Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

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) 625 6 \prime \prime 25 \prime \prime \newline(B) 10×15 10 \prime \prime \times 15 \prime \newline(C) 12×18 12 \prime \times 18 \prime \newline(D) 15×18 15 \prime \times 18 \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) 625 6 \prime \prime 25 \prime \prime \newline(B) 10×15 10 \prime \prime \times 15 \prime \newline(C) 12×18 12 \prime \times 18 \prime \newline(D) 15×18 15 \prime \times 18 \prime \newline(E) None of these
  1. Determine Area Available: Determine the area available for text. Since the margins are 11 inch on the sides and top and 22 inches at the bottom, we need to subtract these margins from the overall dimensions of the paper to get the area available for text. Let's denote the width of the paper as WW and the height as HH. The area for text will then be (W2×1)×(H12)(W - 2 \times 1) \times (H - 1 - 2), because we subtract 11 inch from each side for the width and 11 inch from the top and 22 inches from the bottom for the height.
  2. Set Up Equation: Set up the equation for the area available for text. The equation based on the margins is (W2)×(H3)=150(W - 2) \times (H - 3) = 150 square inches. We need to find the values of WW and HH that satisfy this equation.
  3. Check Answer Choices: Check the answer choices to see which one satisfies the equation.\newlineWe will plug in the dimensions from each answer choice into the equation and see which one gives us an area of 150150 square inches for the text.\newline(A) 6×256'' \times 25'' would give us (62)×(253)=4×22=88(6 - 2) \times (25 - 3) = 4 \times 22 = 88 square inches, which is not enough.
  4. Correct Answer: (B) 10×1510'' \times 15'' would give us (102)×(153)=8×12=96(10 - 2) \times (15 - 3) = 8 \times 12 = 96 square inches, which is also not enough.
  5. Correct Answer: (B) 10×1510'' \times 15'' would give us (102)×(153)=8×12=96(10 - 2) \times (15 - 3) = 8 \times 12 = 96 square inches, which is also not enough.(C) 12×1812'' \times 18'' would give us (122)×(183)=10×15=150(12 - 2) \times (18 - 3) = 10 \times 15 = 150 square inches, which is the correct area.
  6. Correct Answer: (B) 10×1510'' \times 15'' would give us (102)×(153)=8×12=96(10 - 2) \times (15 - 3) = 8 \times 12 = 96 square inches, which is also not enough.(C) 12×1812'' \times 18'' would give us (122)×(183)=10×15=150(12 - 2) \times (18 - 3) = 10 \times 15 = 150 square inches, which is the correct area.(D) 15×1815'' \times 18'' would give us (152)×(183)=13×15=195(15 - 2) \times (18 - 3) = 13 \times 15 = 195 square inches, which is more than needed.
  7. Correct Answer: (B) 10×1510'' \times 15'' would give us (102)×(153)=8×12=96(10 - 2) \times (15 - 3) = 8 \times 12 = 96 square inches, which is also not enough.(C) 12×1812'' \times 18'' would give us (122)×(183)=10×15=150(12 - 2) \times (18 - 3) = 10 \times 15 = 150 square inches, which is the correct area.(D) 15×1815'' \times 18'' would give us (152)×(183)=13×15=195(15 - 2) \times (18 - 3) = 13 \times 15 = 195 square inches, which is more than needed.(E) None of these is not an option because we have found a correct answer in (C).

More problems from Add and subtract decimals: word problems