Fractions and Decimals

34 soru

Soru 1Soru

Arrange the following numerical values in ascending order (from smallest to largest).

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Cevap

The correct ascending order is 25\frac{2}{5}, 0.60.6, 710\frac{7}{10}, and 0.750.75.
When converted to decimals, the values are 0.400.40, 0.600.60, 0.700.70, and 0.750.75. Arranging these from smallest to largest yields the correct sequence: 25\frac{2}{5}, 0.60.6, 710\frac{7}{10}, 0.750.75.

Adım Adım Çözüm

1
Convert the fractions into decimal format to establish a common baseline for comparison.
25=0.4\frac{2}{5} = 0.4 and 710=0.7\frac{7}{10} = 0.7.
Decimals are often much easier to compare directly digit by digit than mixed formats.
2
Append trailing zeros to the decimal values to ensure they all have the same number of decimal places (two places, matching 0.75).
The values to compare are 0.400.40, 0.600.60, 0.700.70, and 0.750.75.
Aligning decimal places prevents magnitude confusion and makes visual comparison straightforward.
3
Order the standardized decimals from smallest to largest.
0.40<0.60<0.70<0.750.40 < 0.60 < 0.70 < 0.75.
The question explicitly requires an ascending order.
4
Map the ordered decimals back to their original given forms.
25<0.6<710<0.75\frac{2}{5} < 0.6 < \frac{7}{10} < 0.75.
The final answer must be presented using the exact numbers provided in the question.

Anahtar Kavram

Converting fractions to decimals to evaluate and compare their magnitudes.
Tahmini Süre:45s
Soru 2Soru

A water storage tank is currently filled to 0.650.65 of its total capacity. If the tank currently contains 130130 liters of water, what is the total capacity of the tank in liters?

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Cevap: 200

Cevap

The total capacity of the tank is 200 liters.
The correct capacity is found by dividing the current volume of water (130 liters) by the decimal that represents the filled portion (0.65). This calculation, 130÷0.65130 \div 0.65, yields 200 liters.

Adım Adım Çözüm

1
Identify the relationship between the filled portion and the total capacity.
Let CC be the total capacity. We establish the equation: 0.65×C=1300.65 \times C = 130.
Translating the word problem into a mathematical equation allows us to solve for the unknown whole amount.
2
Rearrange the equation to solve for the total capacity.
C=1300.65C = \frac{130}{0.65}
Isolating CC on one side of the equation gives us the expression needed to find the total capacity.
3
Perform the division by clearing the decimal in the denominator.
C=1300065=200C = \frac{13000}{65} = 200
Multiplying the numerator and denominator by 100 eliminates the decimal point, making the division straightforward.

Anahtar Kavram

Calculating the total amount when a specific decimal fraction of that amount is known.
Soru 3Soru

Evaluate the following mathematical expression:

1.2+2.4÷45×0.51.2 + 2.4 \div \frac{4}{5} \times 0.5

What is the correct resulting value?

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Cevap: 2.7

Cevap

2.7
The correct sequence of operations is to first standardize the numbers (converting 45\frac{4}{5} to 0.80.8), perform the division (2.4÷0.8=32.4 \div 0.8 = 3), perform the multiplication (3×0.5=1.53 \times 0.5 = 1.5), and finally add the result to the initial value (1.2+1.5=2.71.2 + 1.5 = 2.7).

Adım Adım Çözüm

1
Convert the fraction to a decimal to make all terms uniform.
45=0.8\frac{4}{5} = 0.8. The expression becomes 1.2+2.4÷0.8×0.51.2 + 2.4 \div 0.8 \times 0.5.
Working entirely in decimals (or entirely in fractions) simplifies the arithmetic process.
2
Apply the order of operations (BODMAS/PEMDAS) by resolving division and multiplication from left to right. First, divide.
2.4÷0.8=32.4 \div 0.8 = 3. The expression becomes 1.2+3×0.51.2 + 3 \times 0.5.
Division and multiplication take precedence over addition, and must be performed sequentially from left to right.
3
Perform the multiplication step.
3×0.5=1.53 \times 0.5 = 1.5. The expression becomes 1.2+1.51.2 + 1.5.
Multiplication is the next operation in the left-to-right sequence.
4
Perform the final addition.
1.2+1.5=2.71.2 + 1.5 = 2.7.
Addition is the last operation remaining in the expression.

Anahtar Kavram

Order of Operations (BODMAS) with Mixed Number Formats
Soru 4Soru

Three electronic metronomes are set to tick at regular intervals of 23\frac{2}{3} of a second, 34\frac{3}{4} of a second, and 45\frac{4}{5} of a second, respectively. If they all tick simultaneously at a given moment, after how many seconds will they next tick together simultaneously?

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Cevap: 12

Cevap

12
The correct answer accurately uses the formula for the LCM of fractions, which is the LCM of the numerators (2,3,42, 3, 4) divided by the HCF of the denominators (3,4,53, 4, 5). This results in 12÷1=1212 \div 1 = 12 seconds.

Adım Adım Çözüm

1
Identify the mathematical operation required.
To find when the metronomes tick together next, we must find the Least Common Multiple (LCM) of their interval times: 23\frac{2}{3}, 34\frac{3}{4}, and 45\frac{4}{5}.
The LCM of multiple time intervals gives the smallest total time at which all periodic events align.
2
Apply the formula for the LCM of fractions.
The formula is: LCM=LCM of numeratorsHCF of denominatorsLCM = \frac{\text{LCM of numerators}}{\text{HCF of denominators}}.
This standard formula ensures the resulting value is a multiple of all the given fractional intervals.
3
Calculate the LCM of the numerators.
The numerators are 2,3,2, 3, and 44. Their LCM is 1212.
1212 is the smallest number perfectly divisible by 22, 33, and 44.
4
Calculate the HCF of the denominators.
The denominators are 3,4,3, 4, and 55. Their Highest Common Factor (HCF) is 11.
3,4,3, 4, and 55 are co-prime integers with no common divisor other than 11.
5
Determine the final LCM.
LCM=121=12LCM = \frac{12}{1} = 12 seconds.
Substituting the calculated values into the formula yields the final answer.

Anahtar Kavram

LCM of Fractions
Tahmini Süre:45s
Soru 5Soru

Compute the exact Highest Common Factor (HCF) for the fractions 625\frac{6}{25} and 910\frac{9}{10}. Express your final result as a decimal.

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Cevap: 0.06

Cevap

0.06
The Highest Common Factor (HCF) of a set of fractions is found by dividing the HCF of their numerators by the LCM of their denominators. In this case, HCF(6, 9) is 3, and LCM(25, 10) is 50. This gives the fraction 3/50, which perfectly evaluates to the decimal 0.06.

Adım Adım Çözüm

1
Identify the standard rule for calculating the HCF of fractional numbers.
HCF = (HCF of numerators) / (LCM of denominators)
This is the mathematical formula required to find the greatest common divisor of multiple fractions.
2
Find the Highest Common Factor (HCF) of the two numerators, 6 and 9.
HCF(6, 9) = 3
3 is the largest integer that divides both 6 and 9 without leaving a remainder.
3
Find the Least Common Multiple (LCM) of the two denominators, 25 and 10.
LCM(25, 10) = 50
50 is the smallest positive integer that is a multiple of both 25 and 10.
4
Substitute the results into the formula and convert the fraction to a decimal.
3 / 50 = 0.06
Dividing 3 by 50 yields the terminating decimal 0.06, which is the final required format.

Anahtar Kavram

HCF and LCM of fractions
Soru 6Soru

During a materials testing process, the density of four composite alloys (A, B, C, and D) is recorded in grams per cubic centimeter (g/cm3\text{g/cm}^3):

- Alloy A: 2.152.\overline{15}
- Alloy B: 157\frac{15}{7}
- Alloy C: 2.152.1\overline{5}
- Alloy D: 2813\frac{28}{13}

Arrange the alloys in ascending order based on their density, from the lowest to the highest.

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Cevap

The correct ascending order is Alloy B, Alloy A, Alloy D, Alloy C.
By converting all values to their decimal or pure fractional equivalents, we can observe that Alloy B (2.142...2.142...) is the smallest. The remaining values all begin with 2.152.15, but expanding the subsequent decimal places reveals that Alloy A is 2.1515...2.1515..., Alloy D is 2.1538...2.1538..., and Alloy C is 2.1555...2.1555.... This yields the exact ascending sequence: Alloy B < Alloy A < Alloy D < Alloy C.

Adım Adım Çözüm

1
Separate the whole number from the fractional part for all given values to simplify the comparison.
All values have a whole number part of 2. We only need to compare their fractional parts: A (0.150.\overline{15}), B (17\frac{1}{7}), C (0.150.1\overline{5}), and D (213\frac{2}{13}).
Since all numbers start with 2, isolating the fractional or decimal part reduces computational load.
2
Convert all fractional and decimal parts into a common format (decimal approximation) to establish initial bounds.
A = 0.1515...0.1515..., B 0.1428...\approx 0.1428..., C = 0.1555...0.1555..., D 0.1538...\approx 0.1538...
Decimal expansion allows for rapid estimation and sequencing without finding a large common denominator.
3
Verify the exact order by cross-multiplying the fractional equivalents of the closest values to eliminate any rounding uncertainties.
Fractions to compare: A (1599=533\frac{15}{99} = \frac{5}{33}), B (17\frac{1}{7}), C (1490=745\frac{14}{90} = \frac{7}{45}), D (213\frac{2}{13}). Comparing B and A: 1×33=331 \times 33 = 33, 7×5=357 \times 5 = 35, so B < A. Comparing A and D: 5×13=655 \times 13 = 65, 33×2=6633 \times 2 = 66, so A < D. Comparing D and C: 2×45=902 \times 45 = 90, 13×7=9113 \times 7 = 91, so D < C.
Cross-multiplication of fractions provides absolute mathematical proof of the inequalities: 17<533<213<745\frac{1}{7} < \frac{5}{33} < \frac{2}{13} < \frac{7}{45}.

Anahtar Kavram

Comparing mixed fractions and recurring decimals via decimal expansion and cross-multiplication
Soru 7Soru

A local water reservoir was initially filled to exactly 0.30.\overline{3} of its maximum capacity. After several days of heavy rainfall, an additional 14.414.4 million liters of water flowed into the reservoir, bringing the water level to 0.60.6 of its maximum capacity.

Currently, a nearby town consumes water from the reservoir at a steady rate of 1121\frac{1}{2} million liters per day. Additionally, an undiscovered leak at the base of the reservoir drains water at a constant rate of 0.050.05 million liters per hour.

If the reservoir were completely full and no further water was added, in how many days would it become completely empty?

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Cevap: 20

Cevap

20
The total capacity of the reservoir is found by equating the fractional increase (3513=415\frac{3}{5} - \frac{1}{3} = \frac{4}{15}) to the volume added (14.414.4 million liters), yielding 5454 million liters. The total daily water depletion is the sum of the town's daily usage (1.51.5 million liters) and the leak converted to a daily rate (0.05×24=1.20.05 \times 24 = 1.2 million liters), giving 2.72.7 million liters per day. Dividing 5454 by 2.72.7 gives exactly 2020 days.

Adım Adım Çözüm

1
Convert the decimal fill levels into fractions and compute their difference.
Initial level = 13\frac{1}{3}. Final level = 35\frac{3}{5}. Difference = 3513=415\frac{3}{5} - \frac{1}{3} = \frac{4}{15}.
Working with exact fractions avoids rounding errors from recurring decimals and simplifies finding the proportion of water added.
2
Calculate the maximum capacity of the reservoir.
415×Capacity=14.4    Capacity=14.4×154=54\frac{4}{15} \times \text{Capacity} = 14.4 \implies \text{Capacity} = 14.4 \times \frac{15}{4} = 54 million liters.
The difference in the fractional water level represents exactly the volume of rain added.
3
Calculate the total volume of water lost per day.
Town = 1.51.5 million L/day. Leak = 0.05×24=1.20.05 \times 24 = 1.2 million L/day. Total = 2.72.7 million L/day.
The rates must be in the same time unit (days) before they can be accurately combined.
4
Determine the time required to empty the completely full reservoir.
54÷2.7=2054 \div 2.7 = 20 days.
Dividing the total capacity by the combined daily rate of depletion gives the time in days.

Anahtar Kavram

Fractions and Decimals
Tahmini Süre:3m 0s
Soru 8Soru

A municipal corporation is distributing its annual infrastructure development fund. Exactly 0.50.\overline{5} of the total fund is allocated to urban housing projects. From the remaining fund, 37\frac{3}{7} is assigned to public transportation. The rest of the fund is distributed equally between two sectors: renewable energy and public parks. If the amount allocated to renewable energy is $3.2\$3.2 million, what is the total annual infrastructure development fund?

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Cevap: $25.2\$25.2 million

Cevap

$25.2\$25.2 million
The correct sequence evaluates the recurring decimal 0.50.\overline{5} as 59\frac{5}{9}. This leaves 49\frac{4}{9} of the fund. Transport takes 37\frac{3}{7} of this remainder, which is 1263\frac{12}{63}. The new remainder is 491263=1663\frac{4}{9} - \frac{12}{63} = \frac{16}{63}. This is split equally, meaning renewable energy gets 863\frac{8}{63} of the total. Solving 863x=3.2\frac{8}{63}x = 3.2 gives x=25.2x = 25.2.

Adım Adım Çözüm

1
Convert the recurring decimal to a fraction.
0.5=590.\overline{5} = \frac{5}{9}
Fractional form is required to perform precise subsequent operations without rounding errors.
2
Calculate the remaining fraction of the fund after the housing allocation.
159=491 - \frac{5}{9} = \frac{4}{9}
The next allocation is taken from the remaining fund, not the total fund.
3
Determine the fraction of the total fund allocated to public transportation.
37×49=1263=421\frac{3}{7} \times \frac{4}{9} = \frac{12}{63} = \frac{4}{21}
Multiplying the proportion (37\frac{3}{7}) by the available remainder (49\frac{4}{9}) yields its share of the total.
4
Calculate the new remaining fund for the last two sectors.
49421=28631263=1663\frac{4}{9} - \frac{4}{21} = \frac{28}{63} - \frac{12}{63} = \frac{16}{63}
Subtracting the transportation allocation from the previous remainder gives the final undistributed portion.
5
Determine the fraction of the total fund allocated to renewable energy.
12×1663=863\frac{1}{2} \times \frac{16}{63} = \frac{8}{63}
The problem states the final remainder is split equally between two sectors.
6
Equate the fraction to the given monetary value and solve for the total fund (xx).
863x=3.2    x=3.2×638=0.4×63=25.2\frac{8}{63}x = 3.2 \implies x = 3.2 \times \frac{63}{8} = 0.4 \times 63 = 25.2
This establishes the algebraic relationship to find the initial total.

Anahtar Kavram

Successive Fractional Operations and Recurring Decimals

Alternatif Yöntem

Instead of calculating absolute remaining fractions through subtraction at each step, you can chain the remaining proportions through direct multiplication: Total × (1 - 5/9) × (1 - 3/7) × (1/2) = Total × (4/9) × (4/7) × (1/2) = Total × (8/63). Setting (8/63) equal to 3.2 million quickly yields 25.2 million.
Tahmini Süre:2m 30s
Soru 9Soru

During a system stress test, four data processing modules record their average latency per transaction (in seconds) as given below. Arrange the modules in ascending order based on their latency times (from fastest to slowest).

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Cevap

The correct ascending sequence from fastest to slowest latency is Module Y, followed by Module X, then Module W, and finally Module Z.
By converting all latency values into extended decimals, we find Module Y (0.25656...0.25656...) is the smallest, followed by Module X (0.25714...0.25714...), then Module W (0.25757...0.25757...), and finally Module Z (0.25777...0.25777...). Therefore, the correct ascending order is Module Y, Module X, Module W, Module Z.

Adım Adım Çözüm

1
Calculate the exact value for Module W using the HCF of fractions formula.
HCF = HCF(34,85)LCM(33,22)=1766\frac{\text{HCF}(34, 85)}{\text{LCM}(33, 22)} = \frac{17}{66}. Converted to a decimal, this is 0.25757...0.25757... (0.2570.2\overline{57}).
To standardize the mathematical expression into a comparable decimal format.
2
Convert the value for Module X into a decimal.
9350.2571428...\frac{9}{35} \approx 0.2571428...
Decimals allow for direct digit-by-digit comparison.
3
Expand the recurring decimal for Module Y to several decimal places.
0.256=0.2565656...0.2\overline{56} = 0.2565656...
To provide enough precision to compare digits past the third decimal place.
4
Convert the recurring decimal in Module Z to a fraction, multiply, and convert back to a decimal.
1.28=1281290=11690=58451.2\overline{8} = \frac{128-12}{90} = \frac{116}{90} = \frac{58}{45}. Then, 5845×15=58225\frac{58}{45} \times \frac{1}{5} = \frac{58}{225}. Converting to decimal gives 0.25777...0.25777... (0.2570.25\overline{7}).
To determine its exact decimal value for accurate comparison against the other modules.
5
Compare the decimal values digit by digit from left to right to establish the ascending order.
0.25656...0.25656... (Y) < 0.25714...0.25714... (X) < 0.25757...0.25757... (W) < 0.25777...0.25777... (Z).
The question asks for the fastest to slowest modules, which corresponds to the shortest to longest latency times (ascending numerical order).

Anahtar Kavram

Advanced comparison of fractions, recurring decimals, and the application of HCF/LCM rules for fractions.

Alternatif Yöntem

Find a common denominator for all fractions: 66 for W (17/66), 35 for X (9/35), 990 for Y (254/990), and 225 for Z (58/225). The LCM of 66, 35, 990, and 225 is 34650. Convert all fractions to this denominator to compare the exact numerators, though decimal conversion to the fourth digit is faster.
Tahmini Süre:2m 30s
Soru 10Soru

An industrial chemical plant stores a specialized solvent in three large cylindrical vats. The first vat contains 53.353.\overline{3} liters, the second contains 71.171.\overline{1} liters, and the third contains 26.626.\overline{6} liters of the solvent. The plant manager wants to completely transfer the solvent from all three vats into identical, smaller drums such that every drum is completely filled, no solvent is left over in any vat, and solvents from different vats are not mixed. What should be the maximum possible capacity of each drum?

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Cevap: 8.88.\overline{8} liters

Cevap

The maximum possible capacity of each drum is 8.88.\overline{8} liters.
To find the maximum identical capacity that leaves no remainder, we must calculate the Highest Common Factor (HCF) of the three volumes. By converting the recurring decimals to fractions (1603\frac{160}{3}, 6409\frac{640}{9}, 803\frac{80}{3}) and applying the fraction HCF formula (HCF of numeratorsLCM of denominators\frac{\text{HCF of numerators}}{\text{LCM of denominators}}), we obtain 809\frac{80}{9}, which perfectly translates to 8.88.\overline{8} liters.

Adım Adım Çözüm

1
Convert the given recurring decimals representing the solvent volumes into their simplest fractional forms.
53.3=53+39=160353.\overline{3} = 53 + \frac{3}{9} = \frac{160}{3}, 71.1=71+19=640971.\overline{1} = 71 + \frac{1}{9} = \frac{640}{9}, and 26.6=26+69=80326.\overline{6} = 26 + \frac{6}{9} = \frac{80}{3}.
Fractional forms are required to accurately and properly compute the highest common factor (HCF) of non-integer values.
2
Identify the mathematical operation required based on the physical constraints described in the problem.
We must calculate the Highest Common Factor (HCF) of the three volumes: 1603\frac{160}{3}, 6409\frac{640}{9}, and 803\frac{80}{3}.
The solvent must be divided equally without any remainders, meaning the drum size must be a common factor of all three initial volumes, and the problem asks for the 'maximum possible capacity'.
3
Apply the standard formula for finding the HCF of multiple fractions.
The formula is: HCF=HCF of numeratorsLCM of denominators\text{HCF} = \frac{\text{HCF of numerators}}{\text{LCM of denominators}}.
This formula ensures the resulting fraction will evenly divide each of the original fractions without leaving a remainder.
4
Calculate the HCF of the numerators (160160, 640640, 8080) and the LCM of the denominators (33, 99, 33).
HCF(160,640,80)=80\text{HCF}(160, 640, 80) = 80, and LCM(3,9,3)=9\text{LCM}(3, 9, 3) = 9. Thus, the HCF of the fractions is 809\frac{80}{9}.
8080 is the largest integer dividing 160160, 640640, and 8080. 99 is the smallest integer divisible by 33, 99, and 33.
5
Convert the resulting fraction back into a recurring decimal.
809=8+89=8.8\frac{80}{9} = 8 + \frac{8}{9} = 8.\overline{8} liters.
The final calculated capacity should match the formatting style of the given options.

Anahtar Kavram

Calculating the Highest Common Factor (HCF) of recurring decimals by converting them to fractions and using the fraction HCF rule.
Tahmini Süre:2m 30s
Soru 11Soru

A large agricultural cooperative is dividing a massive tract of land for different crops. They allocate 0.4285710.\overline{428571} of the total land to cultivate sunflowers, 0.160.1\overline{6} of the total land to cultivate maize, and 0.050.0\overline{5} of the total land to cultivate organic vegetables. The remaining land, which measures exactly 110110 hectares, is preserved as a wildlife reserve. What is the total area of the tract of land, in hectares?

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Cevap: 315

Cevap

315
The total area is found by properly converting all recurring decimals into exact fractions (3/73/7, 1/61/6, and 1/181/18), summing them to find the total allocated land (41/6341/63), determining the remaining land fraction (22/6322/63), and setting it equal to the given 110110 hectares. Solving for the whole yields exactly 315315 hectares.

Adım Adım Çözüm

1
Convert the pure recurring decimal 0.4285710.\overline{428571} into a simplified fraction.
The fraction is 37\frac{3}{7}.
Recognizing that 17=0.142857\frac{1}{7} = 0.\overline{142857}, we can multiply by 33 to get 0.4285710.\overline{428571}. Alternatively, using the algebraic method: 428571999999=37\frac{428571}{999999} = \frac{3}{7}.
2
Convert the mixed recurring decimals 0.160.1\overline{6} and 0.050.0\overline{5} into fractions.
0.16=16190=1590=160.1\overline{6} = \frac{16-1}{90} = \frac{15}{90} = \frac{1}{6} and 0.05=5090=590=1180.0\overline{5} = \frac{5-0}{90} = \frac{5}{90} = \frac{1}{18}.
To operate with mixed recurring decimals, subtract the non-repeating part from the entire number, and place it over a denominator consisting of 9s (for repeating digits) followed by 0s (for non-repeating digits after the decimal point).
3
Calculate the total fraction of land allocated to the three crops.
37+16+118=37+3+118=37+418=37+29=27+1463=4163\frac{3}{7} + \frac{1}{6} + \frac{1}{18} = \frac{3}{7} + \frac{3+1}{18} = \frac{3}{7} + \frac{4}{18} = \frac{3}{7} + \frac{2}{9} = \frac{27+14}{63} = \frac{41}{63}.
Finding a common denominator (6363) allows us to sum the individual crop fractions to determine the total proportion of cultivated land.
4
Determine the fraction representing the wildlife reserve and calculate the total land area.
Reserve fraction = 14163=22631 - \frac{41}{63} = \frac{22}{63}. Total Area = 110×6322=315110 \times \frac{63}{22} = 315 hectares.
The unallocated fraction represents the reserve area. Setting this fraction of the total area (TT) equal to 110110 hectares (2263×T=110\frac{22}{63} \times T = 110) gives the final answer.

Anahtar Kavram

Fractions and Decimals
Tahmini Süre:2m 30s
Soru 12Soru

A digital communications system transmits data packets of three specific sizes: 3518\frac{35}{18} MB, 74\frac{7}{4} MB, and 4924\frac{49}{24} MB. To optimize the network buffer, the engineers must define a standardized base unit size. Every transmitted packet must be an exact multiple of this base unit. To maximize efficiency, what is the largest possible size for this base unit?

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Cevap: 772\frac{7}{72} MB

Cevap

The largest possible size for the base unit is 772\frac{7}{72} MB.
To find the largest possible base unit that perfectly divides all three packet sizes, the Highest Common Factor (HCF) of the fractions must be calculated. The formula for the HCF of fractions is HCF(Numerators) / LCM(Denominators). The numerators are 35, 7, and 49, which have an HCF of 7. The denominators are 18, 4, and 24, which have an LCM of 72. Thus, the correct largest base unit is 7/72 MB.

Adım Adım Çözüm

1
Identify the required mathematical operation.
Find the Highest Common Factor (HCF) of the three fractions: 3518\frac{35}{18}, 74\frac{7}{4}, and 4924\frac{49}{24}.
The problem asks for the 'largest possible size' that perfectly divides the sizes of all given packets, which is the definition of HCF.
2
Calculate the HCF of the numerators.
The numerators are 35, 7, and 49. Their HCF is 7.
This value forms the numerator of the final fraction according to the formula.
3
Calculate the Least Common Multiple (LCM) of the denominators.
The denominators are 18, 4, and 24. Their LCM is 72.
This value forms the denominator of the final fraction according to the formula.
4
Combine the results to find the final HCF of the fractions.
772\frac{7}{72} MB.
Applying the formula: HCF of fractions = HCF of NumeratorsLCM of Denominators\frac{\text{HCF of Numerators}}{\text{LCM of Denominators}}.

Anahtar Kavram

Highest Common Factor (HCF) of Fractions
Soru 13Soru

An agronomist is comparing the soil moisture retention rates of four different agricultural plots. The rates, measured in liters per cubic meter per day, are given below. Arrange the plots in ascending order of their moisture retention rates.

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Cevap

The correct ascending order is Plot Beta, Plot Delta, Plot Gamma, and then Plot Alpha.
By converting all rates to extended decimals, we get: Plot Beta (1.444...1.444...), Plot Delta (1.4501.450), Plot Gamma (1.4545...1.4545...), and Plot Alpha (1.4555...1.4555...). Comparing them digit-by-digit from left to right establishes the strictly ascending order: Beta, Delta, Gamma, Alpha.

Adım Adım Çözüm

1
Convert the fraction into a decimal to create a uniform format for comparison.
Plot Beta: 139=1.4444...\frac{13}{9} = 1.4444...
Decimal forms allow for direct digit-by-digit comparison of magnitudes.
2
Expand all values to at least four decimal places to clearly see the repeating patterns.
Plot Alpha: 1.4555...1.4555...
Plot Beta: 1.4444...1.4444...
Plot Gamma: 1.4545...1.4545...
Plot Delta: 1.4500...1.4500...
Truncating or estimating too early can hide the subtle differences between terminating and repeating decimals.
3
Compare the tenths and hundredths places.
All values have a 44 in the tenths place. In the hundredths place, Plot Beta has a 44, while the others have a 55. Thus, Plot Beta is the smallest.
Comparing digits from left to right establishes the overall magnitude.
4
Compare the thousandths place for the remaining three plots.
Plot Delta has 00, Plot Gamma has 44, and Plot Alpha has 55.
This establishes the final order for the remaining values: Delta (1.450...1.450...) < Gamma (1.4545...1.4545...) < Alpha (1.4555...1.4555...).

Anahtar Kavram

Comparing terminating decimals, recurring decimals, and fractions by converting them to expanded decimal forms.
Soru 14Soru

In a highly automated manufacturing plant, three different robotic assembly lines are synchronized by a central control system. Line A completes a specific micro-calibration cycle every 415\frac{4}{15} of a minute. Line B completes its calibration cycle every 625\frac{6}{25} of a minute, and Line C completes its cycle every 835\frac{8}{35} of a minute. If all three lines complete a calibration cycle simultaneously at a given moment, what is the shortest time interval that must elapse before all three lines complete their calibration cycles together again?

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Cevap: 245\frac{24}{5} minutes

Cevap

The shortest time interval is 245\frac{24}{5} minutes.
To find when independent periodic events next occur simultaneously, we must calculate their Lowest Common Multiple (LCM). For fractions, the LCM is found by dividing the LCM of the numerators by the HCF of the denominators. The LCM of the numerators (4,6,84, 6, 8) is 2424, and the HCF of the denominators (15,25,3515, 25, 35) is 55. Therefore, the overall LCM is 245\frac{24}{5}.

Adım Adım Çözüm

1
Identify the mathematical operation required.
We need to find the Lowest Common Multiple (LCM) of the three given time intervals (415\frac{4}{15}, 625\frac{6}{25}, 835\frac{8}{35}) to determine when the periodic events will next coincide.
The LCM represents the smallest common interval of time into which each individual cycle can fit perfectly.
2
Recall the formula for finding 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 standard formula is required when working with periodic fractional amounts.
3
Find the LCM of the numerators.
The numerators are 44, 66, and 88. The lowest common multiple of 4,6, and 84, 6, \text{ and } 8 is 2424.
Multiples of 88 are 8,16,248, 16, 24. 2424 is the smallest multiple evenly divisible by both 44 and 66.
4
Find the Highest Common Factor (HCF) of the denominators.
The denominators are 1515, 2525, and 3535. The highest common factor of 15,25, and 3515, 25, \text{ and } 35 is 55.
The common prime factor shared by 3×53 \times 5, 5×55 \times 5, and 7×57 \times 5 is 55.
5
Apply the values to the fraction LCM formula.
LCM(4,6,8)HCF(15,25,35)=245\frac{\text{LCM}(4, 6, 8)}{\text{HCF}(15, 25, 35)} = \frac{24}{5}.
Combining the calculated numerator and denominator provides the final fractional interval.

Anahtar Kavram

Lowest Common Multiple (LCM) of Fractions
Soru 15Soru

A civil engineering team is installing three different types of sensor cables along a newly constructed bridge. The standard supplier rolls for these cables come in lengths of 552\frac{55}{2} meters, 774\frac{77}{4} meters, and 1215\frac{121}{5} meters. To ensure modular replacement without any waste, the team must cut all the rolls into smaller, equal-length segments. What is the maximum possible length, in meters, of each segment?

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Cevap: 0.55

Cevap

0.55 meters
The problem asks for the maximum equal length that can divide all three roll lengths without any remainder, which mathematically corresponds to the Highest Common Factor (HCF). Using the standard formula for fractions: HCF = HCF(55, 77, 121) / LCM(2, 4, 5) = 11 / 20 = 0.55 meters.

Adım Adım Çözüm

1
Determine the correct mathematical operation based on the problem context.
The requirement to cut multiple rolls into equal segments of maximum possible length with zero waste indicates the need to find the Highest Common Factor (HCF) of the three lengths.
HCF is used to find the largest common quantity that can perfectly divide multiple numbers.
2
State the standard formula for finding the HCF of fractions.
HCF=HCF of numeratorsLCM of denominatorsHCF = \frac{\text{HCF of numerators}}{\text{LCM of denominators}}
This is the mathematical rule for finding the HCF of rational numbers.
3
Calculate the HCF of the numerators.
The numerators are 5555, 7777, and 121121. Their prime factorizations are 5×115 \times 11, 7×117 \times 11, and 11×1111 \times 11. The HCF is 1111.
The largest common prime factor across all three numerators is 11.
4
Calculate the LCM of the denominators.
The denominators are 22, 44, and 55. The lowest common multiple is 2020.
20 is the smallest positive integer divisible by 2, 4, and 5.
5
Compute the final decimal value.
1120=0.55\frac{11}{20} = 0.55
Converting the fraction to a decimal yields the exact required length.

Anahtar Kavram

Highest Common Factor (HCF) of Fractions
Soru 16Soru

At a metropolitan central station, three express commuter trains depart simultaneously at 6:00 AM on Monday. The Green Line completes its round trip every 2142 \frac{1}{4} hours, the Blue Line every 3383 \frac{3}{8} hours, and the Red Line every 4124 \frac{1}{2} hours. Assuming continuous operation without delays, how many hours will it take for all three trains to depart simultaneously from the central station again?

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Cevap: 131213 \frac{1}{2}

Cevap

The correct time is 131213 \frac{1}{2} hours, found by taking the Least Common Multiple (LCM) of the three fractional time intervals.
To find the next simultaneous departure, the Least Common Multiple (LCM) of the round-trip times is needed. Converting the times to improper fractions gives 94\frac{9}{4}, 278\frac{27}{8}, and 92\frac{9}{2}. The LCM of fractions is found by dividing the LCM of the numerators (which is 27) by the Highest Common Factor (HCF) of the denominators (which is 2). This gives 272\frac{27}{2}, or 131213 \frac{1}{2} hours.

Adım Adım Çözüm

1
Convert the mixed numbers to improper fractions.
Green Line: 94\frac{9}{4} hours, Blue Line: 278\frac{27}{8} hours, Red Line: 92\frac{9}{2} hours.
Improper fractions are required to accurately calculate the Least Common Multiple (LCM) of fractions.
2
Identify the formula for finding 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 formula allows us to find the lowest common multiple for non-integer values.
3
Calculate the LCM of the numerators.
The numerators are 9, 27, and 9. Their LCM is 27.
27 is the smallest number perfectly divisible by 9, 27, and 9.
4
Calculate the HCF of the denominators.
The denominators are 4, 8, and 2. Their HCF is 2.
2 is the largest number that perfectly divides 4, 8, and 2.
5
Apply the values to the fraction LCM formula.
272=1312\frac{27}{2} = 13 \frac{1}{2} hours.
Dividing the LCM of numerators by the HCF of denominators yields the final LCM of the fractions.

Anahtar Kavram

Calculating the Least Common Multiple (LCM) of fractions to solve periodic synchronization problems.
Soru 17Soru

Arrange the following real numbers in ascending order (from smallest to largest):

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Cevap

The correct ascending order is 2340<0.57<0.57<712\frac{23}{40} < 0.\overline{57} < 0.5\overline{7} < \frac{7}{12}.
Converting all numbers to decimal expansions gives: 2340=0.5750\frac{23}{40} = 0.5750, 0.57=0.575757...0.\overline{57} = 0.575757..., 0.57=0.577777...0.5\overline{7} = 0.577777..., and 712=0.583333...\frac{7}{12} = 0.583333.... Comparing these digit by digit confirms the correct ascending sequence is 2340<0.57<0.57<712\frac{23}{40} < 0.\overline{57} < 0.5\overline{7} < \frac{7}{12}.

Adım Adım Çözüm

1
Convert each fraction and recurring decimal into its expanded decimal representation.
2340=0.57500...\frac{23}{40} = 0.57500..., 0.57=0.575757...0.\overline{57} = 0.575757..., 0.57=0.577777...0.5\overline{7} = 0.577777..., and 712=0.583333...\frac{7}{12} = 0.583333...
Converting all terms into a common decimal format allows direct place-value comparison.
2
Compare the values digit by digit from left to right starting at the tenths place.
All terms start with 0.570.57, except 712\frac{7}{12} which is 0.58...0.58... (making 712\frac{7}{12} the largest). For the remaining three terms, compare the thousandths digit: 2340\frac{23}{40} has 00 in the ten-thousandths place (0.57500.5750), 0.570.\overline{57} has 77 (0.57570.5757), and 0.570.5\overline{7} has 77 in the thousandths place (0.57770.5777).
Determining place-value differences establishes the relative magnitudes clearly.
3
Sequence the original numbers according to their evaluated decimal order.
0.5750<0.5757...<0.5777...<0.5833...    2340<0.57<0.57<7120.5750 < 0.5757... < 0.5777... < 0.5833... \implies \frac{23}{40} < 0.\overline{57} < 0.5\overline{7} < \frac{7}{12}
Re-substituting the original expressions yields the required ascending order sequence.

Anahtar Kavram

Ordering and comparison of rational numbers, terminating fractions, pure recurring decimals, and mixed recurring decimals.
Soru 18Soru

Three quantities are given as 0.60.\overline{6}, 1225\frac{12}{25}, and 1.41.4. What is the exact ratio of the Least Common Multiple (LCM) of these three quantities to their Highest Common Factor (HCF)?

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Cevap: 6300

Cevap

6300
Converting the given values yields the reduced fractions 23\frac{2}{3}, 1225\frac{12}{25}, and 75\frac{7}{5}. Applying standard fraction formulas, LCM=LCM(2,12,7)HCF(3,25,5)=841=84\text{LCM} = \frac{\text{LCM}(2,12,7)}{\text{HCF}(3,25,5)} = \frac{84}{1} = 84 and HCF=HCF(2,12,7)LCM(3,25,5)=175\text{HCF} = \frac{\text{HCF}(2,12,7)}{\text{LCM}(3,25,5)} = \frac{1}{75}. Taking the ratio LCMHCF\frac{\text{LCM}}{\text{HCF}} gives 84÷175=630084 \div \frac{1}{75} = 6300.

Adım Adım Çözüm

1
Convert all terms to irreducible fractions
0.6=69=230.\overline{6} = \frac{6}{9} = \frac{2}{3}, 1225=1225\frac{12}{25} = \frac{12}{25}, and 1.4=1410=751.4 = \frac{14}{10} = \frac{7}{5}
All numbers must be expressed as simplified fractions ab\frac{a}{b} in lowest terms before applying fraction HCF and LCM formulas.
2
Calculate the Least Common Multiple (LCM) of the fractions
LCM(23,1225,75)=LCM(2,12,7)HCF(3,25,5)=841=84\text{LCM}\left(\frac{2}{3}, \frac{12}{25}, \frac{7}{5}\right) = \frac{\text{LCM}(2, 12, 7)}{\text{HCF}(3, 25, 5)} = \frac{84}{1} = 84
The LCM of a set of fractions is given by LCM of numeratorsHCF of denominators\frac{\text{LCM of numerators}}{\text{HCF of denominators}}.
3
Calculate the Highest Common Factor (HCF) of the fractions
HCF(23,1225,75)=HCF(2,12,7)LCM(3,25,5)=175\text{HCF}\left(\frac{2}{3}, \frac{12}{25}, \frac{7}{5}\right) = \frac{\text{HCF}(2, 12, 7)}{\text{LCM}(3, 25, 5)} = \frac{1}{75}
The HCF of a set of fractions is given by HCF of numeratorsLCM of denominators\frac{\text{HCF of numerators}}{\text{LCM of denominators}}.
4
Find the ratio of the LCM to the HCF
Ratio=84175=84×75=6300\text{Ratio} = \frac{84}{\frac{1}{75}} = 84 \times 75 = 6300
Dividing the computed LCM by the computed HCF yields the required ratio.

Anahtar Kavram

HCF and LCM of Fractions and Recurring Decimals
Soru 19Soru

A district officer allocates a development fund among four sectors: Health, Education, Infrastructure, and Agriculture. Health receives a fraction equal to 0.160.1\overline{6} of the total fund. Education receives 38\frac{3}{8} of the remaining fund after the Health allocation. Infrastructure is directly allocated 0.430.4\overline{3} of the original total fund. The remaining amount is given to Agriculture. If the amount allocated to Infrastructure exceeds the amount allocated to Education by ₹ 87,00087,000, what is the total development fund in Rupees?

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Cevap: 720000

Cevap

The total development fund is ₹ 720,000.
Converting the recurring decimals gives Health's share as 16\frac{1}{6} and Infrastructure's share as 1330\frac{13}{30}. The remaining fund after Health allocation is 56\frac{5}{6}, making Education's share equal to 38×56=516\frac{3}{8} \times \frac{5}{6} = \frac{5}{16} of the total. The difference between Infrastructure's and Education's shares is 1330516=29240\frac{13}{30} - \frac{5}{16} = \frac{29}{240}. Equating 29240\frac{29}{240} of the total fund to ₹ 87,000 yields the total fund as ₹ 720,000.

Adım Adım Çözüm

1
Convert the recurring decimals into simplified vulgar fractions.
0.16=16190=1590=160.1\overline{6} = \frac{16-1}{90} = \frac{15}{90} = \frac{1}{6} and 0.43=43490=3990=13300.4\overline{3} = \frac{43-4}{90} = \frac{39}{90} = \frac{13}{30}.
Converting mixed recurring decimals to fractions allows exact algebraic calculations.
2
Calculate the fraction of the total fund received by Education.
Remaining fund after Health = 116=561 - \frac{1}{6} = \frac{5}{6}. Education's fraction = 38×56=516\frac{3}{8} \times \frac{5}{6} = \frac{5}{16}.
Education receives a fraction of the remaining fund, not the original total fund.
3
Find the difference between Infrastructure's fraction and Education's fraction.
1330516=10475240=29240\frac{13}{30} - \frac{5}{16} = \frac{104 - 75}{240} = \frac{29}{240}.
The LCM of denominators 30 and 16 is 240.
4
Set up the linear equation with the given monetary difference to find the total fund FF.
29240×F=87,000    F=87,000×24029=3,000×240=720,000\frac{29}{240} \times F = 87,000 \implies F = \frac{87,000 \times 240}{29} = 3,000 \times 240 = 720,000.
Solving for FF yields the original total fund allocation.

Anahtar Kavram

Converting mixed recurring decimals to vulgar fractions and evaluating multi-step compound fractional parts of a total quantity.
Soru 20Soru

What is the Highest Common Factor (HCF) of the fractions 34\frac{3}{4} and 910\frac{9}{10}?

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Cevap: 320\frac{3}{20}

Cevap

The correct Highest Common Factor is 320\frac{3}{20}.
To find the Highest Common Factor (HCF) of a set of fractions, the correct formula is to divide the HCF of their numerators by the LCM of their denominators. For the given fractions, the numerators are 3 and 9, and their HCF is 3. The denominators are 4 and 10, and their LCM is 20. Therefore, the HCF of the two fractions is 320\frac{3}{20}.

Adım Adım Çözüm

1
Identify the formula for finding the HCF of fractions.
HCF of fractions = (HCF of Numerators) / (LCM of Denominators)
This is the mathematical rule required to find the greatest common divisor for fractional numbers.
2
Calculate the HCF of the numerators (3 and 9).
The HCF of 3 and 9 is 3.
3 is the largest integer that divides both 3 and 9 without leaving a remainder.
3
Calculate the LCM of the denominators (4 and 10).
The LCM of 4 and 10 is 20.
The multiples of 4 are 4, 8, 12, 16, 20... and the multiples of 10 are 10, 20... The lowest common multiple they share is 20.
4
Divide the result of the numerators' HCF by the denominators' LCM.
320\frac{3}{20}
Applying the values to the formula gives the final fractional answer.

Anahtar Kavram

HCF and LCM of Fractions
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