}T. Since 120T=60 MWh\frac{1}{20}T = 60\text{ MWh}, T=1,200 MWhT = 1,200\text{ MWh}
Bank 3 ended with 60 MWh60\text{ MWh} more than its initial amount.

Anahtar Kavram

Multi-stage ratio rebalancing and conservation of total quantity
Soru 3Soru

In a school club of 4040 students, the ratio of the number of boys to the number of girls is 3:53 : 5. Which of the following statements must be true? Select all such statements.

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Cevap: The number of boys in the club is 1515.; The ratio of the number of boys to the total number of students is 3:83 : 8.; The number of girls in the club exceeds the number of boys by 1010.

Cevap

The correct statements are those indicating that there are 15 boys in the club, that the ratio of boys to total students is 3:8, and that there are 10 more girls than boys in the club.
The total number of ratio parts is 3+5=83 + 5 = 8, representing the entire group of 4040 students. Dividing 4040 by 88 gives 55 students per part. Therefore, the number of boys is 3×5=153 \times 5 = 15, the number of girls is 5×5=255 \times 5 = 25, and the ratio of boys to total students is 15:40=3:815:40 = 3:8. Subtracting the number of boys from girls (2515=1025 - 15 = 10) shows there are 1010 more girls than boys.

Adım Adım Çözüm

1
Calculate total ratio parts and determine the value of one part.
Total ratio parts = 3+5=83 + 5 = 8. Each part corresponds to 408=5\frac{40}{8} = 5 students.
Converting ratio terms into parts allows calculation of actual quantities from the total group size.
2
Compute the total number of boys and girls.
Number of boys = 3×5=153 \times 5 = 15. Number of girls = 5×5=255 \times 5 = 25.
Multiply each component's ratio share by the value of one part.
3
Evaluate the given statements against calculated quantities.
There are 1515 boys; the ratio of boys to total students is 15:40=3:815:40 = 3:8; and the difference between girls and boys is 2515=1025 - 15 = 10.
Determines which options represent true statements about the group.

Anahtar Kavram

Part-to-Part vs. Part-to-Whole Ratios
Soru 4Soru

A pharmaceutical facility uses two liquid compounding lines, Line AA and Line BB, to produce a standardized saline solution. Line AA operates at a constant flow rate of 120120 liters per hour, producing a solution with active compound and purified water in a ratio of 1:31 : 3 by volume. Line BB operates at a constant flow rate of 180180 liters per hour, producing a solution with active compound and purified water in a ratio of 2:32 : 3 by volume. If Line AA runs for 44 hours and Line BB runs for 55 hours, and all output is collected into a single storage tank, what is the ratio of active compound to purified water in the storage tank?

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Cevap: 8:158 : 15

Cevap

The ratio of active compound to purified water in the storage tank is 8:158 : 15.
The correct answer is derived by first converting the flow rates and operating hours into total volumes (480480 L for Line A and 900900 L for Line B). Using the part-to-whole fraction conversion, Line A contributes 120120 L of active compound and 360360 L of water, while Line B contributes 360360 L of active compound and 540540 L of water. Adding these quantities yields 480480 L of active compound and 900900 L of water, which simplifies to the ratio 8:158 : 15.

Adım Adım Çözüm

1
Calculate total output volume and component quantities from Line A
Total Line A Volume = 120 L/hr×4 hrs=480 liters120 \text{ L/hr} \times 4 \text{ hrs} = 480 \text{ liters}. Active Compound = 480×11+3=120 liters480 \times \frac{1}{1+3} = 120 \text{ liters}. Purified Water = 480120=360 liters480 - 120 = 360 \text{ liters}.
Line A produces a total volume of 480 liters, and a 1:31 : 3 ratio means active compound constitutes 14\frac{1}{4} of the total volume.
2
Calculate total output volume and component quantities from Line B
Total Line B Volume = 180 L/hr×5 hrs=900 liters180 \text{ L/hr} \times 5 \text{ hrs} = 900 \text{ liters}. Active Compound = 900×22+3=360 liters900 \times \frac{2}{2+3} = 360 \text{ liters}. Purified Water = 900360=540 liters900 - 360 = 540 \text{ liters}.
Line B produces a total volume of 900 liters, and a 2:32 : 3 ratio means active compound constitutes 25\frac{2}{5} of the total volume.
3
Sum the components from both lines in the storage tank
Total Active Compound = 120+360=480 liters120 + 360 = 480 \text{ liters}. Total Purified Water = 360+540=900 liters360 + 540 = 900 \text{ liters}.
Combining the output of both lines adds the respective volumes of active compound and purified water.
4
Compute and simplify the final ratio of active compound to purified water
Ratio = 480900=815\frac{480}{900} = \frac{8}{15}, expressed as 8:158 : 15.
Dividing both numerator and denominator by their greatest common divisor, 60, yields the simplest integer ratio.

Anahtar Kavram

Weighted mixture ratios combining multi-stage rates and part-to-whole fraction conversions
Soru 5Soru

An artisan bakery produces a signature flour blend by combining wheat, rye, and oat flour. Initially, the ratio of wheat flour to rye flour by weight is 5:35:3, and the ratio of rye flour to oat flour by weight is 4:14:1. The baker prepares an initial batch of this mixture weighing exactly 140140 pounds. To adjust the recipe for a special order, pure oat flour is added to the batch until oat flour accounts for exactly 20%20\% of the total weight of the new mixture. How many pounds of pure oat flour must the baker add?

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Cevap: 20

Cevap

20 pounds
To solve this problem, first express the ratios of wheat, rye, and oat flour with a common term for rye. Since Wheat : Rye = 5 : 3 (or 20 : 12) and Rye : Oat = 4 : 1 (or 12 : 3), the combined ratio Wheat : Rye : Oat is 20 : 12 : 3, giving 35 total parts. In a 140-pound batch, each part equals 4 pounds, meaning the batch contains 12 pounds of oat flour and 128 pounds of non-oat flour. When additional oat flour is added, the non-oat weight stays fixed at 128 pounds. For oat flour to be 20% of the new mixture, non-oat flour must be 80%. Dividing 128 by 0.80 gives a new total weight of 160 pounds. The new oat weight is 20% of 160, which is 32 pounds. Subtracting the initial 12 pounds yields 20 pounds of added oat flour.

Adım Adım Çözüm

1
Combine the two given ratios into a single three-part ratio.
Wheat : Rye = 5:3=20:125 : 3 = 20 : 12, and Rye : Oat = 4:1=12:34 : 1 = 12 : 3. Thus, Wheat : Rye : Oat = 20:12:320 : 12 : 3.
Scaling the ratios so that the common element (Rye) has equal parts allows unification into a single ratio.
2
Calculate the initial weight of each component in the 140-pound batch.
Total ratio parts = 20+12+3=3520 + 12 + 3 = 35 parts. Each part is 140/35=4140 / 35 = 4 pounds. Initial Oat flour = 3×4=123 \times 4 = 12 pounds. Non-oat flour (Wheat + Rye) = (20+12)×4=128(20 + 12) \times 4 = 128 pounds.
Determining the weight of the constant non-oat portion is key to solving mixture adjustment problems.
3
Set up an equation for the new mixture where oat flour makes up 20% of the total weight.
Since non-oat flour remains unchanged at 128128 pounds, it represents 80%80\% of the new total weight. New Total Weight = 128/0.80=160128 / 0.80 = 160 pounds.
If oat flour is 20% of the new total, the non-oat components must constitute the remaining 80%.
4
Calculate the weight of oat flour added.
New Oat Weight = 20%20\% of 160=32160 = 32 pounds. Added Oat Flour = 3212=2032 - 12 = 20 pounds.
Subtracting the initial oat weight from the final required oat weight yields the amount added.

Anahtar Kavram

Combining multi-part ratios and solving ratio adjustment problems by holding unchanged components constant.
Soru 6Soru

In an industrial mechanical system, Gear AA has 1616 teeth, Gear BB has 2424 teeth, and Gear CC has 4040 teeth. Gear AA is meshed directly with Gear BB, and Gear BB is meshed directly with Gear CC. If Gear AA rotates at a constant speed of 150150 revolutions per minute (rpm\text{rpm}), what is the rotational speed, in revolutions per minute (rpm\text{rpm}), of Gear CC?

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Cevap: 60

Cevap

The rotational speed of Gear C is 60 rpm.
For meshed gears, the linear speed of teeth at the point of contact must be identical. Thus, the product of the number of teeth and rotational speed remains constant (NASA=NCSCN_A S_A = N_C S_C). Substituting the known values gives 16×150=40×SC16 \times 150 = 40 \times S_C, yielding 2,400=40SC2,400 = 40 S_C, so SC=60S_C = 60 rpm.

Adım Adım Çözüm

1
Determine the inverse proportional relationship between number of gear teeth and rotational speed.
The product of teeth count and rotational speed is constant across directly meshed gears: NA×SA=NB×SB=NC×SCN_A \times S_A = N_B \times S_B = N_C \times S_C.
Directly meshed gears engage tooth for tooth, meaning they pass the same total number of teeth per unit time.
2
Calculate the total tooth displacement rate per minute from Gear A.
16×150=2,40016 \times 150 = 2,400 teeth per minute.
Gear A has 16 teeth and completes 150 revolutions per minute.
3
Calculate the rotational speed of Gear C.
SpeedC=2,40040=60\text{Speed}_C = \frac{2,400}{40} = 60 rpm.
Gear C has 40 teeth, so dividing the total tooth displacement rate by 40 yields its revolutions per minute.

Anahtar Kavram

Inverse Proportionality in Gear Rates
Tahmini Süre:1m 15s
Soru 7Soru

A bakery uses flour and sugar in a ratio of 5:25:2 by weight to make a cake batter. If a baker needs to prepare a batch using a total of 2828 pounds of flour and sugar combined, how many pounds of flour are needed?

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Cevap: 2020 pounds

Cevap

2020 pounds
To find the amount of flour, determine the fraction of the total mixture that consists of flour. The ratio of flour to sugar is 5:25:2, meaning flour makes up 55 out of 5+2=75 + 2 = 7 total parts. Multiplying this fraction by the total weight gives 57×28=20\frac{5}{7} \times 28 = 20 pounds.

Adım Adım Çözüm

1
Determine the total number of ratio parts
Total ratio parts = 5+2=75 + 2 = 7
The ratio 5:25:2 means that for every 55 parts of flour, there are 22 parts of sugar, giving a total of 77 parts.
2
Calculate the weight of one ratio part
Weight per part = 28 pounds÷7=4 pounds28 \text{ pounds} \div 7 = 4 \text{ pounds}
Dividing the total weight by the total number of parts yields the weight of a single part.
3
Multiply the weight per part by the number of flour parts
Flour weight = 5×4 pounds=20 pounds5 \times 4 \text{ pounds} = 20 \text{ pounds}
Since flour represents 55 of the 77 parts, multiplying 55 by 44 gives the total weight of flour required.

Anahtar Kavram

Part-to-Whole Ratios
Tahmini Süre:45s
Soru 8Soru

Printer X prints at a constant rate of 6060 pages per minute, and Printer Y prints at a constant rate of 4040 pages per minute. If both printers operate simultaneously at their respective constant rates, which of the following statements must be true? Select all that apply.

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Cevap: Working together, the two printers print 100100 pages in 11 minute.; Printer X prints 60%60\% of the total number of pages produced when both printers operate for the same duration.; Working together, the two printers require 55 minutes to print 500500 pages.

Cevap

The correct statements are that working together the printers print 100 pages in 1 minute, Printer X prints 60% of the total pages, and working together they require 5 minutes to print 500 pages.
The combined rate is 60+40=10060 + 40 = 100 pages per minute. In 11 minute, 100100 pages are printed. In 55 minutes, 5×100=5005 \times 100 = 500 pages are printed. Out of every 100100 pages, Printer X produces 6060, which corresponds to 60%60\%.

Adım Adım Çözüm

1
Calculate the combined rate of both printers
Combined rate = 60+40=10060 + 40 = 100 pages per minute
When two entities work together simultaneously, their work rates add directly.
2
Evaluate the proportion of total output printed by Printer X
Proportion = 6060+40=60100=60%\frac{60}{60 + 40} = \frac{60}{100} = 60\%
The fraction of work done by one printer is its rate divided by the total combined rate.
3
Determine the time ratio between Printer Y and Printer X for a fixed job
Time ratio Y : X = 1/401/60=6040=3:2\frac{1/40}{1/60} = \frac{60}{40} = 3:2
Time required for a fixed task is inversely proportional to the work rate.
4
Calculate time needed for 500 pages working together
Time = 500 pages100 pages/min=5 minutes\frac{500\text{ pages}}{100\text{ pages/min}} = 5\text{ minutes}
Time equals total work divided by combined rate.

Anahtar Kavram

Additive work rates and inverse relationship between rate and time in ratio problems
Tahmini Süre:1m 0s
Soru 9Soru

A laboratory technician prepares a cleaning solution by mixing a concentrated formula and distilled water in a ratio of 2:72 : 7 by volume. If the technician uses 1414 liters of the concentrated formula, how many liters of distilled water are required?

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Cevap: 4949

Cevap

4949 liters of distilled water
The ratio of concentrated formula to distilled water is given as 2:72 : 7. Since 1414 liters of concentrated formula represents 22 ratio parts (14=2×714 = 2 \times 7), each ratio part equals 77 liters. Multiplying the 77 ratio parts of distilled water by 77 liters gives 7×7=497 \times 7 = 49 liters.

Adım Adım Çözüm

1
Set up the proportion relating concentrated formula to distilled water
Concentrated FormulaDistilled Water=27=14x\frac{\text{Concentrated Formula}}{\text{Distilled Water}} = \frac{2}{7} = \frac{14}{x}
The given ratio is part-to-part (22 parts formula for every 77 parts water).
2
Solve for the unknown volume xx using cross-multiplication
2x=714    2x=982 \cdot x = 7 \cdot 14 \implies 2x = 98
Cross-multiplying equates the products of the diagonal terms in a proportion.
3
Divide by 22 to isolate xx
x=49x = 49
Isolating xx provides the required amount of distilled water in liters.

Anahtar Kavram

Direct Part-to-Part Proportions
Soru 10Soru

A car travels a total distance of 120120 miles. For the first 6060 miles, the car travels at an average speed of 3030 miles per hour, and for the remaining 6060 miles, the car travels at an average speed of 6060 miles per hour. What is the average speed of the car, in miles per hour, for the entire 120120-mile trip?

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Cevap: 40

Cevap

The average speed of the car for the entire trip is 4040 miles per hour.
The overall average speed is determined by dividing total distance (120120 miles) by total time (33 hours), yielding 4040 miles per hour.

Adım Adım Çözüm

1
Calculate time taken for the first segment
Time = 22 hours
Using Time=DistanceSpeed\text{Time} = \frac{\text{Distance}}{\text{Speed}}, 6030=2\frac{60}{30} = 2 hours.
2
Calculate time taken for the second segment
Time = 11 hour
Using Time=DistanceSpeed\text{Time} = \frac{\text{Distance}}{\text{Speed}}, 6060=1\frac{60}{60} = 1 hour.
3
Determine total distance and total time
Total distance = 120120 miles, Total time = 33 hours
Add distances (60+60=12060 + 60 = 120) and times (2+1=32 + 1 = 3).
4
Calculate overall average speed
Average speed = 4040 miles per hour
Divide total distance by total time: 1203=40\frac{120}{3} = 40.

Anahtar Kavram

Average Speed and Rates
Tahmini Süre:1m 0s
Soru 11Soru

A contractor prepares a concrete mix using gravel and sand in a ratio of 4:74 : 7 by weight. If the contractor uses 2828 tons of sand for a project, what is the total weight, in tons, of the concrete mix?

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Cevap: 4444

Cevap

The total weight of the concrete mix is 4444 tons.
The ratio of gravel to sand is 4:74 : 7. Sand corresponds to 77 parts of the total mixture and weighs 2828 tons, meaning 11 part represents 44 tons. The total mixture comprises 4+7=114 + 7 = 11 parts. Multiplying 1111 parts by 44 tons per part gives a total weight of 4444 tons.

Adım Adım Çözüm

1
Determine the weight per ratio part using the known quantity of sand.
Since 77 parts equal 2828 tons of sand, 11 part =287=4= \frac{28}{7} = 4 tons.
The ratio specifies that sand accounts for 77 equal parts of the total mixture.
2
Calculate the weight of the gravel component.
Gravel weight =4 parts×4 tons/part=16= 4 \text{ parts} \times 4 \text{ tons/part} = 16 tons.
The ratio specifies that gravel accounts for 44 parts of the mixture.
3
Compute the total weight of the concrete mixture.
Total weight =16 tons (gravel)+28 tons (sand)=44= 16 \text{ tons (gravel)} + 28 \text{ tons (sand)} = 44 tons.
The total weight of the mixture is the sum of the weights of all individual ingredients.

Anahtar Kavram

Solving Proportions from Given Component Values
Soru 12Soru

Pipes AA and BB, operating together at their respective constant rates, can fill an empty storage tank in 66 hours. Pipes BB and CC, operating together at their constant rates, can fill the same tank in 88 hours. The rate at which Pipe AA fills the tank is 1.51.5 times the rate at which Pipe CC fills the tank. If all three pipes operate together for 33 hours to fill the empty tank, after which Pipe BB is closed, how many additional hours will it take for Pipes AA and CC together to finish filling the tank?

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Cevap: 65\frac{6}{5}

Cevap

65\frac{6}{5} hours
The rate equations established from the problem yield individual rates of 18\frac{1}{8} tank/hr for Pipe AA, 124\frac{1}{24} tank/hr for Pipe BB, and 112\frac{1}{12} tank/hr for Pipe CC. Working together for 33 hours at a combined rate of 14\frac{1}{4} tank/hr, all three pipes fill 34\frac{3}{4} of the tank, leaving 14\frac{1}{4} of the capacity to be filled. After Pipe BB closes, Pipes AA and CC work at a combined rate of 18+112=524\frac{1}{8} + \frac{1}{12} = \frac{5}{24} tank/hr. Dividing the remaining 14\frac{1}{4} tank by 524\frac{5}{24} tank/hr gives 65\frac{6}{5} hours.

Adım Adım Çözüm

1
Express given conditions in terms of work rates per hour (rA,rB,rCr_A, r_B, r_C).
rA+rB=16r_A + r_B = \frac{1}{6}, rB+rC=18r_B + r_C = \frac{1}{8}, and rA=1.5rC=32rCr_A = 1.5 r_C = \frac{3}{2} r_C.
Filling a full tank in HH hours means completing 1H\frac{1}{H} of the tank per hour.
2
Solve the system of equations for individual rates rA,rB,rCr_A, r_B, r_C.
Substituting rB=18rCr_B = \frac{1}{8} - r_C into rA+rB=16r_A + r_B = \frac{1}{6} yields 32rC+18rC=16    12rC=124    rC=112\frac{3}{2} r_C + \frac{1}{8} - r_C = \frac{1}{6} \implies \frac{1}{2} r_C = \frac{1}{24} \implies r_C = \frac{1}{12}. Consequently, rA=18r_A = \frac{1}{8} and rB=124r_B = \frac{1}{24}.
Finding individual unit rates allows calculating any combined work scenario.
3
Calculate the fraction of the tank filled in the first 3 hours with all three pipes open.
Combined rate rA+rB+rC=18+124+112=3+1+224=624=14r_A + r_B + r_C = \frac{1}{8} + \frac{1}{24} + \frac{1}{12} = \frac{3+1+2}{24} = \frac{6}{24} = \frac{1}{4} of the tank per hour. In 33 hours, 3×14=343 \times \frac{1}{4} = \frac{3}{4} of the tank is filled, leaving 134=141 - \frac{3}{4} = \frac{1}{4} of the tank empty.
Work completed equals rate multiplied by time.
4
Determine the additional time needed for Pipes AA and CC to fill the remaining 14\frac{1}{4} of the tank.
Combined rate of AA and CC is rA+rC=18+112=524r_A + r_C = \frac{1}{8} + \frac{1}{12} = \frac{5}{24} per hour. Additional time t=14524=14×245=65t = \frac{\frac{1}{4}}{\frac{5}{24}} = \frac{1}{4} \times \frac{24}{5} = \frac{6}{5} hours.
Time required equals remaining work divided by active rate.

Anahtar Kavram

Combined Work Rates and Systems of Rate Equations
Soru 13Soru

Three industrial machines, XX, YY, and ZZ, continuously produce liquid chemical mixtures composed solely of Compounds AA and BB.

- Machine XX produces the mixture at a constant rate of 100100 liters per hour, with Compound AA and Compound BB in a ratio of 3:23:2 by volume.
- Machine YY produces the mixture at a constant rate of 150150 liters per hour, with Compound AA and Compound BB in a ratio of 1:41:4 by volume.
- Machine ZZ produces the mixture at a constant rate of RR liters per hour, with Compound AA and Compound BB in a ratio of 7:37:3 by volume.

If all three machines operate simultaneously for 44 hours, the resulting combined mixture contains equal total volumes of Compound AA and Compound BB.

Which of the following statements must be true? Select all that apply.

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Cevap: The operating rate RR of Machine ZZ is 175175 liters per hour.; Machine ZZ accounts for more than 40%40\% of the total volume of the combined mixture produced in 44 hours.; In the 4-hour output, the volume of Compound BB produced by Machine YY is greater than the total volume of Compound BB produced by Machines XX and ZZ combined.

Cevap

The statements asserting that Machine Z's rate is 175 liters per hour, that Machine Z accounts for over 40% of the total output volume, and that Machine Y produces more Compound B than Machines X and Z combined are all true.
The correct options are those stating that Machine Z operates at 175 liters per hour, that Machine Z accounts for over 40% of the 4-hour output, and that Machine Y produces more Compound B than Machines X and Z combined. Machine X produces 240 liters of Compound A and 160 liters of Compound B in 4 hours. Machine Y produces 120 liters of Compound A and 480 liters of Compound B in 4 hours. Machine Z produces 2.8R liters of Compound A and 1.2R liters of Compound B in 4 hours. Equating total Compound A (360 + 2.8R) to total Compound B (640 + 1.2R) gives 1.6R = 280, so R = 175 liters per hour. Machine Z thus contributes 700 liters out of 1700 total liters, which is approximately 41.18% (greater than 40%). Machine Y produces 480 liters of Compound B, which exceeds the 370 liters of Compound B produced by Machines X and Z combined (160 + 210 liters).

Adım Adım Çözüm

1
Calculate the volumes of Compound A and Compound B produced by Machines X and Y in 4 hours.
Machine X produces 400400 L total (240240 L of A, 160160 L of B). Machine Y produces 600600 L total (120120 L of A, 480480 L of B). Combined from X and Y: 360360 L of A and 640640 L of B.
Decompose each machine's total 4-hour output into individual compound amounts using part-to-whole fraction multipliers.
2
Formulate Machine Z's 4-hour output expressions and solve for rate R.
Machine Z produces 4R4R L total (2.8R2.8R L of A, 1.2R1.2R L of B). Setting total A equal to total B: 360+2.8R=640+1.2R    1.6R=280    R=175360 + 2.8R = 640 + 1.2R \implies 1.6R = 280 \implies R = 175 L/hr.
The condition of equal total volumes of Compound A and Compound B forms a single linear equation in terms of R.
3
Determine Machine Z's proportion of the total volume produced by all three machines.
Machine Z volume = 4×175=7004 \times 175 = 700 L. Total volume = 400+600+700=1700400 + 600 + 700 = 1700 L. Percentage = 7001700×100%41.18%>40%\frac{700}{1700} \times 100\% \approx 41.18\% > 40\%.
Evaluate whether Machine Z contributes more than 40% of the complete 1700-liter mixture.
4
Compare Compound B output from Machine Y against the combined Compound B output from X and Z.
Machine Y produces 480480 L of B. Machines X and Z produce 160+(1.2×175)=370160 + (1.2 \times 175) = 370 L of B. 480>370480 > 370 is true.
Direct numerical comparison verifies the inequality statement.
5
Evaluate the remaining ratio claims.
Compound A ratio X to Z is 240:490=24:492:3240 : 490 = 24 : 49 \neq 2 : 3. Combined X and Z ratio of A to B is 730:370=73:377:3730 : 370 = 73 : 37 \neq 7 : 3. Both claims are false.
Check remaining distractors for mathematical accuracy.

Anahtar Kavram

Multi-rate and multi-component ratio mixture balancing
Soru 14Soru

An industrial laboratory prepares a batch of Alloy ZZ weighing 260260 pounds, which consists solely of copper and zinc in a ratio of 7:67 : 6 by weight. A technician then adds 4040 pounds of pure copper to this batch to create Alloy WW. What is the ratio of copper to zinc in Alloy WW?

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Cevap: 3 : 2

Cevap

3 : 2
In Alloy Z, the total weight of 260 pounds is divided into 13 equal parts of 20 pounds each. Thus, copper accounts for 7 parts (140 pounds) and zinc accounts for 6 parts (120 pounds). Adding 40 pounds of pure copper increases the copper content to 180 pounds, while zinc remains at 120 pounds. Comparing copper to zinc yields 180 : 120, which simplifies to 3 : 2.

Adım Adım Çözüm

1
Determine total ratio parts for Alloy Z.
The total number of ratio parts is 7+6=137 + 6 = 13 parts.
To find the weight of each component from a part-to-part ratio, divide the total weight by total ratio parts.
2
Calculate initial weights of copper and zinc in Alloy Z.
Copper weight =260×713=140= 260 \times \frac{7}{13} = 140 pounds; Zinc weight =260×613=120= 260 \times \frac{6}{13} = 120 pounds.
Multiply total weight by each component's fraction of the total.
3
Update component weights after adding pure copper.
New copper weight =140+40=180= 140 + 40 = 180 pounds; Zinc weight remains 120120 pounds.
Only copper is added to the mixture, leaving zinc weight unchanged.
4
Compute and simplify the final ratio of copper to zinc for Alloy W.
Ratio of copper to zinc =180:120=3:2= 180 : 120 = 3 : 2.
Divide both terms by their greatest common divisor, 60.

Anahtar Kavram

Ratios and Component Mixture Adjustment
Soru 15Soru

Three water pumps, P1P_1, P2P_2, and P3P_3, operate at constant individual rates to fill a large reservoir. The ratio of the pumping rate of P1P_1 to that of P2P_2 is 3:43 : 4, and the ratio of the pumping rate of P2P_2 to that of P3P_3 is 3:53 : 5.

At 8:00 AM, all three pumps begin filling an empty reservoir together. At 10:00 AM, pump P1P_1 shuts down, while P2P_2 and P3P_3 continue operating at their original rates. At 11:00 AM, the operating rate of P2P_2 is decreased by 25%25\%, and the operating rate of P3P_3 is increased by 25%25\%. The two remaining pumps continue at these adjusted rates until the reservoir is completely full at 1:00 PM.

If pump P3P_3 were to fill the empty reservoir working alone at its original constant rate, how many hours would it take?

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Cevap: 9.1

Cevap

It would take pump P3P_3 exactly 9.19.1 hours (or 9.19.1 when entered numerically) to fill the empty reservoir alone at its original constant rate.
By unifying the given ratios r1:r2=3:4r_1 : r_2 = 3 : 4 and r2:r3=3:5r_2 : r_3 = 3 : 5, we obtain the relative rates r1=9kr_1 = 9k, r2=12kr_2 = 12k, and r3=20kr_3 = 20k. Summing the work done across the three intervals (8–10 AM at rate 41k41k, 10–11 AM at rate 32k32k, and 11 AM–1 PM at rate 34k34k) yields a total capacity of 182k182k. Dividing total work 182k182k by P3P_3's original rate of 20k20k gives 9.19.1 hours.

Adım Adım Çözüm

1
Unify the individual pumping rate ratios into a single ratio r1:r2:r3r_1 : r_2 : r_3.
r1=9kr_1 = 9k, r2=12kr_2 = 12k, r3=20kr_3 = 20k
Aligning the ratio of P1:P2=3:4=9:12P_1:P_2 = 3:4 = 9:12 and P2:P3=3:5=12:20P_2:P_3 = 3:5 = 12:20 establishes a common scale factor kk.
2
Calculate the work completed in each of the three time intervals.
W1=82kW_1 = 82k, W2=32kW_2 = 32k, W3=68kW_3 = 68k
Multiply the duration of each interval by the sum of active rates during that period.
3
Sum the total work completed to get total capacity and divide by P3P_3's original rate.
Total capacity =182k= 182k; Time =182k20k=9.1= \frac{182k}{20k} = 9.1 hours
The total work equals the sum of work done across all three phases, and time is work divided by rate.

Anahtar Kavram

Multi-stage work rates, compound ratio unification, and percentage rate adjustments
Tahmini Süre:3m 0s
Soru 16Soru

A renewable energy facility stores electricity in three separate battery modules: Module XX, Module YY, and Module ZZ. The ratio of the initial energy stored in Module XX to Module YY is 2:32 : 3, and the ratio of the initial energy stored in Module YY to Module ZZ is 5:85 : 8. Each module discharges its stored energy at a constant individual rate. Operating alone, Module YY can completely discharge its initial stored energy in 55 hours. Module XX discharges energy at a rate 3313%33\frac{1}{3}\% greater than Module YY, and Module ZZ discharges energy at a rate 25%25\% less than the combined discharge rate of Modules XX and YY. If all three modules begin discharging simultaneously, how many hours will it take to completely discharge the total initial energy stored across all three modules?

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Cevap: 4 hours

Cevap

4 hours
To find the total time required, first unify the two given ratios. Since X:Y=2:3X : Y = 2 : 3 and Y:Z=5:8Y : Z = 5 : 8, express both ratios with a common value for YY (1515). This yields X:Y:Z=10:15:24X : Y : Z = 10 : 15 : 24, so total energy is 10k+15k+24k=49k10k + 15k + 24k = 49k. Next, determine rates in terms of kk: Module YY discharges 15k15k in 55 hours, so its rate is 3k3k units/hr. Module XX discharges at a rate 3313%33\frac{1}{3}\% greater than Module YY, giving 3k×43=4k3k \times \frac{4}{3} = 4k units/hr. The combined rate of XX and YY is 4k+3k=7k4k + 3k = 7k units/hr. Module ZZ discharges at 25%25\% less than this combined rate, giving 7k×0.75=5.25k7k \times 0.75 = 5.25k units/hr. The simultaneous discharge rate of all three modules is 4k+3k+5.25k=12.25k4k + 3k + 5.25k = 12.25k units/hr. Dividing total energy 49k49k by 12.25k12.25k gives exactly 4 hours.

Adım Adım Çözüm

1
Determine the unified initial energy ratio among all three modules.
Energy ratio Module XX : Module YY : Module ZZ = 10:15:2410 : 15 : 24, giving a total initial energy of 49k49k units.
Module X:Y=2:3=10:15X : Y = 2 : 3 = 10 : 15 and Module Y:Z=5:8=15:24Y : Z = 5 : 8 = 15 : 24. Combining these gives X:Y:Z=10:15:24X : Y : Z = 10 : 15 : 24 for a common constant kk.
2
Calculate the individual discharge rate of Module Y.
Discharge rate of Module YY = 3k3k energy units per hour.
Module YY has 15k15k energy units and discharges completely in 55 hours, so RateY=15k5=3k\text{Rate}_Y = \frac{15k}{5} = 3k.
3
Determine the discharge rates of Module X and Module Z.
Discharge rate of Module XX = 4k4k units per hour; discharge rate of Module ZZ = 5.25k5.25k units per hour.
Module XX rate is 3313%33\frac{1}{3}\% greater than Module YY: RateX=3k×(1+13)=4k\text{Rate}_X = 3k \times \left(1 + \frac{1}{3}\right) = 4k. Combined rate of XX and YY is 4k+3k=7k4k + 3k = 7k. Module ZZ rate is 25%25\% less than this combined rate: RateZ=7k×0.75=5.25k\text{Rate}_Z = 7k \times 0.75 = 5.25k.
4
Calculate total combined discharge rate and total time required.
Combined rate = 12.25k12.25k units per hour; Total time = 44 hours.
Total rate =4k+3k+5.25k=12.25k= 4k + 3k + 5.25k = 12.25k units per hour. Total time =Total EnergyTotal Rate=49k12.25k=4= \frac{\text{Total Energy}}{\text{Total Rate}} = \frac{49k}{12.25k} = 4 hours.

Anahtar Kavram

Combining multi-part ratios into a unified scale and calculating combined work rates with percentage adjustments
Soru 17Soru

A logistics company operates a fleet of 1414 vans consisting solely of Model XX and Model YY vans.

- The ratio of the cargo volume capacity of one Model XX van to one Model YY van is 3:53 : 5.
- The ratio of the energy consumption rate (in kWh per mile) of one Model XX van to one Model YY van is 4:54 : 5.
- When fully loaded, a Model XX van carries 30 m330\text{ m}^3 of cargo and a Model YY van carries 50 m350\text{ m}^3 of cargo.
- The entire fleet of 1414 vans, when fully loaded, transports a total of 620 m3620\text{ m}^3 of cargo in one trip.
- A Model XX van consumes 2 kWh2\text{ kWh} of energy per mile.

Which of the following statements must be true? Select all such statements.

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Cevap: There are exactly 44 Model XX vans in the fleet.; The total energy consumed by the entire fleet per mile is 33 kWh33\text{ kWh}.

Cevap

The correct statements are that there are exactly 4 Model X vans in the fleet and that the total energy consumed by the entire fleet per mile is 33 kWh.
The statement specifying that there are exactly 4 Model X vans is correct because solving the system nX+nY=14n_X + n_Y = 14 and 30nX+50nY=62030 n_X + 50 n_Y = 620 gives nX=4n_X = 4. The statement asserting that total fleet energy consumption per mile is 33 kWh is also correct because 4(2 kWh)+10(2.5 kWh)=33 kWh4(2\text{ kWh}) + 10(2.5\text{ kWh}) = 33\text{ kWh}.

Adım Adım Çözüm

1
Determine the number of Model X and Model Y vans using fleet totals.
Let nXn_X be the number of Model X vans and nYn_Y be the number of Model Y vans. Given nX+nY=14n_X + n_Y = 14, substitute nY=14nXn_Y = 14 - n_X into the total cargo equation: 30nX+50(14nX)=620    70020nX=620    20nX=80    nX=430 n_X + 50(14 - n_X) = 620 \implies 700 - 20 n_X = 620 \implies 20 n_X = 80 \implies n_X = 4. Thus, nY=10n_Y = 10.
This establishes the exact composition of the fleet.
2
Evaluate the fraction of cargo carried by Model X vans.
Total cargo carried by Model X vans = 4×30=120 m34 \times 30 = 120\text{ m}^3. Fleet total cargo = 620 m3620\text{ m}^3. Fraction = 120620=631\frac{120}{620} = \frac{6}{31}.
Shows that the fleet cargo share depends on the van counts, not just the single-van ratio of 3 to 5.
3
Calculate individual consumption rates and total fleet consumption per mile.
The consumption rate ratio of Model X to Model Y is 4:54 : 5. Given Model X rate is 2 kWh/mile2\text{ kWh/mile}, Model Y rate is 2×54=2.5 kWh/mile2 \times \frac{5}{4} = 2.5\text{ kWh/mile}. Total fleet consumption per mile = 4(2)+10(2.5)=8+25=33 kWh4(2) + 10(2.5) = 8 + 25 = 33\text{ kWh}.
Applies unit rates to the known number of each van model to find total rate.
4
Calculate the true weighted average consumption rate per van.
Average consumption rate per van = Total energy per mileTotal vans=33 kWh14 vans2.36 kWh/mile\frac{\text{Total energy per mile}}{\text{Total vans}} = \frac{33\text{ kWh}}{14\text{ vans}} \approx 2.36\text{ kWh/mile}.
Demonstrates why taking an unweighted mean (2.0+2.5)/2=2.25(2.0 + 2.5)/2 = 2.25 is incorrect when group sizes are unequal.
5
Analyze the impact of a 20% increase in Model Y capacity.
Initial Model Y total capacity = 10×50=500 m310 \times 50 = 500\text{ m}^3. A 20%20\% increase adds 0.20×500=100 m30.20 \times 500 = 100\text{ m}^3. Percent increase in total fleet capacity = 10062016.13%\frac{100}{620} \approx 16.13\%.
A percentage change in a part does not equal the percentage change in the whole.

Anahtar Kavram

Weighted averages and component vs. whole relationships in multi-part ratios and rates.
Soru 18Soru

A chemical synthesis plant uses two reaction vessels, Vessel 1 and Vessel 2, to produce a liquid solution composed entirely of Compound X, Compound Y, and water.

- Vessel 1 produces the solution at a constant rate of 6060 liters per hour, with Compound X, Compound Y, and water in the volume ratio 2:1:32 : 1 : 3, respectively.
- Vessel 2 produces the solution at a constant rate of 9090 liters per hour, with Compound X, Compound Y, and water in the volume ratio 1:3:21 : 3 : 2, respectively.

If both vessels operate simultaneously to fill an initially empty storage tank, which of the following statements must be true regarding the mixture in the storage tank after operating for any duration HH hours (H>0H > 0)? Select all such statements.

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Cevap: Compound Y accounts for exactly 1130\frac{11}{30} of the total volume of solution in the storage tank.; The ratio of the volume of Compound X to the volume of water in the storage tank is 7:127 : 12.

Cevap

The statements confirming that Compound Y accounts for exactly 11/30 of the total solution volume and that the ratio of Compound X to water is 7:12 are both correct.
The total volume rate of solution produced by both vessels combined is 150 liters per hour. Within this total, Compound X is produced at 35 liters per hour, Compound Y at 55 liters per hour, and water at 60 liters per hour. Consequently, Compound Y accounts for 55 / 150 = 11/30 of the total volume, and the ratio of Compound X to water is 35 : 60 = 7 : 12. Both of these statements must be true.

Adım Adım Çözüm

1
Calculate the hourly output rates for each component from Vessel 1.
Total rate = 60 L/hr. Ratio X:Y:Water = 2:1:3 (sum of parts = 6). Compound X = 60 * (2/6) = 20 L/hr; Compound Y = 60 * (1/6) = 10 L/hr; Water = 60 * (3/6) = 30 L/hr.
Decompose the total production rate into component rates using part-to-whole fractions.
2
Calculate the hourly output rates for each component from Vessel 2.
Total rate = 90 L/hr. Ratio X:Y:Water = 1:3:2 (sum of parts = 6). Compound X = 90 * (1/6) = 15 L/hr; Compound Y = 90 * (3/6) = 45 L/hr; Water = 90 * (2/6) = 30 L/hr.
Decompose the second vessel's production rate into component rates.
3
Combine the component rates from both vessels.
Total solution rate = 60 + 90 = 150 L/hr. Compound X = 20 + 15 = 35 L/hr; Compound Y = 10 + 45 = 55 L/hr; Water = 30 + 30 = 60 L/hr.
Find the overall composition delivered to the storage tank per hour.
4
Evaluate each candidate statement against the combined component rates.
Fraction of Y = 55 / 150 = 11/30 (True). Ratio X to Water = 35 : 60 = 7 : 12 (True). X (35 L/hr) > Y (55 L/hr) (False). Water percentage = 60 / 150 = 40% (Not strictly greater than 40%, so False). Fraction of X = 35 / 150 = 7/30 != 7/18 (False).
Determine which statements are mathematically true for any operating duration H > 0.

Anahtar Kavram

Weighted rate aggregation and multi-component ratio analysis
Soru 19Soru

A cyclist completes a journey consisting of three distinct segments: an uphill segment, a flat segment, and a downhill segment. The ratio of the distances of the uphill, flat, and downhill segments is 2:3:52 : 3 : 5, respectively. The cyclist's average speed on the flat segment is twice her average speed on the uphill segment, and her average speed on the downhill segment is three times her average speed on the uphill segment. If the cyclist's overall average speed for the entire journey is 3030 miles per hour, what is her average speed, in miles per hour, on the flat segment?

Cevabı ve açıklamayı göster

Cevap: 31

Cevap

31
The correct average speed on the flat segment is 31 miles per hour. Setting up segment distances as 2x2x, 3x3x, and 5x5x (total distance 10x10x) and segment speeds as vv, 2v2v, and 3v3v, the segment times are t1=2xvt_1 = \frac{2x}{v}, t2=3x2vt_2 = \frac{3x}{2v}, and t3=5x3vt_3 = \frac{5x}{3v}. The total travel time is T=31x6vT = \frac{31x}{6v}. Dividing total distance 10x10x by total time TT yields an overall average speed of 60v31=30\frac{60v}{31} = 30. Solving for vv gives v=15.5v = 15.5 miles per hour. Thus, the average speed on the flat segment is 2v=312v = 31 miles per hour.

Adım Adım Çözüm

1
Define segment distances using ratio multipliers.
Distances are d1=2xd_1 = 2x, d2=3xd_2 = 3x, and d3=5xd_3 = 5x, giving total distance D=10xD = 10x.
The distances of the three segments are in the ratio 2:3:52 : 3 : 5.
2
Express segment speeds relative to the uphill speed vv.
Uphill speed is vv, flat speed is 2v2v, and downhill speed is 3v3v.
The problem states flat speed is twice uphill speed, and downhill speed is three times uphill speed.
3
Calculate the time spent on each segment.
t1=2xvt_1 = \frac{2x}{v}, t2=3x2vt_2 = \frac{3x}{2v}, and t3=5x3vt_3 = \frac{5x}{3v}.
Time equals distance divided by speed (t=dvt = \frac{d}{v}).
4
Calculate total travel time by summing individual segment times.
T=xv(2+32+53)=31x6vT = \frac{x}{v} \left(2 + \frac{3}{2} + \frac{5}{3}\right) = \frac{31x}{6v}.
Combining fractions with a common denominator of 6 gives 12+9+106=316\frac{12 + 9 + 10}{6} = \frac{31}{6}.
5
Relate total distance and total time to the overall average speed.
Average Speed=Total DistanceTotal Time=10x31x6v=60v31=30\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{10x}{\frac{31x}{6v}} = \frac{60v}{31} = 30.
Overall average speed is defined as total distance divided by total time.
6
Solve for vv and calculate the flat segment speed 2v2v.
v=15.5v = 15.5 mph, so flat segment speed =2(15.5)=31= 2(15.5) = 31 mph.
Solving 60v31=30\frac{60v}{31} = 30 yields v=15.5v = 15.5, making 2v=312v = 31.

Anahtar Kavram

Weighted Average Speed and Multi-Segment Distance-Rate-Time Ratios
Soru 20Soru

A commercial coffee roasting facility operates three roasters: XX, YY, and ZZ.

- Roaster XX processes coffee beans at a constant rate of 60 kg/hr60\text{ kg/hr}, producing a blend with an Arabica-to-Robusta ratio of 3:23:2 by weight.
- Roaster YY processes coffee beans at a constant rate of 90 kg/hr90\text{ kg/hr}, producing a blend with an Arabica-to-Robusta ratio of 2:12:1 by weight.
- Roaster ZZ processes coffee beans at a constant rate of 150 kg/hr150\text{ kg/hr}, producing a blend with an Arabica-to-Robusta ratio of 1:41:4 by weight.

All three roasters operate simultaneously for 44 hours to complete a production order. Which of the following statements regarding the production output over this 44-hour period must be true? Select all such statements.

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Cevabı ve açıklamayı göster

Cevap: The total weight of Arabica beans processed across all three roasters is 504 kg504\text{ kg}.; Arabica beans account for exactly 42%42\% of the total weight of coffee beans processed.; The overall ratio of total Arabica weight to total Robusta weight produced is 21:2921:29.

Cevap

The statements confirming that total Arabica weight is 504 kg504\text{ kg}, Arabica accounts for 42%42\% of total weight, and the overall ratio of Arabica to Robusta is 21:2921:29 are all correct.
The total production from Roaster XX is 240 kg240\text{ kg} (144 kg144\text{ kg} Arabica, 96 kg96\text{ kg} Robusta), from Roaster YY is 360 kg360\text{ kg} (240 kg240\text{ kg} Arabica, 120 kg120\text{ kg} Robusta), and from Roaster ZZ is 600 kg600\text{ kg} (120 kg120\text{ kg} Arabica, 480 kg480\text{ kg} Robusta). The total Arabica weight is 144+240+120=504 kg144 + 240 + 120 = 504\text{ kg}, which constitutes 5041,200=42%\frac{504}{1,200} = 42\% of the 1,200 kg1,200\text{ kg} total output. The remaining 696 kg696\text{ kg} is Robusta, giving an Arabica to Robusta ratio of 504:696=21:29504:696 = 21:29.

Adım Adım Çözüm

1
Calculate total output and Arabica quantity for each roaster over 4 hours
Roaster XX: 240 kg240\text{ kg} total, 35×240=144 kg\frac{3}{5} \times 240 = 144\text{ kg} Arabica. Roaster YY: 360 kg360\text{ kg} total, 23×360=240 kg\frac{2}{3} \times 360 = 240\text{ kg} Arabica. Roaster ZZ: 600 kg600\text{ kg} total, 15×600=120 kg\frac{1}{5} \times 600 = 120\text{ kg} Arabica.
Converting hourly rates to total batch weights over 44 hours and applying part-to-whole fractions determined by each ratio.
2
Aggregate total Arabica and Robusta weights across all roasters
Total Arabica = 144+240+120=504 kg144 + 240 + 120 = 504\text{ kg}. Total coffee = 240+360+600=1,200 kg240 + 360 + 600 = 1,200\text{ kg}. Total Robusta = 1,200504=696 kg1,200 - 504 = 696\text{ kg}.
Adding individual component weights to get total aggregate weights.
3
Evaluate percentage and ratio statements
Arabica percentage = 5041,200×100%=42%\frac{504}{1,200} \times 100\% = 42\%. Arabica to Robusta ratio = 504696=2129\frac{504}{696} = \frac{21}{29}.
Verifying overall proportions and simplifying the fraction by dividing by 2424.

Anahtar Kavram

Multi-stage weighted component ratios and rate conversion
Sayfa 1 / 3Sonraki
Ratios, Rates, and Proportions Alıştırma Soruları — GRE General Test | Examkin