Raj is framing a rectangular poster that is 12 wholes and 1 fith

inches long and 9 wholes and 8 thirds
inches wide. What is the perimeter?(1 point)
Responses

2 3340
2 3340

21 2340
21 2340

43 320
43 320

1 answer

To find the perimeter of a rectangular poster, you can use the formula:

\[ \text{Perimeter} = 2 \times (\text{Length} + \text{Width}) \]

First, convert the dimensions of the poster into improper fractions.

  1. Length: \( 12\) wholes and \( 1 \) fifth: \[ 12 + \frac{1}{5} = \frac{12 \times 5 + 1}{5} = \frac{60 + 1}{5} = \frac{61}{5} \]

  2. Width: \( 9 \) wholes and \( 8 \) thirds: \[ 9 + \frac{8}{3} = \frac{9 \times 3 + 8}{3} = \frac{27 + 8}{3} = \frac{35}{3} \]

Now, add the length and width together:

\[ \text{Length} + \text{Width} = \frac{61}{5} + \frac{35}{3} \]

To add these fractions, find a common denominator. The least common multiple of \( 5 \) and \( 3 \) is \( 15 \).

Convert each fraction:

\[ \frac{61}{5} = \frac{61 \times 3}{5 \times 3} = \frac{183}{15} \] \[ \frac{35}{3} = \frac{35 \times 5}{3 \times 5} = \frac{175}{15} \]

Now, add the two fractions:

\[ \frac{183}{15} + \frac{175}{15} = \frac{183 + 175}{15} = \frac{358}{15} \]

Now, substitute this back into the perimeter formula:

\[ \text{Perimeter} = 2 \times \left( \frac{358}{15} \right) = \frac{716}{15} \]

To simplify \(\frac{716}{15}\), perform the division:

\[ 716 \div 15 \approx 47.7333 \text{ which is } 47 \frac{1}{15} \]

This is the total perimeter in mixed fraction form. Since none of the responses seem numerical or are solutions for perimeter, none of the choices provided are correct, as \(47.7333\) is not shown among the options.

Thus, please check the choices or allow me to assist you in another way!