Fractions and Decimals

34 soru

Soru 21Soru

Four identical delivery vans have different amounts of fuel remaining in their tanks. Arrange the vehicles based on their remaining fuel fraction from the least amount to the greatest amount.

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Cevap

The correct ascending sequence of fuel remaining is Van R, Van P, Van S, and Van Q.
By converting all given amounts into decimals (0.600.60, 0.750.75, 0.500.50, and 0.650.65), the values can be directly compared. 0.500.50 is the smallest and 0.750.75 is the largest, resulting in the correct ascending order of Van R, Van P, Van S, and Van Q.

Adım Adım Çözüm

1
Convert all fractional fuel values into decimal form for straightforward comparison.
Van P (35\frac{3}{5}) becomes 0.60, and Van R (12\frac{1}{2}) becomes 0.50.
Converting all amounts to decimals provides a uniform format that is easier to logically order.
2
Order the decimal values from smallest to largest.
0.50<0.60<0.65<0.750.50 < 0.60 < 0.65 < 0.75.
The question specifically requires arranging the vehicles from the least amount of fuel to the greatest amount.

Anahtar Kavram

Converting basic fractions to decimals to compare and order numerical values.
Soru 22Soru

A civil supplies department procures three batches of edible oil weighing 0.60.\overline{6} metric tonnes, 1.31.\overline{3} metric tonnes, and 2.22.\overline{2} metric tonnes. The department intends to repackage the entire procured quantity into smaller, identical standardized containers. To minimize the number of containers used without mixing the batches, what should be the maximum capacity of each standardized container?

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Cevap: 0.20.\overline{2} metric tonnes

Cevap

The maximum capacity of each standardized container should be 0.20.\overline{2} metric tonnes.
The maximum capacity of the standardized containers is determined by finding the Highest Common Factor (HCF) of the oil volumes. First, the given recurring decimals are converted to fractions: 23\frac{2}{3}, 43\frac{4}{3}, and 209\frac{20}{9}. The HCF of these fractions is found by taking the HCF of the numerators (2, 4, 20), which is 2, and dividing it by the LCM of the denominators (3, 3, 9), which is 9. This calculation yields 29\frac{2}{9}, which is equivalent to the recurring decimal 0.20.\overline{2}.

Adım Adım Çözüm

1
Convert the recurring decimals representing the batch weights into fractions in their simplest form.
0.6=69=230.\overline{6} = \frac{6}{9} = \frac{2}{3}, 1.3=1319=129=431.\overline{3} = \frac{13-1}{9} = \frac{12}{9} = \frac{4}{3}, and 2.2=2229=2092.\overline{2} = \frac{22-2}{9} = \frac{20}{9}.
Mathematical operations involving divisors and multiples are most accurate when recurring decimals are expressed exactly as fractions.
2
Identify the mathematical requirement to find the maximum standardized container capacity.
We must determine the Highest Common Factor (HCF) of the three fractional quantities.
To divide all batches completely into identical containers of maximum capacity without leftovers, the container size must be the greatest common divisor of all the given volumes.
3
Apply the standard formula for calculating the HCF of multiple fractions.
HCF of fractions=HCF of numeratorsLCM of denominators\text{HCF of fractions} = \frac{\text{HCF of numerators}}{\text{LCM of denominators}}.
This formula ensures that the resulting fraction will evenly divide each of the original fractional quantities.
4
Calculate the HCF of the numerators and the LCM of the denominators for 23\frac{2}{3}, 43\frac{4}{3}, and 209\frac{20}{9}.
Numerators (2, 4, 20): The HCF\text{HCF} is 2. Denominators (3, 3, 9): The LCM\text{LCM} is 9.
These values provide the numerator and denominator for the final maximum capacity.
5
Form the final fraction and convert it back to a recurring decimal to match the given options.
Maximum capacity=29=0.2\text{Maximum capacity} = \frac{2}{9} = 0.\overline{2} metric tonnes.
Converting back to recurring decimal format answers the problem in the original units.

Anahtar Kavram

Fractions and Decimals - Converting Recurring Decimals and Finding HCF of Fractions
Soru 23Soru

A jeweler is cutting equal-length pieces of gold wire from three different spools to make uniform necklace links without any wastage. The three spools contain gold wire of lengths 272\frac{27}{2} cm, 454\frac{45}{4} cm, and 635\frac{63}{5} cm. What is the maximum possible length of each uniform gold wire piece that can be cut? (Provide your answer as a precise decimal)

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Cevap: 0.45

Cevap

0.45
The maximum uniform length is found by computing the Highest Common Factor (HCF) of the fractional wire lengths. Applying the formula yields an HCF of numerators (9) over the LCM of denominators (20), giving 920\frac{9}{20}, which correctly converts to exactly 0.450.45 cm.

Adım Adım Çözüm

1
Determine the mathematical operation required for the scenario.
Calculate the Highest Common Factor (HCF) of the three fractional lengths.
Cutting multiple lengths into the maximum possible equal segments without wastage is defined by the HCF.
2
Apply the rule for finding the HCF of fractions.
HCF = HCF(Numerators) / LCM(Denominators)
This is the standard formula for finding the greatest common divisor of multiple fractional values.
3
Find the HCF of the numerators 27, 45, and 63.
HCF(27, 45, 63) = 9
9 is the largest integer that divides perfectly into 27, 45, and 63.
4
Find the LCM of the denominators 2, 4, and 5.
LCM(2, 4, 5) = 20
20 is the smallest integer that is a multiple of 2, 4, and 5.
5
Combine the results into the final fraction and convert to a decimal.
920=0.45\frac{9}{20} = 0.45 cm
The question requires the precise decimal representation of the fraction.

Anahtar Kavram

Calculating the Highest Common Factor (HCF) of fractions to solve optimization word problems.
Soru 24Soru

A financial analyst is evaluating the long-term performance of four mutual funds. The average annualized yield rates of the funds are given below. Arrange the funds in descending order based on their yield rates (from highest to lowest).

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Cevap

The correct descending order is Fund X, Fund Y, Fund W, then Fund Z.
By converting all values to their extended decimal forms, we determine their precise magnitudes: Fund X is 8.166...%8.166...\%, Fund Y is 8.155...%8.155...\%, Fund W is 8.151...%8.151...\%, and Fund Z is 8.142...%8.142...\%. Ordering these from highest to lowest yields Fund X, Fund Y, Fund W, then Fund Z.

Adım Adım Çözüm

1
Convert Fund W's yield to extended decimal form for precise comparison.
Fund W = 8.15%=8.1515...%8.\overline{15}\% = 8.1515...\%
Expanding the recurring decimals reveals the sequence of digits beyond the hundredths place.
2
Convert Fund X's yield to decimal form.
Fund X = 816%=8+0.1666...%=8.1666...%8 \frac{1}{6}\% = 8 + 0.1666...\% = 8.1666...\%
Converting the mixed fraction standardizes the format to match the others.
3
Convert Fund Y's yield to extended decimal form.
Fund Y = 8.15%=8.1555...%8.1\overline{5}\% = 8.1555...\%
Properly expanding the specific repeating digit distinguishes it from Fund W.
4
Convert Fund Z's yield to decimal form.
Fund Z = 577%=817%8.1428...%\frac{57}{7}\% = 8 \frac{1}{7}\% \approx 8.1428...\%
Division provides the decimal equivalent needed for a direct comparison.
5
Compare the decimal values digit by digit from left to right.
Highest is 8.166...8.166... (Fund X), then 8.155...8.155... (Fund Y), then 8.151...8.151... (Fund W), and lowest is 8.142...8.142... (Fund Z).
Sorting by the tenths, hundredths, and thousandths digits establishes the definitive descending sequence.

Anahtar Kavram

Converting mixed fractions, improper fractions, and different notations of recurring decimals into standard decimal form for accurate comparison.
Soru 25Soru

A botanical research institute is preparing identical standardized nutrient doses for a plant growth experiment. They have three concentrated liquid nutrient solutions measuring 143\frac{14}{3} liters, 356\frac{35}{6} liters, and 214\frac{21}{4} liters respectively. The researchers need to distribute these solutions into small testing vials such that every vial contains the exact same maximum possible volume of solution, without mixing the different types of nutrients, and leaving no solution behind in the original containers. What should be the volume capacity of each testing vial?

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Cevap: 712\frac{7}{12} liters

Cevap

712\frac{7}{12} liters
To find the maximum identical volume that can evenly divide all three solutions without remainders, we must calculate the Highest Common Factor (HCF) of the fractions 143\frac{14}{3}, 356\frac{35}{6}, and 214\frac{21}{4}. Using the rule that the HCF of fractions equals the HCF of their numerators divided by the LCM of their denominators, we evaluate HCF(14, 35, 21) = 7 and LCM(3, 6, 4) = 12. This results in exactly 712\frac{7}{12} liters per vial.

Adım Adım Çözüm

1
Identify the mathematical operation required based on the scenario.
The problem asks for the maximum identical volume to evenly distribute three different quantities without remainders. This requires finding the Highest Common Factor (HCF) of the given fractional volumes.
HCF is used when dividing varying quantities into the largest possible equal segments.
2
Recall the formula for calculating the HCF of fractions.
HCF=HCF of numeratorsLCM of denominatorsHCF = \frac{\text{HCF of numerators}}{\text{LCM of denominators}}
This is the standard rule for finding the greatest common divisor of fractional values.
3
Calculate the HCF of the numerators (14,35,2114, 35, 21).
The prime factors are: 14=2×714 = 2 \times 7; 35=5×735 = 5 \times 7; 21=3×721 = 3 \times 7. The highest common factor is 77.
To form the numerator of our final fraction answer.
4
Calculate the LCM of the denominators (3,6,43, 6, 4).
The multiples of the denominators meet at 1212. (Since 1212 is divisible by 33, 66, and 44).
To form the denominator of our final fraction answer.
5
Substitute the calculated values back into the fraction HCF formula.
712\frac{7}{12} liters.
Provides the final vial capacity.

Anahtar Kavram

Highest Common Factor (HCF) of Fractions in Word Problems
Soru 26Soru

A civil engineer is evaluating the thickness of four different structural panels, measured in centimeters (cm). To meet safety standards, the panels must be stacked from the thinnest to the thickest. Arrange the following panel thicknesses in ascending order.

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Cevap

The correct ascending order from thinnest to thickest is: Panel Z, Panel X, Panel W, Panel Y.
The correct sequence is determined by converting all measurements into standard decimal forms. Panel Z is exactly 1.3751.375 cm, Panel X is approximately 1.4281.428 cm, Panel W is 1.444...1.444... cm, and Panel Y is exactly 1.4501.450 cm. Sorting these values in ascending order yields Panel Z, Panel X, Panel W, Panel Y.

Adım Adım Çözüm

1
Convert the thickness of Panel Z into a standard decimal format.
Panel Z = 138=1+0.375=1.3751\frac{3}{8} = 1 + 0.375 = 1.375 cm.
Converting a mixed fraction to a decimal allows for direct magnitude comparison.
2
Convert the thickness of Panel X into a standard decimal format.
Panel X = 1071.42857...\frac{10}{7} \approx 1.42857... cm.
An improper fraction must be divided out to identify its exact decimal value.
3
Expand the thickness of Panel W to clearly see its decimal value.
Panel W = 1.4=1.4444...1.\overline{4} = 1.4444... cm.
The overline indicates a recurring decimal that continues infinitely with the digit 4.
4
Compare all four decimal values and arrange them from smallest to largest.
1.375<1.42857...<1.4444...<1.4501.375 < 1.42857... < 1.4444... < 1.450. Therefore, Panel Z < Panel X < Panel W < Panel Y.
The stem requires ordering from thinnest to thickest.

Anahtar Kavram

Converting mixed fractions, improper fractions, and recurring decimals into a common decimal format to accurately compare and order them.
Soru 27Soru

A highway construction company is contracted to pave a new road. In the first month, they complete exactly 0.30.\overline{3} of the total project length. In the second month, they pave 0.6250.625 of the *remaining* length. If there are exactly 2121 kilometers left to pave in the third month to finish the project, what is the total length of the highway in kilometers?

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Cevap: 84

Cevap

The total length of the highway is 84 kilometers.
The total length of the highway is 84 km. In the first month, 1/31/3 is completed, leaving 2/32/3. In the second month, 5/85/8 of this remaining 2/32/3 is completed, which means (15/8)=3/8(1 - 5/8) = 3/8 of the remainder is left for the third month. Calculating this relative to the total: 3/8×2/3=1/43/8 \times 2/3 = 1/4. Since 1/41/4 of the total project equals 21 km, the total length must be 21×4=8421 \times 4 = 84 km.

Adım Adım Çözüm

1
Convert the decimal and recurring decimal into fractions for exact calculation.
0.3=130.\overline{3} = \frac{1}{3} and 0.625=6251000=580.625 = \frac{625}{1000} = \frac{5}{8}.
Fractional calculations prevent rounding errors and simplify multi-step multiplication.
2
Determine the fraction of the highway remaining after the first month.
113=231 - \frac{1}{3} = \frac{2}{3} of the total length remains.
The second month's progress is based on this remaining amount, not the total.
3
Calculate the fraction of the total project paved in the second month.
58×23=1024=512\frac{5}{8} \times \frac{2}{3} = \frac{10}{24} = \frac{5}{12} of the total length.
The second month's work is 0.6250.625 of the remaining 23\frac{2}{3}.
4
Find the final remaining fraction for the third month.
23512=812512=312=14\frac{2}{3} - \frac{5}{12} = \frac{8}{12} - \frac{5}{12} = \frac{3}{12} = \frac{1}{4} of the total project.
Subtracting the second month's progress from what was left after the first month gives the final remainder.
5
Set the remaining fraction equal to the given remaining distance and solve for the total.
14×Total=21Total=21×4=84\frac{1}{4} \times \text{Total} = 21 \Rightarrow \text{Total} = 21 \times 4 = 84 km.
This establishes the relationship between the fractional parts and the actual physical distance.

Anahtar Kavram

Converting recurring and terminating decimals to fractions and computing successive remainders.
Tahmini Süre:1m 30s
Soru 28Soru

A specialized aerospace component is manufactured from a custom metal alloy weighing exactly 160160 kg. The alloy's composition by weight is 0.350.35 aluminum and 516\frac{5}{16} magnesium, with the remaining portion consisting entirely of titanium. What is the exact mass of titanium in this component?

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Cevap: 54

Cevap

54
To find the mass of titanium, the total proportions of the other metals must first be unified into a common format (either fractions or decimals). By converting the magnesium's share of 516\frac{5}{16} to the decimal 0.31250.3125, the combined proportion of aluminum and magnesium is 0.35+0.3125=0.66250.35 + 0.3125 = 0.6625. The titanium makes up the remaining portion of the whole, which is 10.6625=0.33751 - 0.6625 = 0.3375. Finally, multiplying this decimal proportion by the total alloy mass of 160160 kg yields exactly 5454 kg.

Adım Adım Çözüm

1
Convert the fractional part to a decimal.
516=0.3125\frac{5}{16} = 0.3125
Unifying the terms into a single format (decimals) makes addition straightforward.
2
Add the portions of aluminum and magnesium.
0.35+0.3125=0.66250.35 + 0.3125 = 0.6625
This finds the total proportion of the alloy that is NOT titanium.
3
Calculate the proportion of titanium.
10.6625=0.33751 - 0.6625 = 0.3375
The sum of all material proportions in the whole alloy must equal exactly 11.
4
Calculate the final mass of titanium.
0.3375×160=540.3375 \times 160 = 54
Multiplying the component's proportion by the total mass gives the specific weight of that component.

Anahtar Kavram

Converting between fractions and decimals and applying proportional reasoning to find a part of a whole.
Soru 29Soru

Three automated drones patrol a large forest reserve and return to their base station for a rapid data sync at regular intervals. Drone A returns every 3.33.\overline{3} hours, Drone B returns every 2.52.5 hours, and Drone C returns every 2.72.\overline{7} hours. If all three drones depart from the base station simultaneously, how many hours will it take for all three to meet at the base station again for a simultaneous data sync?

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Cevap: 5050

Cevap

It will take 50 hours for all three drones to meet simultaneously at the base station.
The time required for all drones to meet again is the LCM of their individual interval times. Converting the intervals to fractions gives 10/3, 5/2, and 25/9. The LCM of fractions is computed by dividing the LCM of the numerators (10, 5, 25) by the HCF of the denominators (3, 2, 9). The LCM of the numerators is 50 and the HCF of the denominators is 1, resulting in exactly 50 hours.

Adım Adım Çözüm

1
Convert the recurring and terminating decimals into fractions in their simplest form.
Drone A: 3.3=313=1033.\overline{3} = 3 \frac{1}{3} = \frac{10}{3}. Drone B: 2.5=522.5 = \frac{5}{2}. Drone C: 2.7=279=2592.\overline{7} = 2 \frac{7}{9} = \frac{25}{9}.
Fractional forms are required to accurately compute the lowest common multiple of non-integer time intervals.
2
Identify the mathematical operation needed to find the time of the next simultaneous meeting.
We must find the Least Common Multiple (LCM) of the three fractional time intervals.
The LCM represents the smallest time value that is a perfect multiple of all individual drone intervals.
3
State the formula for the LCM of fractions.
LCM of fractions=LCM of numeratorsHCF of denominators\text{LCM of fractions} = \frac{\text{LCM of numerators}}{\text{HCF of denominators}}.
This is the standard arithmetic rule for finding the LCM of multiple fractions.
4
Calculate the LCM of the numerators.
The numerators are 10,5,2510, 5, 25. Their prime factorizations are 10=2×510 = 2 \times 5, 5=55 = 5, and 25=5225 = 5^2. The LCM(10,5,25)=2×52=50\text{LCM}(10, 5, 25) = 2 \times 5^2 = 50.
The LCM must include the highest power of each prime factor present in the numerators.
5
Calculate the HCF of the denominators.
The denominators are 3,2,93, 2, 9. Since they do not share any common prime factors, their HCF(3,2,9)=1\text{HCF}(3, 2, 9) = 1.
The Highest Common Factor is the largest integer that divides all the denominators evenly.
6
Compute the final LCM of the fractions.
LCM=501=50\text{LCM} = \frac{50}{1} = 50 hours.
Applying the values from Steps 4 and 5 into the formula from Step 3 yields the final synchronization time.

Anahtar Kavram

LCM of Fractions and Recurring Decimals
Tahmini Süre:2m 0s
Soru 30Soru

During a physical endurance test, four candidates completed different portions of a 10-kilometer cross-country course before a mandatory rest period. The fraction or decimal representing the portion of the course completed by each candidate is given below. Arrange the candidates in ascending order based on the portion of the course they completed (from least to most).

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Cevap

Candidate Q, Candidate S, Candidate R, Candidate P
To correctly order the candidates, all values must be evaluated in a uniform format. Candidate P is exactly 0.85. Candidate Q is approximately 0.818. Candidate R is approximately 0.844. Candidate S is approximately 0.833. Arranging these numerical values from least to most gives: Q (0.818...), S (0.833...), R (0.844...), P (0.850).

Adım Adım Çözüm

1
Convert the portion completed by Candidate P into a standard decimal format.
1720=17×520×5=85100=0.85\frac{17}{20} = \frac{17 \times 5}{20 \times 5} = \frac{85}{100} = 0.85
Converting all values to standard decimals makes it easier to directly compare their magnitudes.
2
Convert the portion completed by Candidate Q into an expanded decimal format.
0.81=0.8181810.\overline{81} = 0.818181\dots
Expanding the recurring bar notation clarifies the digits beyond the hundredths place.
3
Convert the portion completed by Candidate R into an expanded decimal format.
0.84=0.8444440.8\overline{4} = 0.844444\dots
Expanding the single recurring digit makes its exact magnitude clear for comparison.
4
Convert the portion completed by Candidate S into a decimal format.
56=5÷6=0.833333\frac{5}{6} = 5 \div 6 = 0.833333\dots
Performing the division provides the decimal value needed to compare against the others.
5
Compare the resulting decimal values and arrange them in ascending order.
0.8181<0.8333<0.8444<0.85000.8181\dots < 0.8333\dots < 0.8444\dots < 0.8500
Sorting the decimals from smallest to largest yields the order: Candidate Q, Candidate S, Candidate R, Candidate P.

Anahtar Kavram

Converting mixed fraction types and recurring decimals into a uniform decimal format to accurately compare their relative magnitudes.
Soru 31Soru

An agricultural research center is preparing experimental test plots. The facility has exactly 2582 \frac{5}{8} kg, 211122 \frac{11}{12} kg, and 31163 \frac{1}{16} kg of three different premium organic soil additives. The lead researcher wants to distribute these additives into smaller, identical sample packets such that no material is left over from any of the three original quantities. To ensure efficiency, the packets must be as large as possible. What should be the exact weight capacity of each sample packet?

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Cevap: 748\frac{7}{48} kg

Cevap

The correct packet capacity is 748\frac{7}{48} kg.
To find the largest possible identical packet size that leaves no remainder, we must calculate the Highest Common Factor (HCF) of the three fractional quantities. After converting the mixed numbers to improper fractions (218\frac{21}{8}, 3512\frac{35}{12}, 4916\frac{49}{16}), we apply the required formula: HCF of fractions = (HCF of numerators) / (LCM of denominators). The HCF of the numerators (21,35,49)(21, 35, 49) is 77, and the LCM of the denominators (8,12,16)(8, 12, 16) is 4848. Therefore, the required capacity is 748\frac{7}{48} kg.

Adım Adım Çözüm

1
Convert the mixed numbers representing the quantities of additives into improper fractions.
258=2182 \frac{5}{8} = \frac{21}{8}, 21112=35122 \frac{11}{12} = \frac{35}{12}, and 3116=49163 \frac{1}{16} = \frac{49}{16}.
Calculations involving the HCF and LCM of fractions require the numbers to be in improper fraction format.
2
Identify the mathematical operation required to solve the problem.
Calculate the Highest Common Factor (HCF) of the three fractions.
The problem asks for the largest possible identical packet size that can divide all quantities without a remainder, which maps to finding the HCF.
3
Apply the standard formula for the HCF of fractions.
The formula is: HCF of fractions = (HCF of numerators) / (LCM of denominators).
This formula allows the determination of the greatest common divisor for fractional quantities.
4
Calculate the HCF of the numerators and the LCM of the denominators.
The HCF of the numerators (21,35,49)(21, 35, 49) is 77. The LCM of the denominators (8,12,16)(8, 12, 16) is 4848.
Finding these values provides the numerator and denominator for the final fractional answer.
5
Combine the calculated values to form the final fraction.
The maximum identical capacity is 748\frac{7}{48} kg.
Dividing the numerator HCF by the denominator LCM yields the exact required capacity.

Anahtar Kavram

Calculating the Highest Common Factor (HCF) of fractions using the formula: HCF of numerators / LCM of denominators.
Tahmini Süre:1m 30s
Soru 32Soru

A civil contractor is tasked with upgrading a rural highway. In the first month, the crew successfully upgrades 0.2750.275 of the total highway length. During the second month, they upgrade 512\frac{5}{12} of the remaining length. If the crew has exactly 10.1510.15 kilometers left to upgrade in the third month to complete the project, what is the total length of the highway in kilometers?

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Cevap: 24

Cevap

The total length of the highway is 24 kilometers.
The correct total length is calculated by accurately determining the remaining fraction of the highway step-by-step. Converting the decimal 0.275 to a fraction (11/40) allows for clean, exact operations without repeating decimal rounding errors. By tracking the remainder accurately, we establish that the final 10.15 km represents exactly 203/480 of the total length, resulting in a total of 24 km.

Adım Adım Çözüm

1
Determine the fraction of the highway remaining after the first month.
2940\frac{29}{40} of the total length remains.
The crew upgraded 0.2750.275 of the total. Converting 0.2750.275 to a fraction gives 2751000\frac{275}{1000}, which simplifies to 1140\frac{11}{40}. Subtracting this from the whole gives 11140=29401 - \frac{11}{40} = \frac{29}{40}.
2
Calculate the fraction of the total highway upgraded in the second month.
2996\frac{29}{96} of the total length.
The crew upgraded 512\frac{5}{12} of the remaining length. Therefore, multiply the two fractions: 512×2940=1×2912×8=2996\frac{5}{12} \times \frac{29}{40} = \frac{1 \times 29}{12 \times 8} = \frac{29}{96}.
3
Calculate the total fraction of the highway remaining for the third month.
203480\frac{203}{480} of the total length.
Subtract the second month's progress from the remainder after the first month: 29402996\frac{29}{40} - \frac{29}{96}. The least common multiple of 4040 and 9696 is 480480. Converting to common denominators gives 348480145480=203480\frac{348}{480} - \frac{145}{480} = \frac{203}{480}.
4
Set up the final equation and solve for the total highway length.
2424 km
The remaining fraction equals the given physical distance of 10.1510.15 km. So, 203480×Total=10.15\frac{203}{480} \times \text{Total} = 10.15. Solving for Total yields 10.15×480203\frac{10.15 \times 480}{203}. Since 10.15÷203=0.0510.15 \div 203 = 0.05, the Total is 0.05×480=240.05 \times 480 = 24 km.

Anahtar Kavram

Solving sequential parts-of-a-whole word problems by effectively converting between decimals and fractions to find a remaining proportion.
Soru 33Soru

A specialized textile loom is weaving a highly detailed traditional tapestry. Different specialty threads are inserted at specific periodic intervals along the length of the fabric:
- Gold thread is inserted every 1415\frac{14}{15} of a millimeter.
- Silver thread is inserted every 2120\frac{21}{20} of a millimeter.
- Crimson thread is inserted every 3512\frac{35}{12} of a millimeter.

If all three threads are inserted simultaneously at the very beginning of the pattern, at what exact distance from the start will all three specialty threads be inserted at the exact same position again for the first time?

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Cevap: 210210 mm

Cevap

The three specialty threads will be inserted at the exact same position again for the first time at a distance of 210210 mm.
The correct answer is found by calculating the Least Common Multiple (LCM) of the three fractions. The formula requires dividing the LCM of the numerators (14, 21, 35) by the HCF of the denominators (15, 20, 12). The LCM of the numerators is 210 and the HCF of the denominators is 1, resulting in exactly 210 mm.

Adım Adım Çözüm

1
Identify the mathematical operation required.
The problem asks for the first time three periodic intervals coincide, which requires finding the Least Common Multiple (LCM) of the fractions 1415\frac{14}{15}, 2120\frac{21}{20}, and 3512\frac{35}{12}.
The LCM finds the smallest common multiple of multiple values, representing the first common distance where the intervals align.
2
Calculate the LCM of the numerators.
The numerators are 14, 21, and 35. Their prime factorizations are 14=2×714 = 2 \times 7, 21=3×721 = 3 \times 7, and 35=5×735 = 5 \times 7. The lowest common multiple of these values is 2×3×5×7=2102 \times 3 \times 5 \times 7 = 210.
The formula for the LCM of fractions dictates that the numerator of the result must be the LCM of the individual numerators.
3
Calculate the HCF (Highest Common Factor) of the denominators.
The denominators are 15, 20, and 12. Their prime factorizations are 15=3×515 = 3 \times 5, 20=22×520 = 2^2 \times 5, and 12=22×312 = 2^2 \times 3. Because there is no single prime factor shared across all three denominators, their HCF is 1.
The formula for the LCM of fractions dictates that the denominator of the result must be the HCF of the individual denominators.
4
Compute the final LCM of the fractions.
LCM(14,21,35)HCF(15,20,12)=2101=210\frac{\text{LCM}(14, 21, 35)}{\text{HCF}(15, 20, 12)} = \frac{210}{1} = 210 mm.
Applying the values derived in the previous steps to the fraction LCM formula yields the final coincidence distance.

Anahtar Kavram

Calculating the Least Common Multiple (LCM) of fractions using the formula: LCM of fractions = (LCM of numerators) / (HCF of denominators).
Tahmini Süre:1m 30s
Soru 34Soru

An aerospace engineering facility is calibrating a pressure sensor using four specialized metallic foils. To prevent calibration errors, a technician must stack these foils in strictly ascending order of their thickness (from thinnest to thickest).

The thicknesses of the four foils are:
- Foil P: 0.3450.345 mm
- Foil Q: 1132\frac{11}{32} mm
- Foil R: 0.340.\overline{34} mm
- Foil S: 720\frac{7}{20} mm

Arrange the foils in the correct ascending order of their thickness.

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Cevabı ve açıklamayı göster

Cevap

The correct ascending sequence from thinnest to thickest is Foil R, Foil Q, Foil P, and Foil S.
By converting all foil thicknesses into uniform decimal representations, we establish the values: Foil R (0.3434...0.3434...), Foil Q (0.343750.34375), Foil P (0.3450.345), and Foil S (0.350.35). Comparing these exact decimal values digit-by-digit confirms that the ascending sequence must be R, Q, P, then S.

Adım Adım Çözüm

1
Convert the fraction for Foil S to a decimal.
720=7×520×5=35100=0.35\frac{7}{20} = \frac{7 \times 5}{20 \times 5} = \frac{35}{100} = 0.35 mm
Converting all measurements to a uniform decimal format allows for direct magnitude comparison.
2
Convert the fraction for Foil Q to a decimal.
11÷32=0.3437511 \div 32 = 0.34375 mm
Using long division provides the exact decimal expansion necessary to compare against closely grouped measurements.
3
Expand the recurring decimal for Foil R.
0.34=0.343434...0.\overline{34} = 0.343434... mm
Expanding the repeating digits to at least 4 decimal places makes place-value comparison clear.
4
Compare all four decimal values place-by-place and sequence them from smallest to largest.
0.343434...<0.34375<0.34500<0.350000.343434... < 0.34375 < 0.34500 < 0.35000
Aligning the decimals by their thousandths and ten-thousandths places reveals the exact ascending sequence.

Anahtar Kavram

Converting fractions and recurring decimals into standard decimals to accurately evaluate and compare their magnitudes.
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