Basic Numeracy Practice Questions — State PSC Exam — Page 15

A teacher provides a list of eight numerical values on the board for a classification exercise:

(I) $\sqrt{144}$
(II) $\frac{22}{7}$
(III) $\pi$
(IV) $0.\overline{36}$
(V) $\sqrt{12}$
(VI) $\frac{\sqrt{3}}{\sqrt{27}}$
(VII) $3.14159$
(VIII) $0.121221222...$ (where the number of 2s increases by one each time)

What is the exact count of rational numbers in this list?

Show answer & explanation

Answer: 5

Answer

There are exactly 5 rational numbers in the provided list.

By simplifying each expression and applying the definitions of real numbers, exactly 5 of the 8 values (values I, II, IV, VI, and VII) satisfy the conditions of being a rational number.

Step-by-Step Solution

1

Define the criteria for a rational number.

A rational number is any number that can be expressed as a fraction of two integers (p/q, where q is not zero). This includes all integers, terminating decimals, and repeating decimals.

Establishing the definition is necessary to accurately classify each item.

2

Evaluate the square roots and fractions in the list.

\sqrt{144} = 12

, which is an integer (Rational).

\frac{22}{7}

is a ratio of two integers (Rational).

\sqrt{12} = 2\sqrt{3}

, which contains the root of a non-perfect square (Irrational).

\frac{\sqrt{3}}{\sqrt{27}}

simplifies to

\sqrt{\frac{3}{27}} = \sqrt{\frac{1}{9}} = \frac{1}{3}

(Rational).

Radicals must be simplified to their lowest terms to reveal their true mathematical classification.

3

Evaluate the decimal representations and constants in the list.

0.\overline{36}

is a repeating decimal (Rational).

3.14159

is a terminating decimal (Rational).

\pi

is a transcendental constant with infinite, non-repeating digits (Irrational).

0.121221222...

has a changing pattern, making it non-terminating and non-recurring (Irrational).

Decimals must be analyzed by their termination or repetition properties.

4

Count the total number of items identified as rational.

The rational values are (I), (II), (IV), (VI), and (VII). The total count is 5.

The question asks for the exact numerical count of rational numbers.

Key Concept

Identifying rational and irrational numbers by their fractional, radical, and decimal properties.

Question 282Question →

In an industrial robotics laboratory, three distinct robotic arms perform repetitive cyclic tasks. Arm X completes one full operational cycle every $\frac{16}{15}$ seconds, Arm Y completes a cycle every $\frac{24}{25}$ seconds, and Arm Z completes a cycle every $\frac{32}{35}$ seconds. If all three robotic arms begin their cycles simultaneously from a synchronized home position, what is the minimum time required for all three arms to return to the home position at the exact same instant?

\frac{8}{525}

seconds

\frac{32}{175}

seconds

\frac{96}{5}

seconds

\frac{8}{5}

seconds

Show answer & explanation

Answer: $\frac{96}{5}$ seconds

Answer

The minimum time required for all three robotic arms to synchronize again is

\frac{96}{5}

seconds.

To determine the next simultaneous occurrence of repeating events, we compute the Lowest Common Multiple (LCM) of their cycle durations. For fractions, the LCM is determined by dividing the LCM of the numerators by the Highest Common Factor (HCF) of the denominators. Thus,

\frac{\text{LCM}(16, 24, 32)}{\text{HCF}(15, 25, 35)} = \frac{96}{5}

.

Step-by-Step Solution

1

Identify the mathematical operation required.

Calculate the Lowest Common Multiple (LCM) of the three cycle times.

To find the first instance when multiple independent repeating events synchronize, we must find the LCM of their individual intervals.

2

State the formula for finding the LCM of fractions.

\text{LCM of fractions} = \frac{\text{LCM of Numerators}}{\text{HCF of Denominators}}

This is the standard algebraic rule for determining the lowest common multiple of rational numbers.

3

Calculate the LCM of the numerators: 16, 24, and 32.

The LCM of 16, 24, and 32 is 96.

The prime factorizations are

16 = 2^4

,

24 = 2^3 \times 3

, and

32 = 2^5

. The highest powers of prime factors are

2^5

and

3^1

, so

2^5 \times 3 = 32 \times 3 = 96

.

4

Calculate the HCF of the denominators: 15, 25, and 35.

The HCF of 15, 25, and 35 is 5.

The prime factorizations are

15 = 3 \times 5

,

25 = 5^2

, and

35 = 5 \times 7

. The only common prime factor is 5.

5

Apply the values to the fraction LCM formula.

\frac{96}{5}

Dividing the computed numerator LCM (96) by the computed denominator HCF (5) yields the final answer.

Key Concept

Calculating the Lowest Common Multiple (LCM) of fractional values to synchronize repeating events.

Estimated Time:1m 30s

Question 283Question →

An urban traffic control system manages three independent electronic toll gates. Based on their internal sensor loops, Gate A completes its automated scanning cycle every $\frac{12}{5}$ seconds, Gate B every $\frac{18}{25}$ seconds, and Gate C every $\frac{27}{35}$ seconds. If all three gates reset their cycles simultaneously, how many seconds will it take for all three gates to reset simultaneously again? Express your answer as an exact decimal.

Show answer & explanation

Answer: 21.6

Answer

21.6

Simultaneous repeating events require finding the Lowest Common Multiple (LCM) of their individual periods. For fractional periods, the rule is

\text{LCM} = \frac{\text{LCM of numerators}}{\text{HCF of denominators}}

. The LCM of

12

,

18

, and

27

is

108

, and the HCF of

5

,

25

, and

35

is

5

. Therefore, the LCM of the fractions is

\frac{108}{5}

, which evaluates precisely to

21.6

.

Step-by-Step Solution

1

Identify the mathematical operation required.

Calculate the LCM of the fractions

\frac{12}{5}

,

\frac{18}{25}

, and

\frac{27}{35}

.

Simultaneous cyclic events coincide at the Lowest Common Multiple of their individual time intervals.

2

Determine the LCM of the numerators.

The LCM of

12

,

18

, and

27

is

108

.

The formula for the LCM of fractions requires calculating the LCM of their respective numerators.

3

Determine the HCF of the denominators.

The HCF of

5

,

25

, and

35

is

5

.

The formula for the LCM of fractions requires calculating the HCF of their respective denominators.

4

Apply the fraction LCM formula.

\frac{108}{5} = 21.6

Dividing the LCM of numerators by the HCF of denominators yields the final LCM of the given fractions.

Key Concept

Lowest Common Multiple (LCM) of Fractions

Question 284Question →

A meteorological research station operates three automated data collection buoys in the ocean. The buoys transmit complete environmental telemetry packages to a satellite every $\frac{8}{21}$ hours, $\frac{11}{35}$ hours, and $\frac{15}{49}$ hours, respectively. If all three buoys initiate a transmission simultaneously at a given moment, how many hours will pass before they all initiate a transmission together again?

\frac{1}{735}

\frac{88}{49}

\frac{1320}{7}

\frac{1}{7}

Show answer & explanation

Answer: $\frac{1320}{7}$

Answer

The buoys will transmit together again after

\frac{1320}{7}

hours.

To find the time when all three buoys will transmit together again, we must calculate the Lowest Common Multiple (LCM) of the given time intervals. For fractions, the applicable formula is the LCM of the numerators divided by the highest common factor (HCF) of the denominators. The LCM of the numerators (8, 11, 15) is 1320. The HCF of the denominators (21, 35, 49) is 7. Therefore, the LCM of the fractions is 1320/7.

Step-by-Step Solution

1

Identify that the problem requires finding the Lowest Common Multiple (LCM) of the given fractional time intervals.

The mathematical goal is to compute

\text{LCM}(\frac{8}{21}, \frac{11}{35}, \frac{15}{49})

.

Simultaneous periodic events synchronize again at the least common multiple of their individual periods.

2

Apply the standard formula for the LCM of fractions.

The formula requires calculating

\frac{\text{LCM of numerators}}{\text{HCF of denominators}}

.

This is the established algebraic rule to compute the LCM for rational numbers.

3

Calculate the LCM of the numerators (8, 11, 15).

Since 8, 11, and 15 share no common prime factors, their LCM is simply their product:

8 \times 11 \times 15 = 1320

.

To find the smallest whole number that is evenly divisible by all three numerators.

4

Calculate the HCF of the denominators (21, 35, 49).

Breaking them down into prime factors gives

21=3\times7

,

35=5\times7

, and

49=7\times7

. The highest common factor is 7.

To find the largest whole number that divides evenly into all three denominators.

5

Combine the calculated numerator and denominator results.

\frac{1320}{7}

Substituting the calculated values back into the fraction LCM formula yields the final synchronized interval.

Key Concept

The lowest common multiple (LCM) of multiple fractions is calculated as the LCM of their numerators divided by the highest common factor (HCF) of their denominators.

Question 285Question →

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).

Drag items to arrange them in the correct order

Show answer & explanation

Answer

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...\%

, Fund Y is

8.155...\%

, Fund W is

8.151...\%

, and Fund Z is

8.142...\%

. Ordering these from highest to lowest yields Fund X, Fund Y, Fund W, then Fund Z.

Step-by-Step Solution

1

Convert Fund W's yield to extended decimal form for precise comparison.

Fund W =

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 =

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 =

8.1\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 =

\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...

(Fund X), then

8.155...

(Fund Y), then

8.151...

(Fund W), and lowest is

8.142...

(Fund Z).

Sorting by the tenths, hundredths, and thousandths digits establishes the definitive descending sequence.

Key Concept

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

Question 286Question →

A botanical research institute is preparing identical standardized nutrient doses for a plant growth experiment. They have three concentrated liquid nutrient solutions measuring $\frac{14}{3}$ liters, $\frac{35}{6}$ liters, and $\frac{21}{4}$ 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?

\frac{7}{12}

liters

\frac{1}{210}

liters

210

liters

7

liters

Show answer & explanation

Answer: $\frac{7}{12}$ liters

Answer

\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

\frac{14}{3}

,

\frac{35}{6}

, and

\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

\frac{7}{12}

liters per vial.

Step-by-Step Solution

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 = \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, 21

).

The prime factors are:

14 = 2 \times 7

;

35 = 5 \times 7

;

21 = 3 \times 7

. The highest common factor is

7

.

To form the numerator of our final fraction answer.

4

Calculate the LCM of the denominators (

3, 6, 4

).

The multiples of the denominators meet at

12

. (Since

12

is divisible by

3

,

6

, and

4

).

To form the denominator of our final fraction answer.

5

Substitute the calculated values back into the fraction HCF formula.

\frac{7}{12}

liters.

Provides the final vial capacity.

Key Concept

Highest Common Factor (HCF) of Fractions in Word Problems

Question 287Question →

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.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

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

1.375

cm, Panel X is approximately

1.428

cm, Panel W is

1.444...

cm, and Panel Y is exactly

1.450

cm. Sorting these values in ascending order yields Panel Z, Panel X, Panel W, Panel Y.

Step-by-Step Solution

1

Convert the thickness of Panel Z into a standard decimal format.

Panel Z =

1\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 =

\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.\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... < 1.450

. Therefore, Panel Z < Panel X < Panel W < Panel Y.

The stem requires ordering from thinnest to thickest.

Key Concept

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

Question 288Question →

A highway construction company is contracted to pave a new road. In the first month, they complete exactly $0.\overline{3}$ of the total project length. In the second month, they pave $0.625$ of the *remaining* length. If there are exactly $21$ kilometers left to pave in the third month to finish the project, what is the total length of the highway in kilometers?

84

80

63

504

Show answer & explanation

Answer: 84

Answer

The total length of the highway is 84 kilometers.

The total length of the highway is 84 km. In the first month,

1/3

is completed, leaving

2/3

. In the second month,

5/8

of this remaining

2/3

is completed, which means

(1 - 5/8) = 3/8

of the remainder is left for the third month. Calculating this relative to the total:

3/8 \times 2/3 = 1/4

. Since

1/4

of the total project equals 21 km, the total length must be

21 \times 4 = 84

km.

Step-by-Step Solution

1

Convert the decimal and recurring decimal into fractions for exact calculation.

0.\overline{3} = \frac{1}{3}

and

0.625 = \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.

1 - \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.

\frac{5}{8} \times \frac{2}{3} = \frac{10}{24} = \frac{5}{12}

of the total length.

The second month's work is

0.625

of the remaining

\frac{2}{3}

.

4

Find the final remaining fraction for the third month.

\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.

\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.

Key Concept

Converting recurring and terminating decimals to fractions and computing successive remainders.

Estimated Time:1m 30s

Question 289Question →

A specialized aerospace component is manufactured from a custom metal alloy weighing exactly $160$ kg. The alloy's composition by weight is $0.35$ aluminum and $\frac{5}{16}$ magnesium, with the remaining portion consisting entirely of titanium. What is the exact mass of titanium in this component?

Show answer & explanation

Answer: 54

Answer

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

\frac{5}{16}

to the decimal

0.3125

, the combined proportion of aluminum and magnesium is

0.35 + 0.3125 = 0.6625

. The titanium makes up the remaining portion of the whole, which is

1 - 0.6625 = 0.3375

. Finally, multiplying this decimal proportion by the total alloy mass of

160

kg yields exactly

54

kg.

Step-by-Step Solution

1

Convert the fractional part to a decimal.

\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 = 0.6625

This finds the total proportion of the alloy that is NOT titanium.

3

Calculate the proportion of titanium.

1 - 0.6625 = 0.3375

The sum of all material proportions in the whole alloy must equal exactly

1

.

4

Calculate the final mass of titanium.

0.3375 \times 160 = 54

Multiplying the component's proportion by the total mass gives the specific weight of that component.

Key Concept

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

Question 290Question →

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 $3.\overline{3}$ hours, Drone B returns every $2.5$ hours, and Drone C returns every $2.\overline{7}$ 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?

50

\frac{5}{18}

5

\frac{25}{9}

Show answer & explanation

Answer: $50$

Answer

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.

Step-by-Step Solution

1

Convert the recurring and terminating decimals into fractions in their simplest form.

Drone A:

3.\overline{3} = 3 \frac{1}{3} = \frac{10}{3}

. Drone B:

2.5 = \frac{5}{2}

. Drone C:

2.\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.

\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, 25

. Their prime factorizations are

10 = 2 \times 5

,

5 = 5

, and

25 = 5^2

. The

\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, 9

. Since they do not share any common prime factors, their

\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.

\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.

Key Concept

LCM of Fractions and Recurring Decimals

Estimated Time:2m 0s

Question 291Question →

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).

Drag items to arrange them in the correct order

Show answer & explanation

Answer

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).

Step-by-Step Solution

1

Convert the portion completed by Candidate P into a standard decimal format.

\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.\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.

0.8\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.

\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\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.

Key Concept

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

Question 292Question →

An agricultural research center is preparing experimental test plots. The facility has exactly $2 \frac{5}{8}$ kg, $2 \frac{11}{12}$ kg, and $3 \frac{1}{16}$ 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?

\frac{4}{735}

kg

1 \frac{3}{4}

kg

\frac{7}{48}

kg

183 \frac{3}{4}

kg

Show answer & explanation

Answer: $\frac{7}{48}$ kg

Answer

The correct packet capacity is

\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 (

\frac{21}{8}

,

\frac{35}{12}

,

\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)

is

7

, and the LCM of the denominators

(8, 12, 16)

is

48

. Therefore, the required capacity is

\frac{7}{48}

kg.

Step-by-Step Solution

1

Convert the mixed numbers representing the quantities of additives into improper fractions.

2 \frac{5}{8} = \frac{21}{8}

,

2 \frac{11}{12} = \frac{35}{12}

, and

3 \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)

is

7

. The LCM of the denominators

(8, 12, 16)

is

48

.

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

\frac{7}{48}

kg.

Dividing the numerator HCF by the denominator LCM yields the exact required capacity.

Key Concept

Calculating the Highest Common Factor (HCF) of fractions using the formula: HCF of numerators / LCM of denominators.

Estimated Time:1m 30s

Question 293Question →

A civil contractor is tasked with upgrading a rural highway. In the first month, the crew successfully upgrades $0.275$ of the total highway length. During the second month, they upgrade $\frac{5}{12}$ of the remaining length. If the crew has exactly $10.15$ kilometers left to upgrade in the third month to complete the project, what is the total length of the highway in kilometers?

Show answer & explanation

Answer: 24

Answer

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.

Step-by-Step Solution

1

Determine the fraction of the highway remaining after the first month.

\frac{29}{40}

of the total length remains.

The crew upgraded

0.275

of the total. Converting

0.275

to a fraction gives

\frac{275}{1000}

, which simplifies to

\frac{11}{40}

. Subtracting this from the whole gives

1 - \frac{11}{40} = \frac{29}{40}

.

2

Calculate the fraction of the total highway upgraded in the second month.

\frac{29}{96}

of the total length.

The crew upgraded

\frac{5}{12}

of the remaining length. Therefore, multiply the two fractions:

\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.

\frac{203}{480}

of the total length.

Subtract the second month's progress from the remainder after the first month:

\frac{29}{40} - \frac{29}{96}

. The least common multiple of

40

and

96

is

480

. Converting to common denominators gives

\frac{348}{480} - \frac{145}{480} = \frac{203}{480}

.

4

Set up the final equation and solve for the total highway length.

24

km

The remaining fraction equals the given physical distance of

10.15

km. So,

\frac{203}{480} \times \text{Total} = 10.15

. Solving for Total yields

\frac{10.15 \times 480}{203}

. Since

10.15 \div 203 = 0.05

, the Total is

0.05 \times 480 = 24

km.

Key Concept

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

Question 294Question →

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 $\frac{14}{15}$ of a millimeter.
- Silver thread is inserted every $\frac{21}{20}$ of a millimeter.
- Crimson thread is inserted every $\frac{35}{12}$ 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?

210

mm

\frac{7}{60}

mm

3.5

mm

7

mm

Show answer & explanation

Answer: $210$ mm

Answer

The three specialty threads will be inserted at the exact same position again for the first time at a distance of

210

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.

Step-by-Step Solution

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

\frac{14}{15}

,

\frac{21}{20}

, and

\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 \times 7

,

21 = 3 \times 7

, and

35 = 5 \times 7

. The lowest common multiple of these values is

2 \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 \times 5

,

20 = 2^2 \times 5

, and

12 = 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.

\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.

Key Concept

Calculating the Least Common Multiple (LCM) of fractions using the formula: LCM of fractions = (LCM of numerators) / (HCF of denominators).

Estimated Time:1m 30s

Question 295Question →

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: $0.345$ mm
- Foil Q: $\frac{11}{32}$ mm
- Foil R: $0.\overline{34}$ mm
- Foil S: $\frac{7}{20}$ mm

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

Drag items to arrange them in the correct order

Show answer & explanation

Answer

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...

), Foil Q (

0.34375

), Foil P (

0.345

), and Foil S (

0.35

). Comparing these exact decimal values digit-by-digit confirms that the ascending sequence must be R, Q, P, then S.

Step-by-Step Solution

1

Convert the fraction for Foil S to a decimal.

\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 \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.\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 < 0.35000

Aligning the decimals by their thousandths and ten-thousandths places reveals the exact ascending sequence.

Key Concept

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

Basic Numeracy

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