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30 Sep 2025, 10:21
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
50% (02:30) correct
50% (01:39) wrong based on 2 sessions
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The diagram above represents a rectangular picture (inner region) surrounded by a picture frame of uniform width (darker region). The picture measures 8 inches by 12 inches. The area of the picture frame is equal to the area of the picture itself. What is the width of the frame?
A. 0.5 inches
B. 1 inch
C. 1.5 inches
D. 2 inches
E. 4 inches
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GMAT-Club-Forum-b337fia0.png [ 51.8 KiB | Viewed 142 times ]
04 Oct 2025, 05:31
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Official Explanation
Call the uniform width \(x,\) and the outer dimensions of the picture together with the frame become \(8 + 2x\) and \(12 + 2x,\) because the frame’s width adds to both the top and bottom and the left and right of the picture. The full area of the picture together with frame is then\( (8 + 2x)(12 + 2x) = 96 + 16x + 24x + 4x^2 = 4x^2 + 40x + 96.\) This expression must equal \(2 × 8 × 12 = 192,\) because the frame’s area is equal to the picture’s, implying that the area enclosed by the frame and picture together must equal twice the area of the picture alone. So \(4x^2 + 40x + 96 = 192.\)
Subtract 192 to obtain \(4x^2 + 40x – 96 = 0,\) and divide by 4 on both sides: \(x^2 + 10 x – 24 = 0.\) The left side factors as \((x + 12)(x – 2),\) since \(12x – 2x = 10x,\) and so \(x = 2\) or –12. Only \(x = 2,\) (D), works in context of the problem, because a length cannot be negative. You can also get this result using a trial and error approach with the answer choices. When \(x = 2,\) the outer dimensions become \(12 × 16,\) for a total area of 192. This makes the picture’s area, \(8 × 12 = 96,\) the same as the frame’s area, \(192 – 96 = 96.\)
Answer: D
Call the uniform width \(x,\) and the outer dimensions of the picture together with the frame become \(8 + 2x\) and \(12 + 2x,\) because the frame’s width adds to both the top and bottom and the left and right of the picture. The full area of the picture together with frame is then\( (8 + 2x)(12 + 2x) = 96 + 16x + 24x + 4x^2 = 4x^2 + 40x + 96.\) This expression must equal \(2 × 8 × 12 = 192,\) because the frame’s area is equal to the picture’s, implying that the area enclosed by the frame and picture together must equal twice the area of the picture alone. So \(4x^2 + 40x + 96 = 192.\)
Subtract 192 to obtain \(4x^2 + 40x – 96 = 0,\) and divide by 4 on both sides: \(x^2 + 10 x – 24 = 0.\) The left side factors as \((x + 12)(x – 2),\) since \(12x – 2x = 10x,\) and so \(x = 2\) or –12. Only \(x = 2,\) (D), works in context of the problem, because a length cannot be negative. You can also get this result using a trial and error approach with the answer choices. When \(x = 2,\) the outer dimensions become \(12 × 16,\) for a total area of 192. This makes the picture’s area, \(8 × 12 = 96,\) the same as the frame’s area, \(192 – 96 = 96.\)
Answer: D
