4 bells ring at intervals of 30 minutes, 1 hour, `1(1)/(2)` hour and 1 hour 45 minutes respectively. All the bells ring simultaneously at 12 noon. They will again ring simultaneously at :

Options

A. 12 mid night
B. 2 a.m.
C. 6 a.m.
D. 9 a.m.

Answer

To solve the problem of when the four bells will ring simultaneously again after ringing together at 12 noon, we need to find the least common multiple (LCM) of their ringing intervals. Let's break down the solution step by step.

### Step 1: Convert all intervals to the same unit (minutes)
- Bell 1: 30 minutes
- Bell 2: 1 hour = 60 minutes
- Bell 3: 1.5 hours = 90 minutes
- Bell 4: 1 hour 45 minutes = 105 minutes

### Step 2: Factor each interval into its prime factors
- **30 minutes**:
- \(30 = 2 \times 3 \times 5\)

- **60 minutes**:
- \(60 = 2^2 \times 3 \times 5\)

- **90 minutes**:
- \(90 = 2 \times 3^2 \times 5\)

- **105 minutes**:
- \(105 = 3 \times 5 \times 7\)

### Step 3: Determine the LCM using the highest power of each prime factor
- For the prime factor **2**: The highest power is \(2^2\) from 60.
- For the prime factor **3**: The highest power is \(3^2\) from 90.
- For the prime factor **5**: The highest power is \(5^1\) (common in all).
- For the prime factor **7**: The highest power is \(7^1\) from 105.

### Step 4: Calculate the LCM
Now, we can calculate the LCM:
\[ \text{LCM} = 2^2 \times 3^2 \times 5^1 \times 7^1 \]
Calculating this step by step:
- \(2^2 = 4\)
- \(3^2 = 9\)
- \(5^1 = 5\)
- \(7^1 = 7\)

Now multiply these together:
\[ 4 \times 9 = 36 \]
\[ 36 \times 5 = 180 \]
\[ 180 \times 7 = 1260 \]

### Step 5: Convert LCM back to hours and minutes
The LCM is 1260 minutes. To convert this into hours and minutes:
- Divide 1260 by 60 (since there are 60 minutes in an hour):
\[ 1260 \div 60 = 21 \text{ hours} \]
This means 21 hours after 12 noon.

### Step 6: Calculate the time
Adding 21 hours to 12 noon:
- 12 noon + 21 hours = 9 AM the next day.

### Final Answer
The bells will ring simultaneously again at **9 AM** the next day.

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