Quantitative Comparison, Basic Practice Set: 10 out of 10
Quantitative Comparison, Intermediate Practice Set: 9 out of 10
Quantitative Comparison, Advanced Practice Set: 7 out of 10
About what I expected, given the difficulty levels.
From the Kaplan GRE verbal workbook:
Text Completion (1-blank) Practice Set: 5 out of 5
Text Completion (2- and 3-blank) Practice Set: 5 out of 5
Suspiciously easy, those were.
And now: the math problem that stumped me, but whose solution prompted a Homer Simpson-style "D'oh!" slap to the forehead. Behold:
Answer selections for Quantitative Comparison problems are always as follows:
A. Quantity A is greater
B. Quantity B is greater
C. Quantities A and B are equal
D. The answer cannot be determined from the information provided.
Have fun, and remember that figures are not necessarily drawn to scale. I'll provide the answer in the comments section later, if anyone is interested.
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This excellent Khan Academy video pretty much gives you the answer to the Quant Comparison question. Watch it only if you've given up.
ReplyDeleteI guessed the right answer, but I would not have been able to give the proof in that video. I must have been dredging up something in my subconscious, something I learned long ago. But it's nice to be reminded about the why. Awesome video.
ReplyDeleteA nice proof, I suppose, but he sure did it the hard way. Since he already had the theorem relating inscribed angles and central angles that subtending the same arc, the fact that an inscribed angle that subtends half the circle is perforce a 90 degree angle is a one-stepper.
ReplyDeleteAddofio
Addofio,
ReplyDeleteTrue enough. A diameter is a 180-degree central angle subtending half a circle, so an inscribed angle would indeed have to be half that measure.
If only I'd remembered all that from my old geometry class.
The Kaplan manual's solution assumes the examinee has forgotten those properties of angles and circles. What it does, instead, is construct a line segment, a radius, going from the center-- arbitrarily labeled D-- to point C. At that point, it's just a matter of deducing degree measures based on one's knowledge of isosceles triangles: if angle ABC is 20 degrees, then angle DCB is also 20 degrees, and angle CDB must be 140 degrees, thereby making angle ADC 40 degrees, which makes angles ACD and BAC both 70 degrees.
It's a lot faster if one remembers the relevant geometric properties. I didn't, and originally picked "D" as my answer. Ever since reading the solution, I've been scratching my head and marveling at my own stupidity.