How many `(1)/(6)` of together make `41(2)/(3)` ?

Options

A. 125
B. 150
C. 250
D. 350

Answer

To solve the problem of how many \( \frac{1}{6} \) together make \( 41 \frac{2}{3} \), we can follow these steps:

### Step 1: Convert the mixed number to an improper fraction
First, we need to convert \( 41 \frac{2}{3} \) into an improper fraction.

\[ 41 \frac{2}{3} = 41 + \frac{2}{3} = \frac{41 \times 3 + 2}{3} = \frac{123 + 2}{3} = \frac{125}{3} \]

### Step 2: Set up the equation
Let \( n \) be the number of \( \frac{1}{6} \) fractions. The equation can be set up as follows:

\[ n \times \frac{1}{6} = \frac{125}{3} \]

### Step 3: Solve for \( n \)
To find \( n \), we can multiply both sides of the equation by 6 to eliminate the fraction:

\[ n = \frac{125}{3} \times 6 \]

### Step 4: Simplify the right side
Now, we simplify the right side:

\[ n = \frac{125 \times 6}{3} \]

We can simplify \( \frac{6}{3} \) to 2:

\[ n = 125 \times 2 = 250 \]

### Step 5: Conclusion
Thus, the number of \( \frac{1}{6} \) fractions that make \( 41 \frac{2}{3} \) is \( 250 \).

\[ \text{Final Answer: } 250 \]

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