A piece of paper is to display $150$ square inches of text. If there are to be one-inch margins on the sides and the top and a two-inch margin at the bottom, what are the dimensions of the smallest piece of paper that can be used? $\newline$ Choose $1$ answer: $\newline$ (A) $6 \prime \prime \times 25 \prime \prime$ $\newline$ (B) $10 \prime \prime \times 15 \prime \prime$ $\newline$ (C) $12 \prime \prime \times 18 \prime \prime$ $\newline$ (D) $15 \prime \prime \times 18 \prime \prime$ $\newline$ (E) None of these

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Add and subtract decimals: word problems

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Q. A piece of paper is to display

150

square inches of text. If there are to be one-inch margins on the sides and the top and a two-inch margin at the bottom, what are the dimensions of the smallest piece of paper that can be used?

\newline

Choose

1

answer:

\newline

(A)

6 \prime \prime \times 25 \prime \prime

\newline

(B)

10 \prime \prime \times 15 \prime \prime

\newline

(C)

12 \prime \prime \times 18 \prime \prime

\newline

(D)

15 \prime \prime \times 18 \prime \prime

\newline

(E) None of these

Calculate Text Area: First, let's calculate the area for the text alone, which is given as $150$ square inches.
Account for Margins: Now, we need to account for the margins. There's a one-inch margin on each side and the top, and a two-inch margin at the bottom. So, we add $2$ inches for the sides ( $1$ inch each side), $1$ inch for the top, and $2$ inches for the bottom to both the length and width.
Calculate Paper Dimensions: Let's call the width of the text area $w$ and the length $l$ . The overall width of the paper will be $w + 2$ and the length will be $l + 3$ .
Set Up Equation: The area of the paper is then $(w + 2)(l + 3)$ . We know this must equal the area of the text plus the area of the margins, which is $150$ square inches plus the area of the margins.
Simplify Equation: We don't know the area of the margins yet, but we can calculate it. The margins add $2$ inches to the width and $3$ inches to the length, so the area of the margins is $(2 \times l) + (3 \times w) + (2 \times 3)$ for the corners.
Find Factors of $150$ : We can now set up the equation w + \(2)(l + $3$ ) = $150$ + ( $2$ \times l) + ( $3$ \times w) + $6$ \
Choose Smallest Dimensions: Simplifying the equation, we get $wl + 2l + 3w + 6 = 150 + 2l + 3w + 6$ .
Check Other Pairs: We can cancel out the $2l$ and $3w$ from both sides, and we are left with $wl = 150$ .
Confirm Smallest Size: Now we need to find two numbers that multiply to $150$ and, when $2$ is added to the smaller number and $3$ to the larger number, will give us the smallest possible dimensions for the paper.
Final Answer: The factors of $150$ are $1 \& 150$ , $2 \& 75$ , $3 \& 50$ , $5 \& 30$ , $6 \& 25$ , $10 \& 15$ . We need to choose the pair that, when increased by the margins, results in the smallest dimensions.
Final Answer: The factors of $150$ are $1$ & $150$ , $2$ & $75$ , $3$ & $50$ , $5$ & $30$ , $6$ & $1$ $0$ , $1$ $1$ & $1$ $2$ . We need to choose the pair that, when increased by the margins, results in the smallest dimensions.If we take the pair $1$ $1$ & $1$ $2$ , adding the margins would give us a paper size of $1$ $5$ inches by $1$ $6$ inches.
Final Answer: The factors of $150$ are $1 \& 150$ , $2 \& 75$ , $3 \& 50$ , $5 \& 30$ , $6 \& 25$ , $10 \& 15$ . We need to choose the pair that, when increased by the margins, results in the smallest dimensions.If we take the pair $10 \& 15$ , adding the margins would give us a paper size of $12$ inches by $18$ inches.Checking the other pairs, we see that $12$ inches by $18$ inches is indeed the smallest possible size that can accommodate the text and margins.
Final Answer: The factors of $150$ are $1 \& 150$ , $2 \& 75$ , $3 \& 50$ , $5 \& 30$ , $6 \& 25$ , $10 \& 15$ . We need to choose the pair that, when increased by the margins, results in the smallest dimensions.If we take the pair $10 \& 15$ , adding the margins would give us a paper size of $12$ inches by $18$ inches.Checking the other pairs, we see that $12$ inches by $18$ inches is indeed the smallest possible size that can accommodate the text and margins.So, the correct answer is (C) $1 \& 150$ $2$ .