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26 Apr 2020, 04:37
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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:
45%
(medium)
Question Stats:
65% (03:00) correct
35%
(02:37)
wrong
based on 17
sessions
History
Date
Time
Result
Not Attempted Yet
Attachment:
GMATinsight.jpeg [ 31.09 KiB | Viewed 1787 times ]
A) 93
B) 95
C) 100
D) 105
E) 110
26 Apr 2020, 14:00
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GMATinsight
An educated guess for the answer would be C because it is the only perfect square.
Set \(PX = x\) and \(YR = y\). Then it follows that \(QX = \frac{24}{x}\) and \(YS = \frac{21}{y}\). We have the following equations:
(1) AB = BC, in terms of x and y vertically we have \(x + 3 + y\), and horizontally we have \(\frac{24}{x} + \frac{21}{y} - 3\). These are equal.
(2) XB + XD = YB + YD, then divide the entire equation by \(\sqrt{2}\) to get QX + PX = YR + YS. Which is \(x + \frac{24}{x} = y + \frac{21}{y}\)
It will be hard to isolate one variable at first but if we add the equations we get \(2x + 3 = \frac{42}{y} - 3\) and \(x = \frac{21}{y} - 3 = \frac{21-3y}{y}\).
Plug this in either equation to get an equation with y only, cleaning it up gives \(y^2 + 4y -21 = 0\) and \(y = 3\), \(x = 4\). The side length of the big square is x + y + 3 = 10. Thus the answer is 100, C.
What we can take away from this question is that we would want to use the least amount of variables whenever we can to make solving the system easier.
