What are the dimensions of the viewable portion...?
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A framed picture is shown above. The frame, shown shaded, is 6 inches wide and forms a border of uniform width around the picture. What are the dimensions of the viewable portion of the picture?
1. The area of the shaded region is 24 square inches.
2. The frame is 8 inches tall.
Source: OG Quant Review 2016, pg. 153
Please help. I don't understand the OG explanation. My understanding is to use this technique: Area of everything minus Area of middle = Area of border but I don't know how to apply it to this question. Thanks so much
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Statement 1:
shaded region = total area - picture area.
Thus:
total area - picture area = 24.
Case 1: frame width = 1 inch
In this case:
total area = 6h.
picture area = (4)(h-2).
Since total area - picture area = 24, we get:
6h - (4)(h-2) = 24
6h - 4h + 8 = 24
2h = 16
h = 8.
height of the picture = h-2 = 8-2 = 6.
width of the picture = 4.
Case 2: frame width = 0.5 inch
In this case:
total area = 6h.
picture area = (5)(h-1).
Since total area - picture area = 24, we get:
6h - (5)(h-1) = 24
6h - 5h + 5 = 24
h = 19.
height of the picture = h-1 = 19-1 = 18.
width of the picture = 5.
Since the picture can take on different dimensions, INSUFFICIENT.
Statement 2:
Case 1 also satisfies Statement 2.
In Case 1:
height of the picture = 6.
width of the picture = 4.
Case 3:
In this case:
Height of the picture = 4.
Width of the picture = 2.
Since the picture can take on different dimensions, INSUFFICIENT.
Statements combined:
Only Case 1 satisfies both statements.
In Case 1:
height of the picture = 6.
width of the picture = 4.
SUFFICIENT.
The correct answer is C.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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