Nehru Ji had n chocolates. He distributed them among 4 children in the ratio of `1/2:1/3:1/5:1/8`. If he gave them each one a complete chocolate, the minimum no. of cholocates that he had

Options

A. 139
B. 240
C. 278
D. None of these

Answer

To solve the problem of how many chocolates Nehru Ji had, we will follow these steps:

### Step 1: Understand the Ratio
The chocolates are distributed among 4 children in the ratio of \( \frac{1}{2} : \frac{1}{3} : \frac{1}{5} : \frac{1}{8} \).

### Step 2: Convert the Ratios to a Common Denominator
To work with these fractions, we need to find a common denominator. The denominators are 2, 3, 5, and 8.

### Step 3: Find the LCM of the Denominators
We will find the least common multiple (LCM) of 2, 3, 5, and 8.

- The prime factorization of the numbers:
- \( 2 = 2^1 \)
- \( 3 = 3^1 \)
- \( 5 = 5^1 \)
- \( 8 = 2^3 \)

The LCM is found by taking the highest power of each prime:
- LCM = \( 2^3 \times 3^1 \times 5^1 = 8 \times 3 \times 5 = 120 \)

### Step 4: Calculate the Parts of Each Child
Now we will calculate how many chocolates each child receives based on the LCM:

- For the first child: \( \frac{120}{2} = 60 \)
- For the second child: \( \frac{120}{3} = 40 \)
- For the third child: \( \frac{120}{5} = 24 \)
- For the fourth child: \( \frac{120}{8} = 15 \)

### Step 5: Sum the Parts
Now we will add the parts given to each child to find the total number of chocolates:

\[ 60 + 40 + 24 + 15 = 139 \]

### Conclusion
Thus, the minimum number of chocolates that Nehru Ji had is **139**.

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