Wholesaler of tea has three types of tea that weighs 408 kg, 468 and 516 kg separately. What will be the minimum number of bags of the same size in which all kinds of tea can be kept number unmixed?

Options

A. 15
B. 16
C. 14
D. 12

Answer

To find the minimum number of bags of the same size in which all types of tea can be kept unmixed, we need to determine the Highest Common Factor (HCF) of the weights of the three types of tea: 408 kg, 468 kg, and 516 kg.

### Step-by-Step Solution:

**Step 1: Factorization of the weights.**

1. **Factor 408:**
- Divide by 2: \( 408 \div 2 = 204 \)
- Divide by 2: \( 204 \div 2 = 102 \)
- Divide by 2: \( 102 \div 2 = 51 \)
- Divide by 3: \( 51 \div 3 = 17 \)
- 17 is a prime number.

So, the prime factorization of 408 is:
\[ 408 = 2^3 \times 3^1 \times 17^1 \]

2. **Factor 468:**
- Divide by 2: \( 468 \div 2 = 234 \)
- Divide by 2: \( 234 \div 2 = 117 \)
- Divide by 3: \( 117 \div 3 = 39 \)
- Divide by 3: \( 39 \div 3 = 13 \)
- 13 is a prime number.

So, the prime factorization of 468 is:
\[ 468 = 2^2 \times 3^2 \times 13^1 \]

3. **Factor 516:**
- Divide by 2: \( 516 \div 2 = 258 \)
- Divide by 2: \( 258 \div 2 = 129 \)
- Divide by 3: \( 129 \div 3 = 43 \)
- 43 is a prime number.

So, the prime factorization of 516 is:
\[ 516 = 2^2 \times 3^1 \times 43^1 \]

**Step 2: Find the HCF.**

To find the HCF, we take the lowest power of each common prime factor from the factorizations:

- For \(2\): The minimum power is \(2^2\) (from 468 and 516).
- For \(3\): The minimum power is \(3^1\) (from 408 and 516).
- The prime factors 17, 13, and 43 are not common to all three numbers.

Thus, the HCF is:
\[ HCF = 2^2 \times 3^1 = 4 \times 3 = 12 \]

**Step 3: Calculate the minimum number of bags.**

Since the HCF is 12, this means that the minimum number of bags of the same size that can hold all types of tea without mixing is 12 bags.

### Final Answer:
The minimum number of bags required is **12 bags**.

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