HCF and LCM Practice Questions — State PSC Exam

Question 1Question →

Three automatic signaling beacons in a traffic control system flash at regular time intervals of $\frac{4}{15}$ hours, $\frac{9}{20}$ hours, and $\frac{14}{35}$ hours, respectively. If all three beacons flash simultaneously at 12:00 noon, after how many hours will they all flash together again for the first time?

Show answer & explanation

Answer: 7.2

Answer

7.2 hours

To find when events occurring at fractional time intervals coincide, compute the LCM of the fractions. Crucially, each fraction must first be simplified to its lowest terms.

\frac{14}{35}

reduces to

\frac{2}{5}

. Taking

\frac{\text{LCM}(4, 9, 2)}{\text{HCF}(15, 20, 5)}

gives

\frac{36}{5} = 7.2

hours.

Step-by-Step Solution

1

Simplify all given fractional time intervals to their lowest terms.

The reduced fractions are

\frac{4}{15}

,

\frac{9}{20}

, and

\frac{2}{5}

.

The standard LCM formula for fractions

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

is mathematically valid only when all fractions are reduced to co-prime numerator-denominator pairs.

2

Calculate the LCM of the numerators.

\text{LCM}(4, 9, 2) = 36

.

The least common multiple of

4 = 2^2

,

9 = 3^2

, and

2 = 2^1

is

2^2 \times 3^2 = 36

.

3

Calculate the HCF of the denominators.

\text{HCF}(15, 20, 5) = 5

.

The highest common factor dividing

15

,

20

, and

5

is

5

.

4

Divide the numerator LCM by the denominator HCF to find the simultaneous flashing interval.

\frac{36}{5} = 7.2

hours.

The LCM of the fractional time intervals determines the minimum duration before all events synchronize.

Key Concept

LCM of Fractions with Mandatory Simplification

Estimated Time:2m 0s

Question 2Question →

Three storage vessels in a chemical laboratory contain $\frac{12}{35}$ litres, $\frac{18}{25}$ litres, and $\frac{24}{49}$ litres of different reagents. What is the maximum capacity (in litres) of a single measuring container that can be used to measure the exact quantity of reagent from any of the three vessels without leaving any remainder?

\frac{6}{1225}

litres

\frac{72}{1225}

litres

\frac{1225}{6}

litres

72

litres

Show answer & explanation

Answer: $\frac{6}{1225}$ litres

Answer

The maximum capacity of the measuring container is

\frac{6}{1225}

litres.

To find the largest measuring unit that can measure fractional quantities exactly, we need the HCF of those fractions. The HCF of fractions is defined as

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

. Calculating

\text{HCF}(12, 18, 24) = 6

and

\text{LCM}(35, 25, 49) = 1225

yields

\frac{6}{1225}

litres.

Step-by-Step Solution

1

Identify the mathematical rule required for measuring fractional quantities exactly.

To find the maximum capacity of a container that measures given fractional quantities an exact number of times, compute the HCF of the fractions:

\text{HCF}\left(\frac{a}{b}, \frac{c}{d}, \frac{e}{f}\right) = \frac{\text{HCF}(a, c, e)}{\text{LCM}(b, d, f)}

.

The measuring container's capacity must be a common divisor of all three fractional volumes.

2

Calculate the HCF of the numerators

12

,

18

, and

24

.

\text{HCF}(12, 18, 24) = 6

.

The prime factorizations are

12 = 2^2 \times 3

,

18 = 2 \times 3^2

, and

24 = 2^3 \times 3

. The highest common power of prime factors shared by all three numbers is

2^1 \times 3^1 = 6

.

3

Calculate the LCM of the denominators

35

,

25

, and

49

.

\text{LCM}(35, 25, 49) = 1225

.

The prime factorizations are

35 = 5 \times 7

,

25 = 5^2

, and

49 = 7^2

. The LCM takes the highest power of each prime:

5^2 \times 7^2 = 25 \times 49 = 1225

.

4

Substitute the calculated HCF of numerators and LCM of denominators into the fraction HCF formula.

\text{HCF} = \frac{6}{1225}

litres.

Dividing the HCF of the numerators by the LCM of the denominators gives the exact maximum volume.

Key Concept

HCF of Fractions

Estimated Time:2m 0s

Question 3Question →

Three automated robotic arms in an assembly line complete their respective sorting cycles in $\frac{14}{15}$ minutes, $\frac{21}{20}$ minutes, and $\frac{42}{25}$ minutes. If all three robotic arms start a sorting cycle simultaneously at 9:00:00 AM, after what minimum time interval will all three arms simultaneously begin a new sorting cycle again?

8

minutes

24

seconds

1.4

seconds

42

minutes

51

seconds

42.86

seconds

Show answer & explanation

Answer: $8$ minutes $24$ seconds

Answer

8

minutes

24

seconds

To find the time when events with fractional period lengths coincide again, we compute the LCM of the fraction time intervals. The formula for the LCM of fractions is

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

. Finding

\text{LCM}(14, 21, 42) = 42

and

\text{HCF}(15, 20, 25) = 5

yields

\frac{42}{5} = 8.4

minutes, which equals

8

minutes

24

seconds.

Step-by-Step Solution

1

Identify the mathematical operation required for simultaneous recurrence

The required minimum time interval is the Least Common Multiple (LCM) of the three cycle times:

\text{LCM}\left(\frac{14}{15}, \frac{21}{20}, \frac{42}{25}\right)

minutes.

Simultaneous future events occurring at periodic intervals require finding the LCM of the given time periods.

2

Apply the formula for the LCM of fractions

\text{LCM}\left(\frac{a}{b}, \frac{c}{d}, \frac{e}{f}\right) = \frac{\text{LCM}(a, c, e)}{\text{HCF}(b, d, f)}

For simplified fractions, the LCM is determined by dividing the LCM of all numerators by the HCF of all denominators.

3

Calculate the LCM of the numerators

(14, 21, 42)

14 = 2 \times 7

,

21 = 3 \times 7

,

42 = 2 \times 3 \times 7

. Thus,

\text{LCM}(14, 21, 42) = 2 \times 3 \times 7 = 42

.

The smallest positive integer divisible by 14, 21, and 42 is 42.

4

Calculate the HCF of the denominators

(15, 20, 25)

15 = 3 \times 5

,

20 = 2^2 \times 5

,

25 = 5^2

. Thus,

\text{HCF}(15, 20, 25) = 5

.

The greatest integer that divides 15, 20, and 25 without remainder is 5.

5

Combine the results and convert to minutes and seconds

\text{LCM} = \frac{42}{5} = 8.4

minutes. Converting

0.4

minutes to seconds gives

0.4 \times 60 = 24

seconds. Total time =

8

minutes

24

seconds.

Multiplying the fractional minute by 60 converts it into exact seconds.

Key Concept

LCM of Fractions

Question 4Question →

What is the Highest Common Factor (HCF) of the fractions $\frac{8}{15}$ and $\frac{14}{25}$ ?

\frac{2}{75}

\frac{56}{5}

\frac{2}{5}

\frac{56}{75}

Show answer & explanation

Answer: $\frac{2}{75}$

Answer

The Highest Common Factor (HCF) is

\frac{2}{75}

.

The correct answer is found by taking the Highest Common Factor (HCF) of the numerators (

8

and

14

), which is

2

, and dividing it by the Least Common Multiple (LCM) of the denominators (

15

and

25

), which is

75

. This yields exactly

\frac{2}{75}

.

Step-by-Step Solution

1

Identify the numerators and denominators of the given fractions.

The numerators are

8

and

14

. The denominators are

15

and

25

.

The formula for the HCF of fractions requires processing numerators and denominators separately.

2

Calculate the Highest Common Factor (HCF) of the numerators.

The HCF of

8

and

14

is

2

.

According to the formula, the numerator of the final answer must be the HCF of the original numerators.

3

Calculate the Least Common Multiple (LCM) of the denominators.

The LCM of

15

and

25

is

75

.

According to the formula, the denominator of the final answer must be the LCM of the original denominators.

4

Combine the results into a single fraction.

The resulting fraction is

\frac{2}{75}

.

The HCF of fractions is defined as

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

.

Key Concept

Highest Common Factor (HCF) of Fractions

Question 5Question →

Three automated data backup protocols in a cloud server execute network synchronization pulses at precise intervals. Protocol Alpha pulses every $\frac{18}{5}$ seconds, Protocol Beta every $\frac{24}{25}$ seconds, and Protocol Gamma every $\frac{36}{35}$ seconds. They all initiate their first pulse simultaneously at 08:00:00 AM. Assuming the initial pulse at 08:00:00 AM is excluded from the count, which of the following represents the total number of times all three protocols will pulse together in exactly $12$ minutes?

21000

10

50

51

Show answer & explanation

Answer: $50$

Answer

The data protocols will pulse together

50

times in the

12

-minute period.

To find when all three protocols pulse together, we must determine the Least Common Multiple (LCM) of their cycle times. The LCM of fractions is calculated by dividing the LCM of their numerators (

18, 24, 36

) by the HCF of their denominators (

5, 25, 35

). The LCM of

18, 24,

and

36

is

72

. The HCF of

5, 25,

and

35

is

5

. This gives a synchronized interval of

72/5 = 14.4

seconds. In

12

minutes (

720

seconds), the number of synchronized intervals is

720 \div 14.4 = 50

. Since the initial pulse is excluded as per the instructions, the correct answer remains

50

.

Step-by-Step Solution

1

Identify the mathematical concept required to find when all three protocols pulse simultaneously.

The time of simultaneous pulsing is the Least Common Multiple (LCM) of their individual cycle times:

\frac{18}{5}

,

\frac{24}{25}

, and

\frac{36}{35}

.

Simultaneous occurrence of periodic events always aligns at multiples common to all individual intervals.

2

Apply 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 to find the least common multiple for fractional values.

3

Calculate the LCM of the numerators (

18, 24, 36

) and the HCF of the denominators (

5, 25, 35

).

\text{LCM}(18, 24, 36) = 72

and

\text{HCF}(5, 25, 35) = 5

. Thus, the synchronized interval is

\frac{72}{5}

seconds, or

14.4

seconds.

Prime factorization gives

18 = 2 \times 3^2

,

24 = 2^3 \times 3

,

36 = 2^2 \times 3^2

(LCM =

2^3 \times 3^2 = 72

). For denominators,

5

is the largest common divisor.

4

Convert the given total time window into seconds.

12 \text{ minutes} = 12 \times 60 = 720 \text{ seconds}

.

Units must match the cycle interval (seconds) before division.

5

Calculate the total number of simultaneous pulses.

720 \div 14.4 = 50

. Since the initial pulse is excluded, the final count remains

50

.

Dividing the total time window by the synchronized interval yields the number of subsequent events.

Key Concept

Least Common Multiple (LCM) of fractions and its application to periodic simultaneous events.

Question 6Question →

A logistics manager is arranging shipping containers into equal stacks. When the containers are stacked in groups of $18$ , $24$ , or $32$ , there are always exactly $11$ containers left over. However, when they are stacked in groups of $35$ , there are no containers left over. What is the minimum total number of containers the manager could have?

Show answer & explanation

Answer: 875

Answer

875

The correct answer is derived by recognizing that the total number of containers must take the form of

LCM(18, 24, 32) \times k + 11

. The LCM is

288

, so the number is

288k + 11

. We then find the smallest integer

k

such that

(288k + 11)

is perfectly divisible by

35

. By evaluating the expression for sequential values of

k

(or using modular arithmetic,

8k \equiv 24 \pmod{35}

), we find that

k = 3

is the smallest valid multiplier. Substituting this back yields

288(3) + 11 = 875

.

Step-by-Step Solution

1

Calculate the Least Common Multiple (LCM) of the initial stack group sizes (

18

,

24

, and

32

).

The prime factorizations are

18 = 2 \times 3^2

,

24 = 2^3 \times 3

, and

32 = 2^5

. The LCM is

2^5 \times 3^2 = 32 \times 9 = 288

.

The base cycle for the stacks without remainders requires finding the smallest number perfectly divisible by all three group sizes.

2

Express the total number of containers algebraically.

Let the total number of containers be

N

. We can write

N = 288k + 11

, where

k

is a positive integer.

Since there is always a remainder of

11

containers when divided by these numbers, the total must be

11

more than a multiple of their LCM.

3

Apply the secondary divisibility condition.

The problem states that

N

is exactly divisible by

35

. Therefore,

288k + 11 \equiv 0 \pmod{35}

.

When the containers are grouped by

35

, there is no remainder.

4

Solve the congruence to find the smallest integer

k

.

First, simplify

288 \pmod{35}

: since

35 \times 8 = 280

, we have

288 \equiv 8 \pmod{35}

. The equation becomes

8k + 11 \equiv 0 \pmod{35}

, or

8k \equiv -11 \equiv 24 \pmod{35}

. Dividing both sides by

8

gives

k = 3

.

Finding the smallest valid positive integer

k

will yield the minimum total number of containers.

5

Calculate the final number of containers using

k = 3

.

N = 288(3) + 11 = 864 + 11 = 875

.

Substitute the multiplier back into the original algebraic expression.

Key Concept

Solving for an unknown quantity using the Least Common Multiple (LCM) combined with constant remainder logic and a secondary divisibility condition.

Alternative Method

Instead of using modular arithmetic to solve

288k + 11 \equiv 0 \pmod{35}

, you can manually test integer values for

k

in the formula

N = 288k + 11

. For

k=1

,

N = 288(1) + 11 = 299

(

299 \div 35

leaves remainder

19

). For

k=2

,

N = 288(2) + 11 = 587

(

587 \div 35

leaves remainder

27

). For

k=3

,

N = 288(3) + 11 = 875

(

875 \div 35 = 25

with no remainder). Therefore,

875

is the smallest valid number.

Estimated Time:2m 30s

Question 7Question →

An industrial cutting machine is programmed to divide a rectangular metallic sheet measuring $\frac{24}{5}$ meters in length and $\frac{16}{3}$ meters in width into identical square plates. If the square plates must be of the maximum possible size without leaving any wasted material, what is the area of each resulting square plate?

\frac{64}{225} \text{ m}^2

2304 \text{ m}^2

64 \text{ m}^2

\frac{256}{25} \text{ m}^2

Show answer & explanation

Answer: $\frac{64}{225} \text{ m}^2$

Answer

The area of each resulting square plate is \frac{64}{225} \text{ m}^2.

To cut the rectangle into identical squares of maximum size without waste, the side length of the square must be the Highest Common Factor (HCF) of the rectangle's dimensions. Using the fraction formula, the HCF of

\frac{24}{5}

and

\frac{16}{3}

is

\frac{\text{HCF}(24, 16)}{\text{LCM}(5, 3)} = \frac{8}{15}

meters. The area of the square is the side length squared, which is

(\frac{8}{15})^2 = \frac{64}{225} \text{ m}^2

.

Step-by-Step Solution

1

Determine the required mathematical operation for finding the maximum square size.

The side length of the square must be the Highest Common Factor (HCF) of the length (24/5) and width (16/3).

To divide a rectangle into identical squares without waste, the square's side must perfectly divide both dimensions. The 'maximum size' implies finding the greatest common divisor.

2

Apply the rule for finding the HCF of fractions.

HCF(a/b, c/d) = HCF(a, c) / LCM(b, d). Thus, HCF(24/5, 16/3) = HCF(24, 16) / LCM(5, 3).

The standard formula for the HCF of fractions requires taking the HCF of the numerators divided by the LCM of the denominators.

3

Calculate the specific HCF and LCM values.

HCF(24, 16) = 8 and LCM(5, 3) = 15. The side length of the square is \frac{8}{15} meters.

8 is the largest integer dividing both 24 and 16. 15 is the smallest integer divisible by both 5 and 3.

4

Calculate the area of the square.

Area = (\frac{8}{15})^2 = \frac{64}{225} \text{ m}^2.

The question specifically asks for the area of the plate, which is the square of its side length.

Key Concept

HCF of Fractions and Geometric Application

Question 8Question →

Four navigation buoys in a harbor flash at regular intervals of $3 \frac{1}{3}$ minutes, $4 \frac{1}{6}$ minutes, $5 \frac{5}{9}$ minutes, and $7 \frac{1}{2}$ minutes, respectively. If all four buoys flash simultaneously at exactly 12:00 noon, how many additional times will they all flash together again within the next $12$ hours?

Show answer & explanation

Answer: 4

Answer

4

The correct answer is derived by finding the Least Common Multiple of the fractional intervals to determine the duration between simultaneous flashes (150 minutes). Then, by converting the 12-hour period into 720 minutes, dividing 720 by 150 yields 4.8, indicating the buoys will flash together 4 additional times within that exact timeframe.

Step-by-Step Solution

1

Convert the mixed numbers into improper fractions.

The intervals are

\frac{10}{3}

,

\frac{25}{6}

,

\frac{50}{9}

, and

\frac{15}{2}

minutes.

Mathematical operations involving LCM and HCF of fractions require them to be in proper or improper fractional forms, rather than mixed numbers.

2

Determine the mathematical formula required for simultaneous events.

We must calculate the Least Common Multiple (LCM) of the intervals using the formula:

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

.

Simultaneous cyclic events synchronize at the lowest common multiple of their individual intervals.

3

Calculate the LCM of the numerators.

The numerators are

10, 25, 50

, and

15

. Their LCM is

150

.

150

is the smallest integer divisible by all four numerators (

150 = 10 \times 15 = 25 \times 6 = 50 \times 3 = 15 \times 10

).

4

Calculate the HCF of the denominators.

The denominators are

3, 6, 9

, and

2

. Their HCF is

1

.

The numbers

2

and

3

are coprime (they share no common factors greater than

1

), forcing the overall HCF of the set to be

1

.

5

Calculate the exact time interval between simultaneous flashes.

The LCM of the intervals is

\frac{150}{1} = 150

minutes.

Applying the results from steps 3 and 4 directly into the fractional LCM formula.

6

Determine how many times this 150-minute interval fits into the 12-hour window.

12 \text{ hours} = 720 \text{ minutes}

.

\lfloor \frac{720}{150} \rfloor = \lfloor 4.8 \rfloor = 4

.

Dividing the total available time by the simultaneous interval length gives the exact count of additional flashes.

Key Concept

LCM of fractions in simultaneous cyclic events

Question 9Question →

Three different cryptographic algorithms in a secure server refresh their encryption keys at regular intervals. Algorithm X refreshes every $\frac{8}{3}$ milliseconds, Algorithm Y every $\frac{10}{9}$ milliseconds, and Algorithm Z every $\frac{14}{5}$ milliseconds. They all start a refresh cycle simultaneously. Let $M$ be the minimum number of milliseconds until they all start a refresh cycle together again. Let $H$ be the Highest Common Factor (HCF) of their respective refresh intervals in milliseconds. What is the value of $M \div H$ ?

\frac{1}{6300}

140

78400

6300

Show answer & explanation

Answer: $6300$

Answer

The value of

M \div H

is

6300

.

The value

M

requires finding the LCM of the fractions, which represents the synchronization time.

\text{LCM}(\frac{8}{3}, \frac{10}{9}, \frac{14}{5}) = \frac{\text{LCM}(8, 10, 14)}{\text{HCF}(3, 9, 5)} = \frac{280}{1} = 280

. The value

H

requires finding the HCF of the fractions.

\text{HCF}(\frac{8}{3}, \frac{10}{9}, \frac{14}{5}) = \frac{\text{HCF}(8, 10, 14)}{\text{LCM}(3, 9, 5)} = \frac{2}{45}

. Finally, dividing

M

by

H

gives

280 \div \frac{2}{45} = 280 \times \frac{45}{2} = 140 \times 45 = 6300

.

Step-by-Step Solution

1

Determine the synchronization time

M

by finding the LCM of the three fractional intervals.

M = \text{LCM}(\frac{8}{3}, \frac{10}{9}, \frac{14}{5})

Simultaneous repeating events synchronize at the Lowest Common Multiple of their individual cycle times.

2

Calculate the LCM using the fraction rule:

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

.

Numerators

(8, 10, 14)

have an LCM of

280

. Denominators

(3, 9, 5)

have an HCF of

1

. Thus,

M = \frac{280}{1} = 280

.

This is the mathematical formula required for finding the LCM of rational numbers.

3

Determine the Highest Common Factor

H

of the three fractional intervals.

H = \text{HCF}(\frac{8}{3}, \frac{10}{9}, \frac{14}{5})

The question explicitly requests the HCF of the three given refresh times.

4

Calculate the HCF using the fraction rule:

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

.

Numerators

(8, 10, 14)

have an HCF of

2

. Denominators

(3, 9, 5)

have an LCM of

45

. Thus,

H = \frac{2}{45}

.

This is the mathematical formula required for finding the HCF of rational numbers.

5

Calculate the final required ratio

M \div H

.

280 \div \frac{2}{45} = 280 \times \frac{45}{2} = 140 \times 45 = 6300

.

This satisfies the final computational step requested by the problem stem.

Key Concept

Calculating the Lowest Common Multiple (LCM) and Highest Common Factor (HCF) for fractions using their specific formulas.

Question 10Question →

A wholesale merchant has three batches of specialty syrups measuring $\frac{63}{5}$ liters, $\frac{72}{25}$ liters, and $\frac{81}{10}$ liters in volume. To distribute them in uniform sample sizes, the merchant wants to repackage all the syrup into identical smaller bottles filled exactly to the brim, ensuring no syrup is left over from any of the original batches. What is the maximum possible capacity (in liters) of each sample bottle?

Show answer & explanation

Answer: 0.18

Answer

The maximum possible capacity of each sample bottle is 0.18 liters.

To find the maximum capacity of uniform bottles that can hold the syrup from any batch without leftover, we must find the Highest Common Factor (HCF) of the respective volumes. For fractions, the rule is

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

. The HCF of the numerators (

63, 72, 81

) is

9

, and the LCM of the denominators (

5, 25, 10

) is

50

. The resulting maximum capacity is

\frac{9}{50}

, which equals

0.18

liters.

Step-by-Step Solution

1

Determine the mathematical operation required.

Calculate the HCF of

\frac{63}{5}

,

\frac{72}{25}

, and

\frac{81}{10}

.

Finding the largest equal container size that leaves no remainder requires computing the Highest Common Factor.

2

Apply the rule for finding the HCF of fractions.

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

This standard formula correctly identifies the greatest common divisor for a set of rational numbers.

3

Find the HCF of the numerators.

\text{HCF}(63, 72, 81) = 9

The largest integer that perfectly divides 63, 72, and 81 is 9.

4

Find the LCM of the denominators.

\text{LCM}(5, 25, 10) = 50

The smallest integer that is a multiple of 5, 25, and 10 is 50.

5

Compute the final fraction and convert it to a decimal.

\frac{9}{50} = 0.18

Dividing the HCF of the numerators by the LCM of the denominators yields the exact capacity in liters.

Key Concept

HCF and LCM of fractions

Question 11Question →

An interior designer is planning to cover the lengths of three different hallways with square marble tiles. The lengths of the three hallways are $\frac{15}{4}$ meters, $\frac{25}{6}$ meters, and $\frac{35}{8}$ meters. To avoid cutting any marble tiles, the designer wants to use the largest possible square tiles that can perfectly measure these lengths.

What is the maximum possible side length of each square tile?

\frac{5}{24}

meters

\frac{525}{2}

meters

\frac{5}{2}

meters

\frac{24}{5}

meters

Show answer & explanation

Answer: $\frac{5}{24}$ meters

Answer

\frac{5}{24}

meters

The maximum possible side length that perfectly measures all three hallway lengths is found by calculating the Highest Common Factor (HCF) of the three fractions. According to the mathematical rule for fractions, this is the HCF of the numerators (

15, 25, 35

) divided by the LCM of the denominators (

4, 6, 8

), which yields

\frac{5}{24}

meters.

Step-by-Step Solution

1

Identify the mathematical operation required.

Calculate the Highest Common Factor (HCF) of the three fractions to find the maximum possible side length.

The tiles must perfectly measure the lengths without being cut, meaning the tile length must be a common factor. 'Largest possible' indicates the Highest Common Factor.

2

Recall the formula for finding the HCF of fractions.

HCF of fractions =

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

This is the standard rule for determining the highest common divisor among fractional values.

3

Calculate the HCF of the numerators.

Numerators are

15

,

25

, and

35

. Their HCF is

5

.

The largest integer that perfectly divides

15

,

25

, and

35

is

5

.

4

Calculate the LCM of the denominators.

Denominators are

4

,

6

, and

8

. Their LCM is

24

.

The smallest integer that is a multiple of

4

,

6

, and

8

is

24

.

5

Apply the results to the formula.

\frac{5}{24}

meters.

Substituting the computed values into the fraction formula provides the final answer.

Key Concept

HCF of fractions

Question 12Question →

An inventory analyst is auditing two specific electronic components in a warehouse. He calculates that the product of their exact unit quantities is $32,175$ , and the Highest Common Factor (HCF) of these two quantities is $15$ . If there are more than $100$ units of each component currently in stock, what is the total combined quantity of both components?

Show answer & explanation

Answer: 360

Answer

The correct total combined quantity is 360 units.

By representing the numbers as

15x

and

15y

, we determine that

15x \times 15y = 32,175

, which simplifies to

xy = 143

. Factoring

143

into co-prime pairs gives

(1, 143)

and

(11, 13)

. Multiplying these by the HCF (

15

) gives potential quantities of

(15, 2145)

and

(165, 195)

. Only the pair

(165, 195)

satisfies the condition that both quantities must be greater than

100

. Their sum is

165 + 195 = 360

.

Step-by-Step Solution

1

Express the two unknown quantities mathematically using their HCF.

Let the quantities be

15x

and

15y

, where

x

and

y

are co-prime integers.

Since the HCF of the two numbers is

15

, both numbers must be multiples of

15

. Their remaining factors (

x

and

y

) cannot share any common factors other than

1

.

2

Formulate an equation using the given product of the quantities.

15x \times 15y = 32,175

The problem states the product of the two component quantities is equal to

32,175

.

3

Solve the equation for the product of the co-prime variables

x

and

y

.

225xy = 32,175 \Rightarrow xy = \frac{32,175}{225} = 143

Isolating

xy

simplifies the problem to finding two co-prime factors that multiply to

143

.

4

Identify all co-prime factor pairs of

143

.

The co-prime pairs are

(1, 143)

and

(11, 13)

.

We must list all integer pairs that multiply to

143

and verify they share no common divisors.

5

Calculate the possible original quantities and apply the boundary constraints.

Pair 1 gives

(15, 2145)

. Pair 2 gives

(165, 195)

. Because the problem states both quantities are

>100

, we must choose

(165, 195)

.

The constraint strictly eliminates the first pair, as

15

is not greater than

100

.

6

Sum the valid quantities.

165 + 195 = 360

The question asks for the total combined quantity of both components.

Key Concept

The relationship between the Highest Common Factor (HCF) and the product of two numbers, utilizing co-prime factor pairs.

Question 13Question →

A digital signal processing unit transmits three distinct telemetry data packets in continuous periodic cycles. The transmission cycles for the three packets take $\frac{14}{15}$ seconds, $\frac{7}{5}$ seconds, and $\frac{63}{20}$ seconds, respectively. If all three packets are transmitted simultaneously at the beginning of a diagnostic test, after how many seconds will all three packets be transmitted together again for the first time?

\frac{126}{5}

\frac{7}{60}

\frac{21}{10}

126

Show answer & explanation

Answer: $\frac{126}{5}$

Answer

\frac{126}{5}

To determine when all packets will transmit simultaneously again, we calculate the Least Common Multiple (LCM) of their cycle periods. The LCM of fractions is defined as the LCM of their numerators divided by the HCF of their denominators. Given the numerators (14, 7, 63) and denominators (15, 5, 20), we find LCM(14, 7, 63) = 126 and HCF(15, 5, 20) = 5. Dividing these yields the exact time of

\frac{126}{5}

seconds.

Step-by-Step Solution

1

Identify the mathematical operation required.

The problem asks for the next simultaneous transmission of periodic cycles, which requires finding the Least Common Multiple (LCM) of the given time intervals:

\frac{14}{15}

,

\frac{7}{5}

, and

\frac{63}{20}

.

Repeating simultaneous events align at time intervals that are common multiples of their individual periods. The first alignment corresponds to the least common multiple.

2

Apply the rule for finding the LCM of fractions.

The correct mathematical formula is

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

.

This is the required procedure for finding the LCM of rational numbers expressed as simplified fractions.

3

Determine the LCM of the numerators.

The numerators are

14

,

7

, and

63

. Their prime factorizations are

14 = 2 \times 7

,

7 = 7

, and

63 = 3^2 \times 7

. Thus,

\text{LCM}(14, 7, 63) = 2 \times 3^2 \times 7 = 126

.

The LCM must consist of the highest power of all prime factors present among the numerators.

4

Determine the HCF of the denominators.

The denominators are

15

,

5

, and

20

. Their prime factorizations are

15 = 3 \times 5

,

5 = 5

, and

20 = 2^2 \times 5

. The

\text{Highest Common Factor}(15, 5, 20) = 5

.

The HCF is the largest prime factor shared by all the denominators.

5

Calculate the final LCM of the transmission cycles.

Divide the LCM of the numerators by the HCF of the denominators to obtain exactly

\frac{126}{5}

seconds.

Applying the values derived in the previous steps to the fraction formula yields the exact coincidence time.

Key Concept

Calculating the LCM of fractions to synchronize periodic events.

Question 14Question →

An environmental agency uses three autonomous drones to patrol a protected reserve. The drones fly on continuous looping routes. Drone X completes one full route in $\frac{42}{15}$ hours, Drone Y completes a route in $\frac{35}{20}$ hours, and Drone Z completes a route in $\frac{63}{30}$ hours.

If all three drones depart simultaneously from the base station, after how many hours will they all meet at the base station again for the first time?

Show answer & explanation

Answer: 42

Answer

42

The drones will meet again at a time that is a common multiple of all their individual cycle times. The very first time this happens is represented by the Least Common Multiple (LCM). When calculating the LCM of fractions, it is mathematically required to reduce them to their simplest terms first:

\frac{14}{5}

,

\frac{7}{4}

, and

\frac{21}{10}

. Using the correct formula,

\frac{\text{LCM}(14, 7, 21)}{\text{HCF}(5, 4, 10)}

, we get

\frac{42}{1} = 42

hours.

Step-by-Step Solution

1

Identify the mathematical operation required.

The convergence time is the Least Common Multiple (LCM) of the three cycle times.

The drones will meet again at a time that is a common multiple of all their individual route completion times.

2

Simplify the given fractions to their lowest terms.

\frac{42}{15} \rightarrow \frac{14}{5}

;

\frac{35}{20} \rightarrow \frac{7}{4}

;

\frac{63}{30} \rightarrow \frac{21}{10}

The formula for the LCM of fractions requires all fractions to be in their simplest form to yield the correct result.

3

Calculate the LCM of the simplified numerators.

\text{LCM}(14, 7, 21) = 42

The numerator of the resulting fraction must be divisible by all original numerators.

4

Calculate the HCF of the simplified denominators.

\text{HCF}(5, 4, 10) = 1

The denominator of the resulting fraction must evenly divide all original denominators.

5

Compute the final fraction.

\frac{42}{1} = 42

hours

Applying the formula

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

gives the exact time of the next simultaneous meeting.

Key Concept

Calculating the Least Common Multiple (LCM) of fractions, emphasizing the critical prerequisite of simplifying the fractions first.

Estimated Time:2m 30s

Question 15Question →

A boutique chocolatier is preparing special gift assortments. They have three large blocks of premium cocoa weighing $\frac{21}{2}$ kg, $\frac{35}{4}$ kg, and $\frac{49}{8}$ kg. They must mold these blocks into the largest possible identical solid chocolate bars such that no cocoa is left over from any of the three blocks. What should be the exact weight of each chocolate bar?

\frac{7}{8}

kg

\frac{7}{2}

kg

\frac{735}{2}

kg

7

kg

Show answer & explanation

Answer: $\frac{7}{8}$ kg

Answer

The exact weight of each chocolate bar should be

\frac{7}{8}

kg.

To find the maximum identical weight that can perfectly divide all three cocoa blocks, we must calculate the Highest Common Factor (HCF) of the three fractions. The formula dictates finding the HCF of the numerators (21, 35, 49), which is 7, and dividing it by the LCM of the denominators (2, 4, 8), which is 8. This results in the correct weight of 7/8 kg.

Step-by-Step Solution

1

Identify the mathematical operation required.

The problem asks for the 'largest possible identical' divisions of the blocks with no remainders, meaning we must find the Highest Common Factor (HCF) of the given fractional weights.

HCF provides the maximum uniform size that can perfectly divide a given set of quantities.

2

Recall the formula for finding the HCF of fractions.

HCF of fractions = (HCF of Numerators) / (LCM of Denominators).

This is the standard mathematical rule for calculating the greatest common divisor for fractional values.

3

Calculate the HCF of the numerators: 21, 35, and 49.

The highest common factor for 21, 35, and 49 is 7.

21 = 3 × 7; 35 = 5 × 7; 49 = 7 × 7. The greatest shared prime factor is 7.

4

Calculate the LCM of the denominators: 2, 4, and 8.

The lowest common multiple for 2, 4, and 8 is 8.

8 is a multiple of both 2 and 4, making it the smallest common denominator.

5

Apply the calculated values to the fraction formula.

\frac{7}{8}

kg.

Combining the calculated numerator HCF and denominator LCM yields the final maximum weight.

Key Concept

HCF of Fractions

Question 16Question →

A pharmaceutical laboratory uses three automated dispensers to add chemical reagents into a continuous reaction vessel. Dispenser X adds a drop every $\frac{14}{9}$ seconds, Dispenser Y every $\frac{35}{6}$ seconds, and Dispenser Z every $\frac{49}{12}$ seconds. If all three dispensers release a drop simultaneously, what is the minimum time interval (in seconds) until they all release a drop together again?

\frac{36}{7}

\frac{490}{3}

\frac{7}{36}

\frac{245}{18}

Show answer & explanation

Answer: $\frac{490}{3}$

Answer

The minimum time interval is

\frac{490}{3}

seconds.

The correct answer is found by applying the LCM formula for fractions. Since the events repeat over time, their next simultaneous occurrence happens at a time that is a common multiple of all three periods. By calculating the LCM of the numerators (

14, 35, 49 \rightarrow 490

) and dividing it by the HCF of the denominators (

9, 6, 12 \rightarrow 3

), we get

\frac{490}{3}

seconds.

Step-by-Step Solution

1

Identify the mathematical operation required.

We need to find the Lowest Common Multiple (LCM) of the three fractional time intervals.

Simultaneous periodic events repeat exactly at the LCM of their individual periods.

2

State the formula for the LCM of fractions.

LCM of fractions =

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

.

This standard formula ensures the resulting value is a multiple of each original fraction.

3

Calculate the LCM of the numerators.

The numerators are

14

,

35

, and

49

. Their prime factorizations are

14 = 2 \times 7

,

35 = 5 \times 7

, and

49 = 7^2

. The LCM is

2 \times 5 \times 7^2 = 490

.

The LCM must include the highest power of all prime factors present in the numbers.

4

Calculate the HCF of the denominators.

The denominators are

9

,

6

, and

12

. Their prime factorizations are

9 = 3^2

,

6 = 2 \times 3

, and

12 = 2^2 \times 3

. The highest common factor is

3

.

The HCF is the largest positive integer that divides each of the numbers without leaving a remainder.

5

Apply the formula to find the final LCM.

LCM =

\frac{490}{3}

.

Substituting the calculated numerator LCM and denominator HCF into the fraction formula.

Key Concept

Lowest Common Multiple (LCM) of fractions for periodic events

Estimated Time:1m 30s

Question 17Question →

A city's public transport network features three distinct tram lines that operate on continuous circular routes departing from a central station. Tram Line 1 completes its route every $\frac{45}{4}$ minutes. Tram Line 2 completes its route every $\frac{25}{2}$ minutes, and Tram Line 3 takes $\frac{75}{8}$ minutes per loop. If all three trams depart from the central station simultaneously, how many minutes will it take for them to depart together again for the first time?

Show answer & explanation

Answer: 112.5

Answer

It will take 112.5 minutes for all three trams to depart together again.

The correct answer is found by taking the Least Common Multiple of the fractional times. By finding the LCM of the numerators (225) and dividing it by the Highest Common Factor of the denominators (2), we get 225/2, which evaluates to exactly 112.5 minutes.

Step-by-Step Solution

1

Determine the mathematical operation required to find when the events will synchronize.

Identify the need to calculate the Least Common Multiple (LCM) of the fractions

\frac{45}{4}

,

\frac{25}{2}

, and

\frac{75}{8}

.

The trams will meet again at a time that is a common multiple of their individual loop durations. The 'first time' indicates the least common multiple is needed.

2

Apply the rule for calculating the LCM of fractional values.

Use the formula:

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

.

To synchronize fractional frequencies, the numerators must reach a common multiple while strictly maintaining the largest common baseline unit defined by the denominators.

3

Calculate the LCM of the numerators: 45, 25, and 75.

The LCM of 45, 25, and 75 is 225.

Prime factorization:

45 = 3^2 \times 5

;

25 = 5^2

;

75 = 3 \times 5^2

. Taking the highest powers gives

3^2 \times 5^2 = 9 \times 25 = 225

.

4

Calculate the HCF of the denominators: 4, 2, and 8.

The HCF of 4, 2, and 8 is 2.

2 is the largest integer that can divide 4, 2, and 8 without leaving a remainder.

5

Compute the final synchronized time.

Divide the LCM of numerators by the HCF of denominators:

\frac{225}{2} = 112.5

.

Applying the values to the fraction LCM formula yields the exact time in minutes.

Key Concept

Calculating the Least Common Multiple (LCM) for fractions to solve simultaneous event problems.

Question 18Question →

A metal fabrication company has three long copper rods measuring $\frac{54}{5}$ meters, $\frac{81}{10}$ meters, and $\frac{108}{25}$ meters. The company needs to cut all three rods into smaller segments of equal length such that no copper material is left over. To minimize the total number of segments, what is the maximum possible length of each individual segment?

\frac{162}{25}

meters

\frac{324}{5}

meters

\frac{27}{50}

meters

\frac{27}{5}

meters

Show answer & explanation

Answer: $\frac{27}{50}$ meters

Answer

The maximum possible length of each segment is 27/50 meters.

The maximum possible length for the equal segments is found by calculating the HCF of the given fractions. Using the mathematical rule for fractions, HCF = HCF(numerators) / LCM(denominators). The HCF of 54, 81, and 108 is 27. The LCM of 5, 10, and 25 is 50. Therefore, the maximum length is 27/50 meters.

Step-by-Step Solution

1

Identify the mathematical operation required to find the maximum possible equal length.

The problem requires finding the Highest Common Factor (HCF) of the three fractional lengths.

The rods must be cut into equal pieces without wastage, and the length of each piece must be maximized to minimize the number of segments.

2

Recall the formula for finding the HCF of fractions.

HCF of fractions = (HCF of numerators) / (LCM of denominators).

This formula is necessary to accurately compute the greatest common divisor of non-integer values.

3

Calculate the HCF of the numerators: 54, 81, and 108.

The factors give 54 = 2 × 27, 81 = 3 × 27, and 108 = 4 × 27. The HCF is 27.

We need the greatest integer that divides all the numerators evenly.

4

Calculate the LCM of the denominators: 5, 10, and 25.

The multiples of 25 are 25, 50, 75, etc. Since 50 is divisible by both 5 and 10, the LCM is 50.

We need the smallest common multiple for the denominators to complete the fraction formula.

5

Construct the final fraction.

The maximum length is 27/50 meters.

Dividing the computed numerator HCF by the denominator LCM gives the correct value.

Key Concept

HCF of Fractions

Question 19Question →

An event organizer is arranging chairs for a large conference. When the chairs are arranged in rows of $18$ , $24$ , or $36$ , there are always exactly $5$ chairs left over. However, when the chairs are arranged in rows of $13$ , all chairs are perfectly accommodated with none left over. What is the minimum possible total number of chairs the organizer has?

Show answer & explanation

Answer: 221

Answer

The minimum possible total number of chairs is 221.

The correct answer is derived by first establishing that any number leaving a remainder of

5

when divided by

18

,

24

, and

36

must be of the form

72k + 5

, where

72

is the LCM of the divisors. By systematically checking values of

k

, we find that

k=3

is the smallest integer that makes the expression

(72k + 5)

perfectly divisible by

13

, resulting in

72(3) + 5 = 221

.

Step-by-Step Solution

1

Set up the conditions for the total number of chairs mathematically.

Let the total number of chairs be

N

. We are given

N \equiv 5 \pmod{18}

,

N \equiv 5 \pmod{24}

,

N \equiv 5 \pmod{36}

, and

N \equiv 0 \pmod{13}

.

Translating the word problem into modular arithmetic helps systematically apply the concepts of LCM and divisibility.

2

Find the Least Common Multiple (LCM) of the first set of divisors.

The divisors are

18

,

24

, and

36

. Their prime factorizations are

18 = 2 \times 3^2

,

24 = 2^3 \times 3

, and

36 = 2^2 \times 3^2

. The LCM is

2^3 \times 3^2 = 8 \times 9 = 72

.

Any number that leaves the same remainder when divided by multiple divisors must be a multiple of their LCM plus that remainder.

3

Express

N

using the LCM and the common remainder.

Since

N

leaves a remainder of

5

when divided by

18

,

24

, or

36

, we can write

N = 72k + 5

, where

k

is a non-negative integer (

k = 0, 1, 2, \dots

).

This general formula captures all possible numbers of chairs that satisfy the first condition.

4

Apply the final divisibility condition to find

k

.

We require

N

to be perfectly divisible by

13

, meaning

72k + 5 \equiv 0 \pmod{13}

.

This guarantees the solution satisfies the second condition where arranging chairs in rows of 13 leaves no remainder.

5

Simplify the congruence modulo 13 and solve for

k

.

Divide

72

by

13

to find the remainder:

72 = 13 \times 5 + 7

. So,

72k \equiv 7k \pmod{13}

. The equation becomes

7k + 5 \equiv 0 \pmod{13}

. Testing values for

k

: if

k=1

,

7(1)+5=12

(not divisible); if

k=2

,

7(2)+5=19

(not divisible); if

k=3

,

7(3)+5=26

(divisible by

13

, since

26 = 13 \times 2

). The smallest valid

k

is

3

.

Finding the smallest non-negative integer

k

ensures we find the minimum possible number of chairs.

6

Calculate the final value of

N

.

N = 72(3) + 5 = 216 + 5 = 221

.

Substituting

k=3

back into our general formula gives the final answer.

Key Concept

Solving simultaneous remainder and divisibility conditions using the Least Common Multiple (LCM).

Question 20Question →

A gourmet coffee roaster has three batches of rare coffee beans weighing $\frac{15}{2}$ kg, $\frac{25}{4}$ kg, and $\frac{35}{8}$ kg respectively. The roaster wants to package all the beans into smaller, equal-sized bags such that each bag contains the maximum possible weight of coffee beans without mixing the batches, and no beans are left over. What should be the weight of each bag?

\frac{525}{2}

kg

\frac{525}{8}

kg

\frac{5}{8}

kg

\frac{5}{2}

kg

Show answer & explanation

Answer: $\frac{5}{8}$ kg

Answer

The weight of each bag should be

\frac{5}{8}

kg.

The problem asks for the maximum possible equal capacity for the bags, which requires calculating the Highest Common Factor (HCF) of the given fractional weights. The HCF of a set of fractions is found by calculating the HCF of their numerators (

15

,

25

,

35

) and dividing it by the LCM of their denominators (

2

,

4

,

8

). The HCF of the numerators is

5

, and the LCM of the denominators is

8

. Therefore, the correct weight is

\frac{5}{8}

kg.

Step-by-Step Solution

1

Determine the mathematical operation required.

We need to find the Highest Common Factor (HCF) of the given fractional weights.

The bags must be of equal size, hold the maximum possible weight, and leave no remainder, which matches the definition of HCF.

2

Recall the formula for finding the HCF of fractions.

The formula is: (HCF of numerators) / (LCM of denominators).

This formula ensures the resulting fraction correctly divides all the given fractions into integers.

3

Calculate the HCF of the numerators.

The numerators are

15

,

25

, and

35

. Their HCF is

5

.

5

is the largest integer that divides

15

,

25

, and

35

without leaving a remainder.

4

Calculate the LCM of the denominators.

The denominators are

2

,

4

, and

8

. Their LCM is

8

.

8

is the smallest integer that is a multiple of

2

,

4

, and

8

.

5

Combine the results to find the HCF of the fractions.

The final HCF is

\frac{5}{8}

.

Substituting the calculated HCF and LCM into the fraction formula yields

\frac{5}{8}

.

Key Concept

Finding the Highest Common Factor (HCF) of fractions.

HCF and LCM

Answer

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