A wine seller had three types of wine, 403 litres of 1 st kind, 434 litres of 2nd kind and 465 litres of 3rd kind. Find the least possible number of casks of equal size in which different types of wine can be filled without mixing

Options

A. 46
B. 44
C. 42
D. 48

Answer

To find the least possible number of casks of equal size in which different types of wine can be filled without mixing, we need to determine the highest common factor (HCF) of the quantities of the three types of wine: 403 litres, 434 litres, and 465 litres.

### Step-by-Step Solution:

**Step 1: Find the prime factorization of each quantity.**

1. **403:**
- Check divisibility by 2 (not divisible).
- Check divisibility by 3 (not divisible).
- Check divisibility by 5 (not divisible).
- Check divisibility by 7 (not divisible).
- Check divisibility by 11 (not divisible).
- Check divisibility by 13:
\[ 403 \div 13 = 31 \quad (\text{So, } 403 = 13 \times 31) \]

2. **434:**
- Check divisibility by 2:
\[ 434 \div 2 = 217 \quad (\text{So, } 434 = 2 \times 217) \]
- Check divisibility of 217:
- Check divisibility by 7 (not divisible).
- Check divisibility by 11 (not divisible).
- Check divisibility by 13:
\[ 217 \div 13 = 17 \quad (\text{So, } 217 = 13 \times 17) \]
- Therefore, \(434 = 2 \times 13 \times 17\).

3. **465:**
- Check divisibility by 3:
\[ 465 \div 3 = 155 \quad (\text{So, } 465 = 3 \times 155) \]
- Check divisibility of 155:
- Check divisibility by 5:
\[ 155 \div 5 = 31 \quad (\text{So, } 155 = 5 \times 31) \]
- Therefore, \(465 = 3 \times 5 \times 31\).

**Step 2: Identify the common factors.**
- The prime factorizations are:
- \(403 = 13 \times 31\)
- \(434 = 2 \times 13 \times 17\)
- \(465 = 3 \times 5 \times 31\)

The common prime factor among all three quantities is \(31\).

**Step 3: Calculate the HCF.**
- The HCF is the product of the lowest powers of all common prime factors:
- HCF = \(31\)

**Step 4: Calculate the number of casks.**
- To find the least possible number of casks, we divide each quantity by the HCF:
- For 403 litres:
\[ \frac{403}{31} = 13 \]
- For 434 litres:
\[ \frac{434}{31} = 14 \]
- For 465 litres:
\[ \frac{465}{31} = 15 \]

**Step 5: Calculate the total number of casks.**
- The total number of casks required is the sum of the individual casks:
\[ 13 + 14 + 15 = 42 \]

### Final Answer:
The least possible number of casks of equal size in which different types of wine can be filled without mixing is **42**.