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My comprehensive Quant Flashcards!
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Updated on: 26 Apr 2018, 00:47
Hello, You will find attached my flashcards. There are a few mistakes, so take care. They cover all the quant topics and are filled with examples. If you enjoy them, please don't forget to award me Kudos! Thanks for that!FOR MORE FLASHCARDS SEE THIS POST.
Originally posted by miguelmick on 22 Feb 2011, 08:46.
Last edited by Bunuel on 26 Apr 2018, 00:47, edited 1 time in total.
Edited.



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My comprehensive Quant Flashcards!
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23 Feb 2011, 18:17



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Re: My comprehensive Quant Flashcards!
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07 Jul 2011, 14:30
Well organized.... But for weighted average, I feel an alternative approach will be easy Quote: Percents and weighted averages Cereal K is 10% sugar and Cereal B is 2% sugar. What should be the ratio of them to produce a 4% sugar cereal? Technique: Pick a smart number for one of the quantities and call the other quantity x. For example, picking 100 grams for Cereal K: 100*0.1 + 0.02*x = 0.04(100 + x) 10 + 0.02x = 4 + 0.04x 0.02x = 6 x = 300 So, the ratio is 3 parts of Cereal B to each part of Cereal K, or 1:3
10k+2b=4(k+b) ==>6k=2b ==> k:b = 1:3Quote: Percent Change and Weighted Averages The revenue from pen sales was up 5%, but the revenue from pencil sales declined 13%. If the overall revenue was down 1%, what was the ratio of pen and pencil revenues? 105 + 0.87x = 0.99(100 + x) 6 = 0.12x x = 50 So, the ratio is 2:1 105Pen+87Pencil=99(Pen+Pencil) ==> 6Pen = 12Pencil ==> Pen : Pencil = 2:1Let me know your thoughts... In fact this approach can be applied to any mixture problem : add/replace/remove combinations
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Re: My comprehensive Quant Flashcards!
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20 Jul 2011, 11:10
These flashcards are very useful. I have updated file with a table of content. Hope you find it helpful. Thanks.



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Re: My comprehensive Quant Flashcards!
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16 Nov 2014, 10:02
For the many of you who asked for verbal flashcards, take a loot at this: ultimatelycomprehensivegmatsentencecorrectionflashcardsnotes185033.html#p1416768
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Re: My comprehensive Quant Flashcards!
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08 Jun 2016, 07:52
Hi, guys! Such a great material! So, I put the entire content to an "actual" flashcard program and improved it with visual material to illustrate parts of it. It is already helping me anywhere I go since it is available for Android, IOs, and desktop version. http://www.cram.com/search?query=felippeoliveria&search_in%5B%5D=username&image_filter=all&period=any&sm=1Soon I will prepare a Verbal also based on the available SC and CR material available in the forum Hope you guys like it



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Re: My comprehensive Quant Flashcards!
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23 Nov 2017, 18:20
This is a nice comprehensive guide  thanks for putting it together.
Two things I noticed:
1) Slide 26 (Geometry  Coordinate Plane): The solution to the equations y = 4x + 10 and 2x + 3y = 26 is x = 2/7 and y = 62/7. The stated solution of x = 4 and y = 6 does not satisfy the equation y = 4x + 10.
2) Slide 27 (Geometry  Coordinate Plane): What are the coordinates for the point on line AB that is 3 times as far from A as from B and that is between points A and B, knowing that A = (5, 6) and B = (2, 0)? My solution (below) is different from that in the powerpoint. The methodology in the powerpoint makes an incorrect assumption that the target point, which lays 1/4th of the way between B and A, has an xcoordinate and ycoordinate, each of which are also 1/4th of the way between B and A.
Let point C represent the solution to the question.
Slope and yintercept of the line connecting A and B: slope: rise/run = (60)/(52) = 6 / 3 = 2 yintercept: y = 2x + b. substituting in x and y coordinates of B gives: 0 = (2)(2) + b => b = 4 Therefore equation of the line connecting A and B is y = 2x  4
The distance between A and B, using pythagorean theorem: Length of side 1 = absolute value of 5(2) = 3 Length of side 2 = absolute value of 6  0 = 6 Length of hypotenuse = sqrt(9+36) = sqrt(45) = 3 sqrt(5)
Therefore we know that the distance between B and C (i.e., the hypotenuse of the right triangle with vertices at B and C) is: 3 sqrt(5) / 4 Given the slope of the line containing A, B and C is 2, we know that the right triangle with vertices at B and C has legs in a 1:2 ratio.
Let x = the length of the horizontal leg of the right triangle with vertices at B and C. Let y = the length of the vertical leg. We know that y = 2x. Therefore: (x)^2 + (2x)^2 = 3 sqrt(5) / 4 5x^2 = 3 sqrt(5) / 4 x^2 = 3 sqrt(5) / 20 x = sqrt(3 sqrt(5)) / (2 sqrt(5)), which is approximately equal to 1.15829
Therefore the coordinates of point C can be represented in relation to point B: C = (2  x, 0 + 2x), in which x is the irrational number which we calculated above.



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Re: My comprehensive Quant Flashcards!
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09 Jul 2018, 22:08
This is so helpful..Thanks !!



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27 Oct 2018, 07:38
Can someone explain why on slide 97, the least possible value of k includes two 2's, and not one?
"The prime factorization of k³ results AT LEAST on 2,2,2,2,3,5. So, since k³ = k.k.k, each k has at least two 2’s, one 3 and one 5. So, k’s least possible value is 2.2.3.5 = 60"




Re: My comprehensive Quant Flashcards! &nbs
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27 Oct 2018, 07:38






