A person distributes his pens among four friends A, B, C, D in the ratio `1/3:1/4 : 1/5: 1/6`. What is the minimum number of pens that the person should have?

Options

A. 57
B. 65
C. 75
D. 45

Answer

To solve the problem of distributing pens among four friends A, B, C, and D in the ratio \( \frac{1}{3} : \frac{1}{4} : \frac{1}{5} : \frac{1}{6} \), we will follow these steps:

### Step 1: Identify the Ratios
The given ratios are:
- A: \( \frac{1}{3} \)
- B: \( \frac{1}{4} \)
- C: \( \frac{1}{5} \)
- D: \( \frac{1}{6} \)

### Step 2: Find the LCM of Denominators
To combine these fractions into a common ratio, we need to find the least common multiple (LCM) of the denominators (3, 4, 5, and 6).

- The prime factorization of each number is:
- \( 3 = 3^1 \)
- \( 4 = 2^2 \)
- \( 5 = 5^1 \)
- \( 6 = 2^1 \times 3^1 \)

- The LCM is obtained by taking the highest power of each prime:
- \( 2^2 \) (from 4)
- \( 3^1 \) (from 3 or 6)
- \( 5^1 \) (from 5)

Thus, \( \text{LCM}(3, 4, 5, 6) = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60 \).

### Step 3: Convert Each Ratio to Have the Same Denominator
Now, we convert each ratio to have a denominator of 60:
- For A: \( \frac{1}{3} = \frac{20}{60} \)
- For B: \( \frac{1}{4} = \frac{15}{60} \)
- For C: \( \frac{1}{5} = \frac{12}{60} \)
- For D: \( \frac{1}{6} = \frac{10}{60} \)

So the ratios become:
- A : B : C : D = 20 : 15 : 12 : 10

### Step 4: Simplify the Ratio
The simplified ratio is:
- A : B : C : D = 20 : 15 : 12 : 10

### Step 5: Calculate Total Parts
Now, we calculate the total parts of the ratio:
- Total = \( 20 + 15 + 12 + 10 = 57 \)

### Step 6: Determine Minimum Number of Pens
The total number of pens must be a multiple of the total parts (57). The minimum number of pens that can be distributed among A, B, C, and D is the least multiple of 57, which is:
- \( 57 \times 1 = 57 \)

### Conclusion
Thus, the minimum number of pens that the person should have is **57**.

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