A catering contractor delivered a bag of rice in each of his five kitchens. At the end of the day he collected the balance of rice `1/3` kg, `2/5` kg, `3/7` kg, `1/2` kg and `2/3 kg` respectively. How much was the total quantity of rice he collected ?

Options

A. `3(21/70)` kg
B. `2(23/70)` kg
C. `3(24/50)` kg
D. `2(22/50)` kg

Answer

To find the total quantity of rice collected by the catering contractor from his five kitchens, we need to add the quantities of rice collected from each kitchen. The quantities are given in fractions:

1. \( \frac{1}{3} \) kg
2. \( \frac{2}{5} \) kg
3. \( \frac{3}{7} \) kg
4. \( \frac{1}{2} \) kg
5. \( \frac{2}{3} \) kg

### Step 1: Find the Least Common Multiple (LCM) of the denominators
The denominators are 3, 5, 7, 2, and 3. We need to find the LCM of these numbers to add the fractions.

- The prime factorization of each number is:
- \( 3 = 3^1 \)
- \( 5 = 5^1 \)
- \( 7 = 7^1 \)
- \( 2 = 2^1 \)

The LCM is found by taking the highest power of each prime number:
- LCM = \( 2^1 \times 3^1 \times 5^1 \times 7^1 = 210 \)

### Step 2: Convert each fraction to have a common denominator
Now we convert each fraction to have the denominator of 210.

1. \( \frac{1}{3} = \frac{1 \times 70}{3 \times 70} = \frac{70}{210} \)
2. \( \frac{2}{5} = \frac{2 \times 42}{5 \times 42} = \frac{84}{210} \)
3. \( \frac{3}{7} = \frac{3 \times 30}{7 \times 30} = \frac{90}{210} \)
4. \( \frac{1}{2} = \frac{1 \times 105}{2 \times 105} = \frac{105}{210} \)
5. \( \frac{2}{3} = \frac{2 \times 70}{3 \times 70} = \frac{140}{210} \)

### Step 3: Add the fractions
Now we can add the fractions:

\[ \frac{70}{210} + \frac{84}{210} + \frac{90}{210} + \frac{105}{210} + \frac{140}{210} \]

Combine the numerators:

\[ 70 + 84 + 90 + 105 + 140 = 489 \]

So, we have:

\[ \frac{489}{210} \]

### Step 4: Simplify the fraction if possible
Now we simplify \( \frac{489}{210} \).

Both numbers can be divided by 3:

\[ \frac{489 \div 3}{210 \div 3} = \frac{163}{70} \]

### Final Answer
Thus, the total quantity of rice collected is \( \frac{163}{70} \) kg, which is approximately 2.33 kg.

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