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20 Sep 2014, 23:43
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
80% (02:41) correct
20%
(03:00)
wrong
based on 577
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An art gallery has only paintings and sculptures. Currently, 1/3 of the pieces of art are displayed, and 1/6 of the pieces on display are sculptures. If 1/3 of the pieces not on display are paintings, and 200 sculptures are not on display, how many pieces of art does the gallery have?
A. 360
B. 450
C. 540
D. 640
E. 720
A. 360
B. 450
C. 540
D. 640
E. 720
22 Sep 2014, 01:51
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Solved using matrix method:
scul.png [ 5.14 KiB | Viewed 18064 times ]
1. Let total art = x
2. \(\frac{1}{3}\) are on display = \(\frac{x}{3}\)
3. So total not on display\(= x - \frac{x}{3} = \frac{2x}{3}\)
4. 1/3 NOT on display are paintings \(= \frac{2x}{3} * \frac{1}{3} = \frac{2x}{9}\)
5. Sculptures NOT on display = 200
Setting up the equation:
\(\frac{2x}{9} + 200 = \frac{2x}{3}\)
x = 450
Answer = B
Attachment:
scul.png [ 5.14 KiB | Viewed 18064 times ]
1. Let total art = x
2. \(\frac{1}{3}\) are on display = \(\frac{x}{3}\)
3. So total not on display\(= x - \frac{x}{3} = \frac{2x}{3}\)
4. 1/3 NOT on display are paintings \(= \frac{2x}{3} * \frac{1}{3} = \frac{2x}{9}\)
5. Sculptures NOT on display = 200
Setting up the equation:
\(\frac{2x}{9} + 200 = \frac{2x}{3}\)
x = 450
Answer = B
General Discussion
21 Sep 2014, 01:12
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gmatquant25
Too many words and redundant info there.
(i) 1/3 of the pieces of art are displayed, hence 2/3 of the pieces of art are NOT displayed.
(ii) 1/6 of the pieces on display are sculptures, hence 5/6 of the pieces on display are paintings.
(iii) 1/3 of the pieces NOT on display are paintings, hence 2/3 of the pieces NOT on display are sculptures.
200 sculptures are not on display, so according to (iii) 2/3*{not on display} = 200 --> {not on display} = 300.
According to (i) 2/3*{total} = 300 --> {total} = 450.
Answer: B.
