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25 Jul 2017, 22:45
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
45% (02:16) correct
55%
(02:12)
wrong
based on 110
sessions
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In the diagram below, what is the length of the line segment AC?
(1) The coordinates of P and Q are (0, 6) and (2, 0), respectively.
(2) The length of the line segment AB is 2.
2017-07-26_0944.png [ 7.01 KiB | Viewed 7885 times ]
(1) The coordinates of P and Q are (0, 6) and (2, 0), respectively.
(2) The length of the line segment AB is 2.
Attachment:
2017-07-26_0944.png [ 7.01 KiB | Viewed 7885 times ]
26 Jul 2017, 07:03
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Bunuel
It should be C
Before we solve this, consider that this diagram may not be drawn to scale as is the case with many GMAT questions.
Statement 1: Insufficient
It gives us nothing about the projections \(PA\) and \(QC\). It does however give us the slope of the line at \(-3\).
Statement 2: Insufficient
This gives us no information about point the line \(PQ\).
Statement 1 + 2: Sufficient
Since we have a fixed slope and angles of the triangles, if we slide the line \(AB\) down where it meets P, this would not change the dimensions of the triangle as we have kept the length \(AB\) same.
Now, we can find the point of intersection of two lines \(AC\) and \(BC\) by setting \(x = 2\) (This may be hard to digest but note that we have slid the line \(AB\) down to the point \(P\) on Y axis. So draw a horizontal line at P that goes moves right to the tune of 2).
\(y = mx + c\)
\(y = -3*2 + 6\) -- (we already know y intersect is 6).
\(y = 0\)
This means \(C\) point is actually the same as \(Q\). (moving 2 right makes y = 0, same coordinates as \(Q\)) Hence,
\(AC = PQ = \sqrt{2^2 + 6^2}\)
26 Jul 2017, 13:46
Kudos
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Ans C:
St. 1: Provides us with the slope of line AC = line PQ = -3, using Coordinates of P and Q.
And, Slope = Rise/Run = BC/AB = -3
St. 2: gives AB=2
Using St 1+2: BC=6 and Hence, we can calculate AC using Pythagoras.
St. 1: Provides us with the slope of line AC = line PQ = -3, using Coordinates of P and Q.
And, Slope = Rise/Run = BC/AB = -3
St. 2: gives AB=2
Using St 1+2: BC=6 and Hence, we can calculate AC using Pythagoras.
