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# My comprehensive Quant Flashcards!

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Intern
Joined: 23 Feb 2013
Posts: 8
Re: My comprehensive Quant Flashcards! [#permalink]

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11 Nov 2013, 14:31
Thanks very much for sharing this. I appreciate your efforts.
Intern
Joined: 08 May 2014
Posts: 1
Concentration: Accounting
Re: My comprehensive Quant Flashcards! [#permalink]

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14 May 2014, 15:43
This is great! Thank you so much for this! Greatly appreciated
Intern
Joined: 21 May 2014
Posts: 1
Re: My comprehensive Quant Flashcards! [#permalink]

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22 May 2014, 11:27
Thanks!! very helpful, quick review is what i need! Ill check if I see any mistakes and ask
Current Student
Joined: 04 Jul 2014
Posts: 306
Location: India
GMAT 1: 640 Q47 V31
GMAT 2: 640 Q44 V34
GMAT 3: 710 Q49 V37
GPA: 3.58
WE: Analyst (Accounting)
Re: My comprehensive Quant Flashcards! [#permalink]

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16 Nov 2014, 10:51
Best quant flashcard! :D
_________________

Cheers!!

JA
If you like my post, let me know. Give me a kudos!

Current Student
Joined: 04 Jul 2014
Posts: 306
Location: India
GMAT 1: 640 Q47 V31
GMAT 2: 640 Q44 V34
GMAT 3: 710 Q49 V37
GPA: 3.58
WE: Analyst (Accounting)
Re: My comprehensive Quant Flashcards! [#permalink]

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16 Nov 2014, 11:02
1
KUDOS
For the many of you who asked for verbal flashcards, take a loot at this: ultimately-comprehensive-gmat-sentence-correction-flash-cards-notes-185033.html#p1416768
_________________

Cheers!!

JA
If you like my post, let me know. Give me a kudos!

Intern
Joined: 24 Nov 2015
Posts: 1
Re: My comprehensive Quant Flashcards! [#permalink]

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28 Nov 2015, 02:39
AWESOME thanks a lot
Intern
Joined: 06 Mar 2015
Posts: 29
Re: My comprehensive Quant Flashcards! [#permalink]

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11 May 2016, 08:55
Thanks a ton.

They are indeed helpful. Great Work.
Manager
Joined: 23 Jun 2009
Posts: 197
Location: Brazil
GMAT 1: 470 Q30 V20
GMAT 2: 620 Q42 V33
Re: My comprehensive Quant Flashcards! [#permalink]

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08 Jun 2016, 08:52
1
This post was
BOOKMARKED
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.

Soon I will prepare a Verbal also based on the available SC and CR material available in the forum

Hope you guys like it
Intern
Joined: 15 Jun 2016
Posts: 49
Location: India
Concentration: Technology, Strategy
GMAT 1: 730 Q50 V39
Re: My comprehensive Quant Flashcards! [#permalink]

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10 Jul 2016, 04:41
Adding the 500th kudo Nice collection of quants concepts....
Intern
Joined: 25 Dec 2011
Posts: 1
Re: My comprehensive Quant Flashcards! [#permalink]

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21 Jul 2016, 19:49
And it's a ppt file which can be edited. Kudos!
Intern
Joined: 06 Sep 2016
Posts: 1
Re: My comprehensive Quant Flashcards! [#permalink]

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14 Sep 2016, 15:01
Excellent work.keep it up
Manager
Joined: 26 Jun 2013
Posts: 91
Location: India
Schools: ISB '19, IIMA , IIMB
GMAT 1: 590 Q42 V29
GPA: 4
WE: Information Technology (Retail Banking)
Re: My comprehensive Quant Flashcards! [#permalink]

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04 Feb 2017, 05:15
miguelmick wrote:
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!

Edit: Also, please see this post for the GMAT Club Flashcards:
http://gmatclub.com/forum/gmat-flashcards-108651.html

Thanks a lot....these are so useful
_________________

Remember, if it is a GMAT question, it can be simplified elegantly.

Intern
Joined: 21 Sep 2016
Posts: 14
Location: Zambia
Schools: Duke '20 (A)
GPA: 3
Re: My comprehensive Quant Flashcards! [#permalink]

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20 Jul 2017, 06:43
miguelmick wrote:

Intern
Joined: 16 Dec 2016
Posts: 12
Location: United States (DC)
GMAT 1: 720 Q47 V42
WE: Consulting (Consulting)
Re: My comprehensive Quant Flashcards! [#permalink]

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20 Jul 2017, 18:02
These flash cards were a real help. Thanks!!
Intern
Joined: 19 Nov 2017
Posts: 3
Re: My comprehensive Quant Flashcards! [#permalink]

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22 Nov 2017, 12:08
Dear Miguel
Thank you very much for the flashcards, they will definitively help me.
Please, I have a question for the flascard related to Geometry:coordinate plan, Slide 27:
" Consider that the distance between A and B is 4x, and we need a point which is 3x distant from A and x distance from B.
Therefore, 3x = 4 --- x = 0.75. The point is located on x-axis -2.75"

How Do I get 3x=4? I dont understand that part
Intern
Joined: 09 Nov 2017
Posts: 5

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23 Nov 2017, 19:20
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 x-coordinate and y-coordinate, 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 y-intercept of the line connecting A and B:
slope: rise/run = (6-0)/(-5--2) = 6 / -3 = -2
y-intercept: 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.
Manager
Joined: 21 Jul 2014
Posts: 69
GMAT Date: 07-30-2015
Re: My comprehensive Quant Flashcards! [#permalink]

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10 Feb 2018, 10:47
1
KUDOS
They are indeed one of the best flash cards available. You can just breeze though the cards.
Re: My comprehensive Quant Flashcards!   [#permalink] 10 Feb 2018, 10:47

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