Question

$151\text{oz}$ of mixed nuts is packed into 20 packets such that each packet has equal number of mixed nuts. Find the $oz$ of nuts present in each pack.

• A. $\frac{151}{20}\text{oz}$
• B. $7.55\text{oz}$
  
• C. $7\frac{11}{20}\text{oz}$
• D. $\frac{7550}{1000}\text{oz}$

Answer 1

Answer: (A), (B), (C), (D)

$\frac{7550}{1000}\text{oz}$

20 packets consist of total $151\text{oz}$ of mixed nuts.
Each pack will consist of $\frac{151}{20}\text{oz}$ of mixed nuts.

We know, $20\times 5=100$
$\therefore$ Multiply numerator and denominator by $5$ to convert the denominator a power of $10.$

$\frac{151\times 5}{20\times 5}\text{oz}=\frac{755}{100}\text{oz}=7.55\text{oz}$

Also, $\frac{151}{20}\text{oz}$ can be expressed as a mixed fraction.

$\frac{151}{20}\text{oz}=\frac{20\times 7+11}{20}\text{oz}=\frac{20\times 7}{20}+\frac{11}{20}\text{oz}=7+\frac{11}{20}\text{oz}=7\frac{11}{20}\text{oz}$

$\therefore$ Each pack will contain $7.55\text{oz}$ or $7\frac{11}{20}\text{oz}$ of mixed nuts.