A rectangular photograph is surrounded by a border that is 1 inch wide on each side. The total area of the photograph and the border is M square inches. If the border had been 2 inches wide on each side, the total area would have been (M + 52) square inches. What is the perimeter of the photograph, in inches?
A. 34
B. 36
C. 38
D. 40
E. 42
OA D
Source: GMAT Prep
A rectangular photograph is surrounded by a border that is 1 inch wide on each side. The total area of the photograph an
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Let Length and Breadth of photograph be \(L\) and \(B\) respectively.BTGmoderatorDC wrote: ↑Mon Oct 03, 2022 5:01 pmA rectangular photograph is surrounded by a border that is 1 inch wide on each side. The total area of the photograph and the border is M square inches. If the border had been 2 inches wide on each side, the total area would have been (M + 52) square inches. What is the perimeter of the photograph, in inches?
A. 34
B. 36
C. 38
D. 40
E. 42
OA D
Source: GMAT Prep
Perimeter is given by \(2*(L+b) \qquad (1)\)
According to the question:
\((L+2)(B+2) = m \qquad (2)\)
and
\((L+4)(B+4) = m + 52 \longrightarrow (L+4)(B+4) - 52 = m \qquad (3)\)
Equating \((2)\) and \((3)\)
\((L+2)(B+2) = (L+4)(B+4) - 52\)
\(LB + 2L + 2B + 4 = LB + 4L + 4B + 16 -52\)
Simplify
\(2L + 2B = 40 \longrightarrow 2(L+B) = 40\) (Check eq \((1)\))
Therefore, answer is D
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Solution:BTGmoderatorDC wrote: ↑Mon Oct 03, 2022 5:01 pmA rectangular photograph is surrounded by a border that is 1 inch wide on each side. The total area of the photograph and the border is M square inches. If the border had been 2 inches wide on each side, the total area would have been (M + 52) square inches. What is the perimeter of the photograph, in inches?
A. 34
B. 36
C. 38
D. 40
E. 42
OA D
Source: GMAT Prep
We are given that a rectangular photograph is surrounded by a border that is 1 inch wide on each side and that the total area of the photograph and border is M square inches. If we let L = the length of the photograph and W = the width of the photograph, since the border surrounds the length and width on two sides, the length of the photograph and border is L + 2 and the width of the photograph and border is W + 2.
We can now represent the area of the border and photograph:
area = length x width
M = (L + 2)(W + 2)
M = WL + 2W + 2L + 4
We are also given that if the border had been 2 inches wide on each side, the total area would have been (M + 52) square inches. Thus, the new length of the border and photograph would be L + 4 and the new width of the border and photograph would be W + 4.
The new area of the border and photograph is:
M + 52 = (L + 4)(W + 4)
M + 52 = WL + 4W + 4L + 16
M = WL + 4W + 4L – 36
We have two equations for M. Let’s equate them and simplify:
WL + 2W + 2L + 4 = WL + 4W + 4L – 36
2W + 2L + 4 = 4W + 4L – 36
2W + 2L = 4W + 4L – 40
2W + 2L = 40
Since perimeter = 2L + 2W, the perimeter of the photograph is 40 inches.
Answer: D
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