Fractions and Rational Numbers

32 soru

Soru 21Soru

A water storage tank is initially filled to 45\frac{4}{5} of its total capacity with water. First, 14\frac{1}{4} of the water in the tank is drained for irrigation. Next, an amount of water equal to 13\frac{1}{3} of the remaining water in the tank is added back. Finally, 38\frac{3}{8} of the water currently in the tank is removed for domestic use. Which of the following statements about the volume of water in the tank must be true? Select all such statements.

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Cevap: After the second operation, the volume of water in the tank is equal to the initial volume of water before any operations.; The final volume of water in the tank is equal to 58\frac{5}{8} of the initial volume of water.; The total volume of water removed from the tank during the first and third operations combined is equal to 12\frac{1}{2} of the total capacity of the tank.

Cevap

The correct statements are: (1) After the second operation, the volume of water in the tank is equal to the initial volume of water before any operations; (2) The final volume of water in the tank is equal to 5/8 of the initial volume of water; and (3) The total volume of water removed from the tank during the first and third operations combined is equal to 1/2 of the total capacity of the tank.
The statement regarding the volume after the second operation is correct because adding 1/3 of the remaining 3/5 capacity adds 1/5 capacity, returning the total volume to 4/5 capacity. The statement comparing the final volume to the initial volume is correct because (1/2) divided by (4/5) equals 5/8. The statement regarding total water removed in the first and third operations is correct because 1/5 capacity plus 3/10 capacity equals 1/2 capacity.

Adım Adım Çözüm

1
Define total tank capacity as TT and find initial water volume.
Initial volume V0=45TV_0 = \frac{4}{5}T.
The problem states the tank is initially 45\frac{4}{5} full.
2
Calculate water remaining and removed after Operation 1.
Removed = 14×45T=15T\frac{1}{4} \times \frac{4}{5}T = \frac{1}{5}T. Remaining V1=45T15T=35TV_1 = \frac{4}{5}T - \frac{1}{5}T = \frac{3}{5}T.
Draining 14\frac{1}{4} of existing water leaves 34\frac{3}{4} of the existing water.
3
Calculate water remaining after Operation 2.
Added = 13×35T=15T\frac{1}{3} \times \frac{3}{5}T = \frac{1}{5}T. New volume V2=35T+15T=45TV_2 = \frac{3}{5}T + \frac{1}{5}T = \frac{4}{5}T.
Adding 13\frac{1}{3} of the remaining volume increases it by a factor of 1+13=431 + \frac{1}{3} = \frac{4}{3}.
4
Calculate water remaining and removed after Operation 3.
Removed = 38×45T=310T\frac{3}{8} \times \frac{4}{5}T = \frac{3}{10}T. Final volume V3=45T×(138)=45T×58=12TV_3 = \frac{4}{5}T \times \left(1 - \frac{3}{8}\right) = \frac{4}{5}T \times \frac{5}{8} = \frac{1}{2}T.
Removing 38\frac{3}{8} of current water leaves 58\frac{5}{8} of that volume.
5
Evaluate each given statement against calculated values.
V2=V0=45TV_2 = V_0 = \frac{4}{5}T (True). V3V0=12T45T=58\frac{V_3}{V_0} = \frac{\frac{1}{2}T}{\frac{4}{5}T} = \frac{5}{8} (True). Total removed = 15T+310T=12T\frac{1}{5}T + \frac{3}{10}T = \frac{1}{2}T (True).
Direct comparison with calculated step outcomes confirms these three statements are valid.

Anahtar Kavram

Sequential Fraction Multiplication and Part-to-Whole Relationships
Tahmini Süre:1m 45s
Soru 22Soru

In a manufacturing facility, three automated machines—Machine A, Machine B, and Machine C—produce components at constant individual rates. Machine A working alone completes 310\frac{3}{10} of a standard daily order in 33 hours. Machine B working alone completes 25\frac{2}{5} of the order in 66 hours. Machine C working alone completes 14\frac{1}{4} of the order in 55 hours.

Initially, all three machines work together for 33 hours. At the end of 33 hours, Machine A malfunctions and stops, while Machines B and C continue working together without interruption until the daily order is finished. How many total hours does it take from the start to complete the entire daily order?

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Cevap: 6

Cevap

The total time required from the start to complete the entire daily order is 6 hours.
Each machine's rate per hour is found by dividing the given fraction by hours: Machine A is 110\frac{1}{10}, B is 115\frac{1}{15}, and C is 120\frac{1}{20}. Summing these gives an initial combined rate of 1360\frac{13}{60} per hour. In the first 3 hours, the three machines complete 3×1360=13203 \times \frac{13}{60} = \frac{13}{20} of the order, leaving 11320=7201 - \frac{13}{20} = \frac{7}{20} of the order remaining. When Machine A stops, the remaining combined rate of B and C is 115+120=760\frac{1}{15} + \frac{1}{20} = \frac{7}{60} per hour. Dividing the remaining 720\frac{7}{20} by 760\frac{7}{60} yields 3 additional hours. Adding the initial 3 hours gives a total duration of 6 hours.

Adım Adım Çözüm

1
Determine individual hourly work rates for each machine.
Machine A rate = 110\frac{1}{10} order/hr, Machine B rate = 115\frac{1}{15} order/hr, Machine C rate = 120\frac{1}{20} order/hr.
Divide the fraction of the job completed by the duration in hours to find the unit rate for each machine.
2
Calculate the combined rate of all three machines and the fraction completed in the first 3 hours.
Combined rate = 1360\frac{13}{60} order/hr; Work completed in 3 hours = 1320\frac{13}{20}, leaving 720\frac{7}{20} of the order unfinished.
Sum the three rates using a common denominator of 60, then multiply by 3 hours to find the total work done during the first stage.
3
Calculate the combined rate of Machines B and C, and find the additional time required to complete the remaining fraction.
Combined rate of B and C = 760\frac{7}{60} order/hr; Additional time needed = 3 hours.
Divide the remaining fraction 720\frac{7}{20} by the combined rate 760\frac{7}{60} to determine the remaining hours needed.
4
Add the initial duration and additional duration to obtain the total time.
Total time = 3+3=63 + 3 = 6 hours.
The question asks for the total elapsed time from the start of the job.

Anahtar Kavram

Adding rational fractions with different denominators to solve multi-stage work and rate problems.
Soru 23Soru

A technology company allocates its quarterly research budget among three key projects: Project Alpha, Project Beta, and Project Gamma. Initially, Project Alpha receives 25\frac{2}{5} of the total budget. Of the remaining budget, Project Beta is allocated 47\frac{4}{7}, and Project Gamma receives the rest. During a mid-quarter review, 16\frac{1}{6} of Project Alpha's allocated budget and 14\frac{1}{4} of Project Beta's allocated budget are transferred to Project Gamma. After these transfers, Project Gamma's final budget is what fraction of the total initial quarterly research budget?

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Cevap: 43105\frac{43}{105}

Cevap

43105\frac{43}{105}
Project Alpha receives 25\frac{2}{5} of the total budget, leaving 35\frac{3}{5}. Project Beta gets 47\frac{4}{7} of 35\frac{3}{5}, which is 1235\frac{12}{35}. Project Gamma's initial share is 351235=935\frac{3}{5} - \frac{12}{35} = \frac{9}{35}. Transferring 16\frac{1}{6} of Alpha's share contributes 115\frac{1}{15}, and transferring 14\frac{1}{4} of Beta's share contributes 335\frac{3}{35}. Adding these to Gamma's initial share yields 935+335+115=1235+115=36+7105=43105\frac{9}{35} + \frac{3}{35} + \frac{1}{15} = \frac{12}{35} + \frac{1}{15} = \frac{36 + 7}{105} = \frac{43}{105}.

Adım Adım Çözüm

1
Determine Project Alpha's initial allocation and the remaining budget fraction.
Project Alpha gets 25\frac{2}{5} of the total budget BB. The remaining fraction is 125=35B1 - \frac{2}{5} = \frac{3}{5} B.
Subtractions from the whole yield the remaining unallocated portion.
2
Calculate the initial allocations for Project Beta and Project Gamma.
Project Beta receives 47×35B=1235B\frac{4}{7} \times \frac{3}{5} B = \frac{12}{35} B. Project Gamma receives the rest of the remaining budget: 35B1235B=2135B1235B=935B\frac{3}{5} B - \frac{12}{35} B = \frac{21}{35} B - \frac{12}{35} B = \frac{9}{35} B.
Project Beta's share is a fraction of the remaining budget, not the total budget.
3
Calculate the transferred amounts from Project Alpha and Project Beta to Project Gamma.
Transfer from Alpha = 16×25B=115B\frac{1}{6} \times \frac{2}{5} B = \frac{1}{15} B. Transfer from Beta = 14×1235B=335B\frac{1}{4} \times \frac{12}{35} B = \frac{3}{35} B.
Transfers are fractional parts of each project's individual initial allocation.
4
Sum Project Gamma's initial share and the two transferred amounts.
Gamma's final fraction = 935+115+335=1235+115=36105+7105=43105\frac{9}{35} + \frac{1}{15} + \frac{3}{35} = \frac{12}{35} + \frac{1}{15} = \frac{36}{105} + \frac{7}{105} = \frac{43}{105}.
Finding the least common denominator (105105) allows exact addition of the rational numbers.

Anahtar Kavram

Multi-step sequential operations with fractions and rational numbers
Soru 24Soru

A charity foundation allocates its annual grant budget among three sectors: Healthcare, Education, and Environmental Conservation. First, 25\frac{2}{5} of the total annual budget is allocated to Healthcare. Next, 59\frac{5}{9} of the remaining budget is allocated to Education. The final remainder of $120,000\$120,000 is allocated to Environmental Conservation. What was the foundation's total annual grant budget?

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Cevap: $450,000\$450,000

Cevap

$450,000\$450,000
The option specifying $450,000\$450,000 is correct because allocating 25\frac{2}{5} leaves 35\frac{3}{5} of the budget. Taking 59\frac{5}{9} of 35\frac{3}{5} gives 13\frac{1}{3} of the total. The combined allocations leave 1(25+13)=4151 - (\frac{2}{5} + \frac{1}{3}) = \frac{4}{15} of the total budget, which equals $120,000\$120,000. Dividing $120,000\$120,000 by 415\frac{4}{15} yields $450,000\$450,000.

Adım Adım Çözüm

1
Determine the fraction of the total budget remaining after the Healthcare allocation.
Since Healthcare receives 25\frac{2}{5} of the total budget, the remaining fraction is 125=351 - \frac{2}{5} = \frac{3}{5}.
Sequential fraction operations require updating the remaining whole after each step.
2
Calculate the fraction of the total budget allocated to Education.
Education receives 59\frac{5}{9} of the remaining 35\frac{3}{5}, which is 59×35=1545=13\frac{5}{9} \times \frac{3}{5} = \frac{15}{45} = \frac{1}{3} of the total budget.
To find a fraction of a remaining fraction, multiply the two fractions together.
3
Find the final remaining fraction allocated to Environmental Conservation.
Subtract both allocations from the whole: 1(25+13)=11115=4151 - \left(\frac{2}{5} + \frac{1}{3}\right) = 1 - \frac{11}{15} = \frac{4}{15}.
The final dollar amount corresponds to the fraction of the total budget left over.
4
Solve for the total annual grant budget.
415×Total=$120,000    Total=$120,000×154=$450,000\frac{4}{15} \times \text{Total} = \$120,000 \implies \text{Total} = \$120,000 \times \frac{15}{4} = \$450,000.
Dividing the part by its corresponding fractional proportion yields the total whole.

Anahtar Kavram

Sequential Fraction Operations and Finding the Whole from a Fractional Part
Soru 25Soru

In a community library, 415\frac{4}{15} of the total book collection consists of fiction books, 25\frac{2}{5} consists of non-fiction books, and the remaining 240 books are children's books. If 38\frac{3}{8} of the fiction books are hardcovers and the rest are softcovers, how many softcover fiction books are in the library?

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Cevap: 120

Cevap

The total number of softcover fiction books in the library is 120.
To find the number of softcover fiction books, first calculate the total number of books in the library. The combined fraction of fiction and non-fiction books is 415+25=415+615=1015=23\frac{4}{15} + \frac{2}{5} = \frac{4}{15} + \frac{6}{15} = \frac{10}{15} = \frac{2}{3}. Therefore, children's books represent the remaining 123=131 - \frac{2}{3} = \frac{1}{3} of the collection. Since 13\frac{1}{3} of the total equals 240 books, the total collection is 240×3=720240 \times 3 = 720 books. Next, compute the total number of fiction books: 415×720=192\frac{4}{15} \times 720 = 192. Finally, since 38\frac{3}{8} of fiction books are hardcovers, the remaining fraction of softcovers is 138=581 - \frac{3}{8} = \frac{5}{8}. Multiplying 58×192\frac{5}{8} \times 192 gives 120 softcover fiction books.

Adım Adım Çözüm

1
Find the combined fraction of fiction and non-fiction books, then determine the fraction of children's books.
Combined fraction = 415+615=1015=23\frac{4}{15} + \frac{6}{15} = \frac{10}{15} = \frac{2}{3}. Children's books fraction = 123=131 - \frac{2}{3} = \frac{1}{3}.
The sum of all non-overlapping fractions comprising the entire collection must equal 1.
2
Calculate the total number of books in the library.
Total collection = 240÷13=720240 \div \frac{1}{3} = 720 books.
The 240 children's books represent exactly 13\frac{1}{3} of the whole collection.
3
Determine the total count of fiction books.
Fiction books = 415×720=192\frac{4}{15} \times 720 = 192 books.
Fiction books make up 415\frac{4}{15} of the total 720 books.
4
Calculate the number of softcover fiction books.
Softcover fiction books = (138)×192=58×192=120\left(1 - \frac{3}{8}\right) \times 192 = \frac{5}{8} \times 192 = 120 books.
If 38\frac{3}{8} of the fiction subset are hardcovers, the remaining 58\frac{5}{8} of that subset are softcovers.

Anahtar Kavram

Solving multi-step rational number word problems by combining fractions and determining fractional parts of a subset.
Tahmini Süre:1m 30s
Soru 26Soru

An agricultural research station divides an experimental plot of land among three crops: Wheat, Rye, and Barley. Initially, Wheat occupies 49\frac{4}{9} of the total plot area, Rye occupies 13\frac{1}{3} of the total area, and Barley occupies the remaining area. First, 13\frac{1}{3} of the Wheat area is converted to Rye. Next, 25\frac{2}{5} of the new total Rye area is converted to Barley. Finally, 14\frac{1}{4} of the new total Barley area is converted back to Wheat. If the final area occupied by Wheat is 126126 hectares, what was the total area, in hectares, of the experimental plot?

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Cevap: 315

Cevap

315 hectares
By tracking the dynamic changes to each crop's area step-by-step, the final Wheat area is 54135=25\frac{54}{135} = \frac{2}{5} of the total land area. Setting 25T=126\frac{2}{5} T = 126 yields T=315T = 315 hectares.

Adım Adım Çözüm

1
Determine initial fraction allocations for each crop
Wheat: 49\frac{4}{9}, Rye: 13=39\frac{1}{3} = \frac{3}{9}, Barley: 1(49+39)=291 - \left(\frac{4}{9} + \frac{3}{9}\right) = \frac{2}{9}
The sum of all three crop areas must equal the total area (11).
2
Apply the first conversion (13\frac{1}{3} of Wheat area converted to Rye)
Wheat transferred: 13×49=427\frac{1}{3} \times \frac{4}{9} = \frac{4}{27}. New Wheat: 49427=827\frac{4}{9} - \frac{4}{27} = \frac{8}{27}. New Rye: 13+427=1327\frac{1}{3} + \frac{4}{27} = \frac{13}{27}. Barley: 29=627\frac{2}{9} = \frac{6}{27}.
Wheat loses 13\frac{1}{3} of its current area, and Rye gains that exact amount.
3
Apply the second conversion (25\frac{2}{5} of new Rye area converted to Barley)
Rye transferred: 25×1327=26135\frac{2}{5} \times \frac{13}{27} = \frac{26}{135}. New Rye: 132726135=39135\frac{13}{27} - \frac{26}{135} = \frac{39}{135}. New Barley: 627+26135=56135\frac{6}{27} + \frac{26}{135} = \frac{56}{135}. Wheat remains 827=40135\frac{8}{27} = \frac{40}{135}.
Rye loses 25\frac{2}{5} of its updated area, and Barley gains that exact amount.
4
Apply the third conversion (14\frac{1}{4} of new Barley area converted to Wheat)
Barley transferred: 14×56135=14135\frac{1}{4} \times \frac{56}{135} = \frac{14}{135}. New Wheat: 40135+14135=54135=25\frac{40}{135} + \frac{14}{135} = \frac{54}{135} = \frac{2}{5}.
Barley loses 14\frac{1}{4} of its updated area, and Wheat gains that exact amount.
5
Solve for the total plot area TT
25T=126    T=126×52=315\frac{2}{5} T = 126 \implies T = \frac{126 \times 5}{2} = 315 hectares.
Set the final calculated fraction of Wheat equal to the given area of 126126 hectares.

Anahtar Kavram

Sequential fractional updates of non-disjoint sub-amounts
Tahmini Süre:2m 30s
Soru 27Soru

At a financial trading firm, incoming trade orders pass through a three-stage automated verification pipeline:

1. In Stage 1, 16\frac{1}{6} of all incoming orders are rejected as invalid, and 15\frac{1}{5} of the remaining valid orders are flagged for manual review.
2. In Stage 2, of the orders that were cleared directly without manual review, 38\frac{3}{8} are executed immediately, while the rest are placed in a waiting queue.
3. In Stage 3, 13\frac{1}{3} of the queued orders are processed, leaving the remaining queued orders unprocessed.

If exactly 250250 queued orders remain unprocessed at the end of Stage 3, how many incoming orders were flagged for manual review in Stage 1?

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Cevap: 150

Cevap

150 orders were flagged for manual review in Stage 1.
Let NN represent the total number of incoming trade orders. In Stage 1, after 16N\frac{1}{6}N orders are rejected, 56N\frac{5}{6}N remain. Flagging 15\frac{1}{5} of these remaining valid orders means 15×56N=16N\frac{1}{5} \times \frac{5}{6}N = \frac{1}{6}N orders are flagged for manual review, leaving 56N16N=23N\frac{5}{6}N - \frac{1}{6}N = \frac{2}{3}N orders cleared directly. In Stage 2, 58\frac{5}{8} of the cleared orders enter the queue, which equals 58×23N=512N\frac{5}{8} \times \frac{2}{3}N = \frac{5}{12}N. In Stage 3, 23\frac{2}{3} of the queued orders remain unprocessed, giving 23×512N=518N\frac{2}{3} \times \frac{5}{12}N = \frac{5}{18}N. Setting 518N=250\frac{5}{18}N = 250 yields N=900N = 900. The number of orders flagged for manual review is 16×900=150\frac{1}{6} \times 900 = 150.

Adım Adım Çözüm

1
Determine the fraction of orders cleared directly in Stage 1.
The fraction of valid orders remaining after rejection is 116=561 - \frac{1}{6} = \frac{5}{6}. The fraction flagged for manual review is 15×56=16\frac{1}{5} \times \frac{5}{6} = \frac{1}{6} of the total. Thus, the fraction cleared directly is 5616=46=23\frac{5}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3} of total incoming orders.
Sequential fractional deductions must be applied to the remaining amounts at each sub-stage.
2
Determine the fraction of orders placed in the queue during Stage 2.
Of the 23\frac{2}{3} directly cleared orders, 38\frac{3}{8} are executed immediately, so 138=581 - \frac{3}{8} = \frac{5}{8} are placed in the queue. The fraction of total incoming orders queued is 58×23=512\frac{5}{8} \times \frac{2}{3} = \frac{5}{12}.
Fraction of a fraction requires multiplying the operational fraction by the current subtotal.
3
Determine the fraction of total incoming orders that remain unprocessed at the end of Stage 3.
Since 13\frac{1}{3} of queued orders are processed, 113=231 - \frac{1}{3} = \frac{2}{3} of the queued orders remain unprocessed. The fraction of total orders unprocessed is 23×512=518\frac{2}{3} \times \frac{5}{12} = \frac{5}{18}.
Multiply the remaining fraction in Stage 3 by the total queued fraction from Stage 2.
4
Calculate the total number of incoming orders (NN) and the target flagged orders.
Setting 518N=250\frac{5}{18}N = 250 gives N=250×185=900N = 250 \times \frac{18}{5} = 900 total incoming orders. The number of orders flagged for manual review is 16×900=150\frac{1}{6} \times 900 = 150.
Equate the given numerical quantity to its fractional equivalent to find the total, then compute the required part.

Anahtar Kavram

Multi-step fraction calculations involving fractions of remaining quantities and working backward from a known part to the whole.
Soru 28Soru

On Monday, a community kitchen starts with a bulk container of rice. First, 13\frac{1}{3} of the initial supply of rice is used for lunch. Next, 14\frac{1}{4} of the remaining rice in the container is used for dinner. After preparing both meals, exactly 6060 kilograms of rice remain in the container. How many kilograms of rice were in the container initially?

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Cevap: 120 kg

Cevap

120 kg
The correct option is 120 kg. After using 13\frac{1}{3} of the initial supply for lunch, 23\frac{2}{3} of the supply remains. Using 14\frac{1}{4} of this remaining amount for dinner leaves 34\frac{3}{4} of the 23\frac{2}{3} remaining. Multiplying these fractions gives 34×23=12\frac{3}{4} \times \frac{2}{3} = \frac{1}{2} of the original supply. Since 12\frac{1}{2} of the original supply equals 6060 kg, the initial total supply was 60×2=12060 \times 2 = 120 kg.

Adım Adım Çözüm

1
Determine the fraction of rice remaining after lunch.
Since 13\frac{1}{3} of the initial supply is used for lunch, 113=231 - \frac{1}{3} = \frac{2}{3} of the initial supply remains.
The remaining portion after lunch is the base for the next fractional reduction.
2
Determine the fraction of rice used for dinner and the fraction remaining overall.
Dinner uses 14\frac{1}{4} of the remaining 23\frac{2}{3}, which is 14×23=16\frac{1}{4} \times \frac{2}{3} = \frac{1}{6} of the initial total supply. The total fraction used is 13+16=12\frac{1}{3} + \frac{1}{6} = \frac{1}{2}. Thus, the fraction remaining after dinner is 112=121 - \frac{1}{2} = \frac{1}{2} of the initial total.
Sequential fractional reductions require multiplying the second fraction by the remaining fraction from the first step.
3
Solve for the initial total mass of rice.
Let RR be the initial mass in kilograms. Then 12R=60    R=120\frac{1}{2} R = 60 \implies R = 120 kg.
Setting the calculated remaining fraction equal to the given numerical remaining quantity yields the starting total.

Anahtar Kavram

Sequential Fraction of Remaining Quantities
Tahmini Süre:1m 30s
Soru 29Soru

A specialty baking recipe consists of three ingredients: flour, sugar, and butter. By weight, 12\frac{1}{2} of the recipe is flour and 13\frac{1}{3} of the recipe is sugar, with the remaining portion consisting entirely of butter. Which of the following statements about the ingredients in the recipe must be true? Select all such statements.

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Cevap: The butter accounts for 16\frac{1}{6} of the total weight of the recipe.; The ratio of the weight of sugar to the weight of butter is 2:12 : 1.

Cevap

The statement that butter accounts for 1/6 of the total weight of the recipe and the statement that the ratio of the weight of sugar to the weight of butter is 2:1 are both correct.
The statement asserting that butter accounts for 1/6 of the total weight is correct because subtracting the combined fraction of flour and sugar (5/6) from the total (1) yields 1/6. The statement asserting that the ratio of sugar to butter is 2:1 is correct because dividing 1/3 by 1/6 equals 2.

Adım Adım Çözüm

1
Calculate the fraction of the recipe that is butter.
Butter fraction = 1(12+13)=156=161 - \left(\frac{1}{2} + \frac{1}{3}\right) = 1 - \frac{5}{6} = \frac{1}{6}.
The total of all constituent fractions must sum to 1.
2
Calculate the ratio of sugar to butter.
Ratio=1/31/6=13×61=21=2:1\text{Ratio} = \frac{1/3}{1/6} = \frac{1}{3} \times \frac{6}{1} = \frac{2}{1} = 2 : 1.
Divide the fraction representing sugar by the fraction representing butter.
3
Evaluate the statement regarding combined flour and sugar weight.
Combined fraction = 12+13=36+26=5625\frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \neq \frac{2}{5}.
Adding fractions requires finding a common denominator (6).
4
Evaluate the statement regarding percentage difference between sugar and flour.
Percent less = 1/21/31/2×100%=1/61/2×100%=13×100%=33.33%50%\frac{1/2 - 1/3}{1/2} \times 100\% = \frac{1/6}{1/2} \times 100\% = \frac{1}{3} \times 100\% = 33.33\% \neq 50\%.
Percent decrease must use the original reference quantity (flour) as the denominator.
5
Evaluate the ratio of flour to the combined remaining ingredients.
Remaining ingredients fraction = 1/3+1/6=1/21/3 + 1/6 = 1/2. Ratio of flour (1/2) to remaining (1/2) = 1:11:21 : 1 \neq 1 : 2.
Compare the part (flour) to the other part (sugar + butter), not to the whole.

Anahtar Kavram

Operations on rational numbers, fraction subtraction/addition using common denominators, part-to-part ratio comparisons, and percentage change base selection.
Soru 30Soru

At the beginning of a dry season, a municipal water reservoir contains an initial volume of VV gallons of water. During the first month, 14\frac{1}{4} of the initial volume is released for agricultural irrigation. During the second month, 15\frac{1}{5} of the remaining volume evaporates, after which 30,00030,000 gallons of water are pumped into the reservoir from an underground aquifer. If the final volume of water in the reservoir is equal to 34\frac{3}{4} of the initial volume VV, what was the initial volume VV, in gallons?

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Cevap: 200,000

Cevap

200,000 gallons
Subtracting the first month's release of 14V\frac{1}{4}V leaves 34V\frac{3}{4}V. Evaporating 15\frac{1}{5} of this remaining amount leaves 45×34V=35V\frac{4}{5} \times \frac{3}{4}V = \frac{3}{5}V. Adding 30,00030,000 gallons gives a final volume of 35V+30,000\frac{3}{5}V + 30,000. Setting this equal to 34V\frac{3}{4}V results in 34V35V=320V=30,000\frac{3}{4}V - \frac{3}{5}V = \frac{3}{20}V = 30,000, which solves to V=200,000V = 200,000.

Adım Adım Çözüm

1
Determine the volume remaining after the first month's irrigation release.
The volume remaining is V14V=34VV - \frac{1}{4}V = \frac{3}{4}V.
Releasing 14\frac{1}{4} of the initial volume leaves 114=341 - \frac{1}{4} = \frac{3}{4} of the initial volume.
2
Calculate the volume remaining after the second month's evaporation.
The volume remaining after evaporation is (115)×34V=45×34V=35V\left(1 - \frac{1}{5}\right) \times \frac{3}{4}V = \frac{4}{5} \times \frac{3}{4}V = \frac{3}{5}V.
Evaporating 15\frac{1}{5} of the remaining water leaves 45\frac{4}{5} of that remaining amount.
3
Formulate the linear equation including the pumped volume and solve for VV.
\frac{3}{5}V + 30,000 = \frac{3}{4}V \implies 30,000 = \frac{3}{4}V - \frac{3}{5}V = \frac{15 - 12}{20}V = \frac{3}{20}V \implies V = \frac{30,000 \times 20}{3} = 200,000$.
Equating the expression for the final water volume to 34V\frac{3}{4}V yields a single linear equation in VV.

Anahtar Kavram

Sequential fractional reduction of remaining amounts and solving linear equations involving rational numbers.
Soru 31Soru

At a research institute, an annual equipment budget is distributed among three departments. First, 25\frac{2}{5} of the total budget is allocated to the Biology department. Next, 38\frac{3}{8} of the remaining budget is allocated to the Chemistry department. The Physics department receives the remaining balance of $45,000\$45,000. What was the institute's total annual equipment budget?

Cevabı ve açıklamayı göster

Cevap: $120,000

Cevap

$120,000
Subtracting the Biology department's share of 25\frac{2}{5} leaves 35\frac{3}{5} of the total budget. The Chemistry department receives 38\frac{3}{8} of this remainder, leaving 58\frac{5}{8} of the remainder for the Physics department. Multiplying these fractions yields 58×35=38\frac{5}{8} \times \frac{3}{5} = \frac{3}{8} of the total budget for Physics. Equating 38\frac{3}{8} of the total budget to $45,000\$45,000 gives a total budget of $120,000\$120,000.

Adım Adım Çözüm

1
Determine the fraction of the budget remaining after the Biology allocation.
The remaining fraction is 125=351 - \frac{2}{5} = \frac{3}{5} of the total budget.
The Biology department receives 25\frac{2}{5} of the overall budget.
2
Calculate the fraction of the total budget allocated to Physics.
The Physics department receives 138=581 - \frac{3}{8} = \frac{5}{8} of the remaining budget, which equals 58×35=38\frac{5}{8} \times \frac{3}{5} = \frac{3}{8} of the total budget.
The Chemistry department receives 38\frac{3}{8} of the remaining budget, leaving 58\frac{5}{8} of that remainder for Physics.
3
Solve for the total budget BB.
\frac{3}{8}B = 45,000 \implies B = 45,000 \times \frac{8}{3} = 120,000.
The Physics allocation of $45,000\$45,000 represents 38\frac{3}{8} of the total budget BB.

Anahtar Kavram

Sequential Fraction of Remainder Calculations
Tahmini Süre:1m 30s
Soru 32Soru

Let rr and ss be rational numbers such that 0<r<120 < r < \frac{1}{2} and 1<s<21 < s < 2. Which of the following inequalities MUST be true? Select all that apply.

Geçerli olan tümünü seçin

Cevabı ve açıklamayı göster

Cevap: rs<1r \cdot s < 1; \frac{s}{r} > 2

Cevap

The correct inequalities that must be true are rs<1r \cdot s < 1 and sr>2\frac{s}{r} > 2.
The inequality stating that the product of rr and ss is less than 11 is guaranteed because multiplying the maximum possible bounds (12\frac{1}{2} and 22) gives 11. The inequality stating that the ratio of ss to rr is greater than 22 is guaranteed because the numerator is strictly greater than 11 and the denominator is strictly less than 12\frac{1}{2}.

Adım Adım Çözüm

1
Analyze rs<1r \cdot s < 1
Since r<12r < \frac{1}{2} and s<2s < 2, and both are positive rational numbers, multiplying their upper limits gives rs<(12)(2)=1r \cdot s < \left(\frac{1}{2}\right)(2) = 1. This statement MUST be true.
Properties of inequality multiplication for positive rational numbers.
2
Analyze sr>2\frac{s}{r} > 2
Since s>1s > 1 and r<12r < \frac{1}{2}, taking the ratio gives sr>11/2=2\frac{s}{r} > \frac{1}{1/2} = 2. This statement MUST be true.
Dividing a larger positive number by a fraction less than 1/21/2 amplifies the quotient.
3
Test counterexamples for remaining options
For sr>1s - r > 1, pick s=1.2s = 1.2 and r=0.4sr=0.81r = 0.4 \Rightarrow s - r = 0.8 \le 1 (False). For rs>14\frac{r}{s} > \frac{1}{4}, pick r=0.1r = 0.1 and s=1.6rs=116<14s = 1.6 \Rightarrow \frac{r}{s} = \frac{1}{16} < \frac{1}{4} (False). For r+s>32r + s > \frac{3}{2}, pick r=0.1r = 0.1 and s=1.1r+s=1.2<1.5s = 1.1 \Rightarrow r + s = 1.2 < 1.5 (False).
A single valid counterexample disproves a 'must be true' statement.

Anahtar Kavram

Properties of Rational Numbers and Inequalities
ÖncekiSayfa 2 / 2
Fractions and Rational Numbers Alıştırma Soruları — GRE General Test — Sayfa 2 | Examkin