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Question

At a telephone exchange, three phones ring at intervals of 20 sec, 24 sec and 30 seconds. If they ring together at 11:25 a.m, when will they next ring together?

This question was previously asked in
RRB NTPC 2015 CBT 1 Question Paper (29-Mar-2016) (Shift 1)
The correct answer is
11:27 a.m.

Find Next Simultaneous Ring Time

This problem involves finding the Least Common Multiple (LCM) of the ringing intervals to determine when the phones will ring together again.

1. Identify Ringing Intervals

  • Phone 1 rings every 20 seconds.
  • Phone 2 rings every 24 seconds.
  • Phone 3 rings every 30 seconds.

2. Calculate the LCM

To find when they will ring together, we need the LCM of 20, 24, and 30.

  • Prime factorization of 20: $2^2 \times 5$
  • Prime factorization of 24: $2^3 \times 3$
  • Prime factorization of 30: $2 \times 3 \times 5$
  • The LCM is found by taking the highest power of each prime factor present: $2^3 \times 3^1 \times 5^1 = 8 \times 3 \times 5 = 120$.

The LCM is $120$ seconds.

3. Convert LCM to Minutes

Convert the interval back into minutes:

$ 120 \text{ seconds} = \frac{120}{60} \text{ minutes} = 2 \text{ minutes} $

4. Determine Next Ring Time

Add this interval to the time they last rang together:

Last ring time: 11:25 a.m.

Interval: 2 minutes

Next ring time: $ 11:25 \text{ a.m.} + 2 \text{ minutes} = 11:27 \text{ a.m.} $

Therefore, the phones will next ring together at 11:27 a.m.

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Important Questions from LCM and HCF

  1. 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?

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

  3. 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.

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

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