The weights of three packets are `2""3/4` kg, `3""1/3` kg and `5""2/5` kg. Find the total weight of all the three packets.

Answer

To find the total weight of the three packets, we need to add the weights given in mixed fractions. The weights are:

1. \(2 \frac{3}{4}\) kg
2. \(3 \frac{1}{3}\) kg
3. \(5 \frac{2}{5}\) kg

### Step 1: Convert the mixed fractions to improper fractions

1. For \(2 \frac{3}{4}\):
\[ 2 \frac{3}{4} = \frac{2 \times 4 + 3}{4} = \frac{8 + 3}{4} = \frac{11}{4} \]

2. For \(3 \frac{1}{3}\):
\[ 3 \frac{1}{3} = \frac{3 \times 3 + 1}{3} = \frac{9 + 1}{3} = \frac{10}{3} \]

3. For \(5 \frac{2}{5}\):
\[ 5 \frac{2}{5} = \frac{5 \times 5 + 2}{5} = \frac{25 + 2}{5} = \frac{27}{5} \]

### Step 2: Add the improper fractions

Now we need to add \(\frac{11}{4}\), \(\frac{10}{3}\), and \(\frac{27}{5}\). To do this, we first find the least common multiple (LCM) of the denominators 4, 3, and 5.

### Step 3: Find the LCM of the denominators

The LCM of 4, 3, and 5:
- The prime factorization is:
- \(4 = 2^2\)
- \(3 = 3^1\)
- \(5 = 5^1\)

The LCM is:
\[ 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60 \]

### Step 4: Convert each fraction to have the same denominator

1. Convert \(\frac{11}{4}\):
\[ \frac{11}{4} = \frac{11 \times 15}{4 \times 15} = \frac{165}{60} \]

2. Convert \(\frac{10}{3}\):
\[ \frac{10}{3} = \frac{10 \times 20}{3 \times 20} = \frac{200}{60} \]

3. Convert \(\frac{27}{5}\):
\[ \frac{27}{5} = \frac{27 \times 12}{5 \times 12} = \frac{324}{60} \]

### Step 5: Add the fractions

Now we can add the fractions:
\[ \frac{165}{60} + \frac{200}{60} + \frac{324}{60} = \frac{165 + 200 + 324}{60} = \frac{689}{60} \]

### Step 6: Convert the improper fraction back to a mixed number

To convert \(\frac{689}{60}\) to a mixed number:
1. Divide 689 by 60:
- \(689 \div 60 = 11\) remainder \(29\)

Thus, we can write:
\[ \frac{689}{60} = 11 \frac{29}{60} \]

### Final Answer

The total weight of all three packets is:
\[ 11 \frac{29}{60} \text{ kg} \]