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22 Mar 2024, 08:30
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Difficulty:
75%
(hard)
Question Stats:
66% (02:54) correct
34%
(02:49)
wrong
based on 155
sessions
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An artist painted a series of 5 landscapes, titled Midnight 1-5, which were then displayed on two walls in an art gallery, with no other paintings hung on either wall. The table shows the value of each Midnight paninitg as a fraction of the value of the complete series. If the total value of the paintings displayed on the first wall was greater than \( \frac{ 2}{3 } \) of the total value of the series, was Midnight 3 displayed on the first wall ?
(1) Midnight 1 and Midnight 5 were displayed on the second wall.
(2) Midnight 2 and Midnight 4 were displayed on the first wall.
23 Mar 2024, 00:25
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Gangadhar111990
Assume that the value of the complete series is $30. The value of each painting in the series is as shown below -
Attachment:
Screenshot 2024-03-23 124247.png [ 83.34 KiB | Viewed 3526 times ]
If the total value of the paintings displayed on the first wall was greater than \( \frac{ 2}{3 } \) of the total value of the series, was Midnight 3 displayed on the first wall
We can infer that the total value of the paintings displayed on the first wall is greater than $20 and that the total value of the paintings displayed on the second wall is less than $10.
Statement 1
(1) Midnight 1 and Midnight 5 were displayed on the second wall.
If Midnight 1 and Midnight 5 were displayed on the second wall, the total value of the paintings would be at least $9. From the question premise, we know that the total value of the paintings on the second wall doesn't exceed $10. Hence, we can conclude that Midnight 3 was displayed on the first wall.
We can eliminate B, C, and E.
Attachment:
Screenshot 2024-03-23 124328.png [ 22.65 KiB | Viewed 3423 times ]
Statement 2
(2) Midnight 2 and Midnight 4 were displayed on the first wall.
If Midnight 2 and Midnight 4 were displayed on the first wall, the total value of the paintings would be at least $6. From the question premise, we know that the total value of the paintings on the first wall is at least $20. Hence, we can have two possible cases.
Case 1: Midnight 3 was displayed on the first wall.
Attachment:
Screenshot 2024-03-23 124435.png [ 24.95 KiB | Viewed 3328 times ]
Case 2: Midnight 3 was displayed on the second wall.
Attachment:
Screenshot 2024-03-23 124523.png [ 24.93 KiB | Viewed 3297 times ]
This statement alone is not sufficient. We can eliminate D.
Option A
23 Mar 2024, 02:08
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Gangadhar111990
Take LCM :
Values become : \(\frac{6}{30},\frac{7}{30},\frac{5}{30},\frac{9}{30},\frac{3}{30}\)
Convert into decimal \(.2 , .23, .16, .3, .1\)
Value of paintings on first wall is greater than \(.66\)
(1) Midnight 1 and Midnight 5 were displayed on the second wall.
Thus Midnight 2 and 4 will be on first wall with total value of \(.23 +.3= .53\)
Without midnight \(3\) the total value on first wall will not exceed \(.66\)
Thus midnight \(3\) has to be on first wall.
SUFF.
(2) Midnight 2 and Midnight 4 were displayed on the first wall.
Value of 1st wall with midnight \(2\) and \( 4 = .23 + .3 =.53\)
With Midnight \(2,4 \) and \(3 =.23 +.3 +.16= .69 \) Yes to the question
With Midnight \(2,4 \) and \(1 =.23 +.3 + .2 = .73 \) No to the question
etc.
INSUFF.
Ans A
Hope it helped.
