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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?
Choose 1 answer:
(A) 
6''×25''
(B) 
10'×15'
(C) 
12'×18''
(D) 
15'×18'
(E) None of these

A piece of paper is to display 150150 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?\newlineChoose 11 answer:\newline(A) 6×25 6 \prime \prime \times 25 \prime \prime \newline(B) 10×15 10 \prime \prime \times 15 \prime \prime \newline(C) 12×18 12 \prime \prime \times 18 \prime \prime \newline(D) 15×18 15 \prime \prime \times 18 \prime \prime \newline(E) None of these

Full solution

Q. A piece of paper is to display 150150 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?\newlineChoose 11 answer:\newline(A) 6×25 6 \prime \prime \times 25 \prime \prime \newline(B) 10×15 10 \prime \prime \times 15 \prime \prime \newline(C) 12×18 12 \prime \prime \times 18 \prime \prime \newline(D) 15×18 15 \prime \prime \times 18 \prime \prime \newline(E) None of these
  1. Calculate Text Area: First, let's calculate the area for the text alone, which is given as 150150 square inches.
  2. 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 22 inches for the sides (11 inch each side), 11 inch for the top, and 22 inches for the bottom to both the length and width.
  3. Calculate Paper Dimensions: Let's call the width of the text area ww and the length ll. The overall width of the paper will be w+2w + 2 and the length will be l+3l + 3.
  4. Set Up Equation: The area of the paper is then (w+2)(l+3)(w + 2)(l + 3). We know this must equal the area of the text plus the area of the margins, which is 150150 square inches plus the area of the margins.
  5. Simplify Equation: We don't know the area of the margins yet, but we can calculate it. The margins add 22 inches to the width and 33 inches to the length, so the area of the margins is (2×l)+(3×w)+(2×3)(2 \times l) + (3 \times w) + (2 \times 3) for the corners.
  6. Find Factors of 150150: We can now set up the equation w + \(2)(l + 33) = 150150 + (22 \times l) + (33 \times w) + 66\
  7. Choose Smallest Dimensions: Simplifying the equation, we get wl+2l+3w+6=150+2l+3w+6wl + 2l + 3w + 6 = 150 + 2l + 3w + 6.
  8. Check Other Pairs: We can cancel out the 2l2l and 3w3w from both sides, and we are left with wl=150wl = 150.
  9. Confirm Smallest Size: Now we need to find two numbers that multiply to 150150 and, when 22 is added to the smaller number and 33 to the larger number, will give us the smallest possible dimensions for the paper.
  10. Final Answer: The factors of 150150 are 1&1501 \& 150, 2&752 \& 75, 3&503 \& 50, 5&305 \& 30, 6&256 \& 25, 10&1510 \& 15. We need to choose the pair that, when increased by the margins, results in the smallest dimensions.
  11. Final Answer: The factors of 150150 are 11 & 150150, 22 & 7575, 33 & 5050, 55 & 3030, 66 & 1100, 1111 & 1122. We need to choose the pair that, when increased by the margins, results in the smallest dimensions.If we take the pair 1111 & 1122, adding the margins would give us a paper size of 1155 inches by 1166 inches.
  12. Final Answer: The factors of 150150 are 1&1501 \& 150, 2&752 \& 75, 3&503 \& 50, 5&305 \& 30, 6&256 \& 25, 10&1510 \& 15. We need to choose the pair that, when increased by the margins, results in the smallest dimensions.If we take the pair 10&1510 \& 15, adding the margins would give us a paper size of 1212 inches by 1818 inches.Checking the other pairs, we see that 1212 inches by 1818 inches is indeed the smallest possible size that can accommodate the text and margins.
  13. Final Answer: The factors of 150150 are 1&1501 \& 150, 2&752 \& 75, 3&503 \& 50, 5&305 \& 30, 6&256 \& 25, 10&1510 \& 15. We need to choose the pair that, when increased by the margins, results in the smallest dimensions.If we take the pair 10&1510 \& 15, adding the margins would give us a paper size of 1212 inches by 1818 inches.Checking the other pairs, we see that 1212 inches by 1818 inches is indeed the smallest possible size that can accommodate the text and margins.So, the correct answer is (C) 1&1501 \& 15022.

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