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17 Dec 2017, 23:19
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
67% (02:25) correct
33%
(02:24)
wrong
based on 203
sessions
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In the figure above, a line segment joining the point (3,10) and which of the following points on the square will separate the square into two regions whose areas are in the ratio 7 to 1?
(A) (11,3)
(B) (11,4)
(C) (11,6)
(D) (11,7)
(E) (11,8)
Attachment:
2017-12-18_1011.png [ 6.73 KiB | Viewed 3863 times ]
17 Dec 2017, 23:26
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E
Area of the square = 8*8
=64
To divide the ratio into 7:1 we have to find a point on the vertical side of the square opposite to (3,10) such that the area of the triangle is 8.
Area of triangle is 1/2*b*8=8
Hence solving for b we get b=2. This point is (11,8). Hence E
Sent from my iPhone using GMAT Club Forum
Area of the square = 8*8
=64
To divide the ratio into 7:1 we have to find a point on the vertical side of the square opposite to (3,10) such that the area of the triangle is 8.
Area of triangle is 1/2*b*8=8
Hence solving for b we get b=2. This point is (11,8). Hence E
Sent from my iPhone using GMAT Club Forum
18 Dec 2017, 04:57
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First instinct is that there will be 2 such points - one lying on x = 11 and the other lying on y = 2.
Looking at the options, we see that we are looking for the point on x = 11. It will be much closer to (11, 10) than to (11, 2) so of the given options, (11, 8) makes the most sense. Let's check.
Area of square = 8*8 = 64
Area of right triangle made by (3, 10), (11, 8) and (11, 10) = (1/2)*2*8 = 8
This is 1/8th of the total area of the square and hence it satisfies our condition.
Answer (E)
Looking at the options, we see that we are looking for the point on x = 11. It will be much closer to (11, 10) than to (11, 2) so of the given options, (11, 8) makes the most sense. Let's check.
Area of square = 8*8 = 64
Area of right triangle made by (3, 10), (11, 8) and (11, 10) = (1/2)*2*8 = 8
This is 1/8th of the total area of the square and hence it satisfies our condition.
Answer (E)
