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Question

Six bells begin to toll together and toll, respectively, at intervals of 3, 4, 6, 7, 8 and 12 seconds. After how many seconds, will they toll together again?

The correct answer is

168

Understanding the Bells Tolling Together Problem

This question asks us to find the time when six bells, which start tolling together and then toll at regular intervals of 3, 4, 6, 7, 8, and 12 seconds, will next toll simultaneously. This type of problem requires finding the moment in time when all their individual cycles align again.

The time when all the bells will toll together again will be the smallest multiple that is common to all their individual tolling intervals. In mathematical terms, this is the Least Common Multiple (LCM) of the given intervals.

Calculating the LCM of Tolling Intervals

The given tolling intervals are 3, 4, 6, 7, 8, and 12 seconds. To find the LCM, we can use the prime factorization method. We find the prime factors of each number:

  • Prime factors of 3: \(3^1\)
  • Prime factors of 4: \(2 \times 2 = 2^2\)
  • Prime factors of 6: \(2 \times 3 = 2^1 \times 3^1\)
  • Prime factors of 7: \(7^1\)
  • Prime factors of 8: \(2 \times 2 \times 2 = 2^3\)
  • Prime factors of 12: \(2 \times 2 \times 3 = 2^2 \times 3^1\)

To find the LCM, we take the highest power of each prime factor that appears in any of the factorizations:

  • Highest power of 2: \(2^3\) (from the number 8)
  • Highest power of 3: \(3^1\) (from the numbers 3, 6, and 12)
  • Highest power of 7: \(7^1\) (from the number 7)

Now, we multiply these highest powers together to find the LCM:

LCM = \(2^3 \times 3^1 \times 7^1 = 8 \times 3 \times 7\)

LCM = \(24 \times 7\)

LCM = \(168\)

Conclusion: When Bells Toll Together Again

The Least Common Multiple of 3, 4, 6, 7, 8, and 12 is 168. This means that after 168 seconds, all six bells will have completed a whole number of their respective tolling cycles and will, therefore, toll together again simultaneously.

The intervals are 3, 4, 6, 7, 8, and 12 seconds. After 168 seconds:

  • Bell 1 (3s interval) tolls \(168 \div 3 = 56\) times.
  • Bell 2 (4s interval) tolls \(168 \div 4 = 42\) times.
  • Bell 3 (6s interval) tolls \(168 \div 6 = 28\) times.
  • Bell 4 (7s interval) tolls \(168 \div 7 = 24\) times.
  • Bell 5 (8s interval) tolls \(168 \div 8 = 21\) times.
  • Bell 6 (12s interval) tolls \(168 \div 12 = 14\) times.

Since 168 is a multiple of every interval, they all toll at exactly 168 seconds, marking the first time they toll together after the initial moment.

Revision Table: Key Concepts

Concept Explanation Relevance to Problem
Multiple A number that can be divided by another number without a remainder. The time when bells toll is a multiple of their interval.
Common Multiple A number that is a multiple of two or more numbers. The time they toll together is a common multiple of all intervals.
Least Common Multiple (LCM) The smallest positive common multiple of two or more numbers. The first time they toll together again (after the start) is the LCM of their intervals.
Prime Factorization Breaking down a number into its prime factors. A method used to calculate the LCM efficiently.

Additional Information: Applications of LCM

The concept of LCM is widely used in various real-life scenarios and mathematical problems besides bells tolling together. Here are a few examples:

  • Scheduling Events: Finding when recurring events (like buses arriving, lights flashing, or tasks repeating) will happen simultaneously.
  • Fractions: Finding the least common denominator (LCD) when adding or subtracting fractions, which is the LCM of the denominators.
  • Cycling Problems: Determining when objects moving in cycles (like gears meshing or planets aligning) will return to a starting configuration.
  • Retail: Calculating when promotions on different products that run on cycles might overlap.

Understanding how to find the LCM is a fundamental skill in number theory and has practical applications in coordinating events that occur at regular, repeating intervals.

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Important Questions from LCM and HCF

  1. A and B are two prime numbers such that A > B and their LCM is 209. The value of A 2 - B is:

  2. Find the least number which when divided by 12, 18, 24 and 30 leaves 4 as remainder in each case, but when divided by 7 leaves no remainder.

  3. Calculate the HCF of \(\frac{12}{5}\) \(\frac{14}{15}\)  and  \(\frac{16}{17}\) .

  4. Three numbers are in the proportion of 3 : 8 : 15 and their LCM is 8280. What is their HCF?

  5. A and B are two prime numbers such that A > B and their LCM is 209. The value of B 2 –  A is:

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