A milk vendor has 21 lt of cow milk, 42 lt of toned milk and 63 lt of double toned milk. If he wants so pack them in cans so that each can contains same lt of milk and does not want to mix any two kinds of milk in a can, then the least number of cans required is

Options

A. 3
B. 6
C. 9
D. 12

Answer

To solve the problem of how many cans the milk vendor needs, we will follow these steps:

### Step 1: Identify the quantities of milk
The vendor has:
- 21 liters of cow milk
- 42 liters of toned milk
- 63 liters of double toned milk

### Step 2: Find the Highest Common Factor (HCF)
To ensure that each can contains the same amount of milk without mixing types, we need to find the HCF of the three quantities (21, 42, and 63).

**Finding the factors:**
- Factors of 21: 1, 3, 7, 21
- Factors of 42: 1, 2, 3, 6, 7, 14, 21, 42
- Factors of 63: 1, 3, 7, 9, 21, 63

**Common factors:**
The common factors among 21, 42, and 63 are 1, 3, 7, and 21. The highest of these is 21.

### Step 3: Calculate the number of cans for each type of milk
Now, we will divide each quantity of milk by the HCF (21 liters) to find out how many cans are needed for each type of milk.

- **Cow Milk:**
\[ \text{Number of cans} = \frac{21 \text{ liters}}{21 \text{ liters/can}} = 1 \text{ can} \]

- **Toned Milk:**
\[ \text{Number of cans} = \frac{42 \text{ liters}}{21 \text{ liters/can}} = 2 \text{ cans} \]

- **Double Toned Milk:**
\[ \text{Number of cans} = \frac{63 \text{ liters}}{21 \text{ liters/can}} = 3 \text{ cans} \]

### Step 4: Total number of cans
Now, we add the number of cans required for each type of milk:
\[ \text{Total number of cans} = 1 + 2 + 3 = 6 \text{ cans} \]

### Conclusion
The least number of cans required is **6**.

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