Share your best tricks/workarounds/techniques

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(I hope a similar topic doesn't exist yet, otherwise please merge it.)

I recently found a technique how to tackle Data Sufficiency questions which helped me to boost my score in that area. I am looking for more such techniques, tricks, and workarounds for specific question types (or whole areas) and thought it would be useful if everyone shared his/her own and we put it together in one place so people can cherry-pick what's useful for them.

It doesn't matter whether it's your own technique or whether you found/learned it somewhere else. It might even be just general knowledge or common sense. It also doesn't matter whether the technique is helpful for everyone or only you. There might be someone who can benefit from it.

I will start...

1-2-TEN/1-2-C Technique for Data Sufficiency questions

Again, this is not my own technique but it's based on Kaplan and BeatTheGMAT, so many of you will know it already. However, I found it highly valuable for me so I will list it here. It's actually 2 techniques combined: Encoding the answer choices (Kaplan) and using a structured elimination method (BeatTheGMAT), and giving it an easy to remember name so it can easily me memorized for the exam.

The technique addresses the following issues of DS questions:
1. The five standard answer choices A-E are lengthy and hard to remember; practicing DS questions I regularly found myself flipping back in the book to see what answer a letter stands for.
2. While everyone knows eliminating answers is good, having a structured approach to eliminate answers is better, because a) you might be able to eliminate several answers in a bunch and b) you can make sure to not miss any potential combination (I sometimes found myself too early deciding for eg. answer A) because the first statement was sufficient, forgetting to check for answer D) each statement is sufficient).
3. Some elimination approaches are not easy to remember, such as the decision tree in the OG.

The technique
  • Step 1-2-TEN: Simply write down these five characters as an easier to remember coding for the answer choices A-E:
    1: Statement 1 alone is sufficient
    2: Statement 2 alone is sufficient
    T: Both statements together are sufficient but neither alone is sufficient
    E: Each statement alone is sufficient
    N: Neither statement is sufficient

    (Of course, you would only write down "1-2-T-E-N" but not the explanations)
  • Step 1-2-C: Check the first statement, check the second statement, check the combined statement (if necessary):
    a) For both statements, check if they are sufficient. If the statement is sufficient, you can strike from your list the other statement (1 or 2 respectively), as well as "T" and "N". If the statement is insufficient, you can strike from your list that statement as well as "E". In many cases, this will already lead to a solution.
    b) If both statements alone are insufficient, check the combination of the two statements. If the combination is sufficient, the answer is "T" (decoded it's "C", of course), otherwise it's "N" (decoded "E")
For better understanding, watch the BeatTheGMAT video on YouTube (note though that in the video they are not using the easier-to-remember coding but the letters A-E).

Elimination Technique for Roman Numeral questions

This simple technique is so super drop-dead logical that I almost hate myself for not getting the idea myself. Instead, I found it in this Kaplan video on YouTube.

Roman Numeral questions ask to identify statements as true or false based on the information given in the question stem. Example:
In which of the following pairs are the two numbers equivalent?

I. (3^0.5)^2 and 1/81
II. 7^-1 and -(1/7)
III. 3^3 x 3^3 and 243

(A) None
(B) II only
(C) III only
(D) I and II
(E) I, II, and III
Same as with the Data Sufficiency questions, while eliminating answers is good, having a structured approach to eliminate answers is better, because you might be able to eliminate several answers in a bunch.

The technique
Rather than testing all statements one after another, start with the statement that occurs most frequently in the answer choices. This way, for many questions, you will be able to eliminate several answer choices at once by only testing one statement, sometimes even narrowing it down to only one answer choice left---which must be the correct one then. Otherwise, if you need to continue eliminating, proceed in the same way or (if all remaining statement occur equally frequently in the remaining answer choices, proceed with the statement that is most easy to solve).

A good example can be found in this Kaplan video on YouTube.
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by Brent@GMATPrepNow » Sat Nov 10, 2012 8:25 am
Nice idea for a thread.

We have some free videos that cover some techniques that allow you to quickly perform certain calculations in your head:

- Calculating combinations: https://www.gmatprepnow.com/module/gmat-counting?id=789
- Dividing by 5: https://www.gmatprepnow.com/module/gener ... es?id=1112
- Squaring numbers ending in 5: https://www.gmatprepnow.com/module/gmat- ... ts?id=1024

Cheers,
Brent
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by Brent@GMATPrepNow » Sat Nov 10, 2012 8:32 am
One of my favorite strategies involves Data Sufficiency (DS) questions featuring geometric figures. These questions typically require you to determine whether or not the given statements force the geometric figure into having one fixed measurement. One strategy for this question type is to mentally grab points and lines on the diagram and try to move them, without breaking any of the restrictions you have been given.

This concept/strategy is covered in greater detail in our free video: https://www.gmatprepnow.com/module/gmat- ... cy?id=1103

The above video is part of a free set of videos that cover everything you need to know about Data Sufficiency questions: https://www.gmatprepnow.com/module/gmat-data-sufficiency

Cheers,
Brent
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by talueng » Sat Nov 10, 2012 10:47 am
Divisibility Rules

This is very basic primary school knowledge but can be very handy for certain questions. The simple rules help to identify whether a (potentially very large) integer can be divided (ie. divided with a remainder of 0) by 2, 3, etc.:

The technique
Rather than having to actually divide the integer by the given (or all the given) number, certain properties of the integer can tell whether it can be divided by certain basic numbers:

An integer can be divided (without a remainder) by...
  • 0: never
  • 1: always
  • 2: if the last digit can be divided by 2 (ie. if the last digit is a 0,2,4,6,8)
  • 3: if the sum of digits can be divided by 3
  • 4: if the integer composed of the last 2 digits can be divided by 4
  • 5: if the last digit is a 0 or 5
  • 6: if the integer can be divided by 2 and 3 (ie. if the last digit can be divided by 2 and the sum of digits can be divided by 3)
  • 8: if the integer composed of the last 3 digits can be divided by 8
  • 9: if the sum of digits can be divided by 9
  • 10: if the last digit is a 0
  • 11: if the alternating sum of digits can be divided by 11 (whereas the alternating sum of digits AS=d0-d1+d2-d3+... ie. subtracting and adding the digits alternately starting with the last digit)
Rules for 7, 13, and other prime numbers exist but are more complex (and, thus, harder to remember). I will write down those for 7 and 13:

An integer can be divided (without a remainder) by...
  • 7: if the special sum of digits SS = d0+3d1+2d2-d3-3d4-2d5+d6+3d7+... can be divided by 7 (di being the digits of the integer)
  • 13: if the special sum of digits SS = d0-3d1+9d2-d3+3d4-9d5+d6-3d7+... can be divided by 7 (di being the digits of the integer)
(I myself am memorizing the rules for 7 as the "1-3-2 rule" and for 13 as the "1-3-9" rule)

Example:
Can n=5,702,400 be divided by...?

2: yes, because the last digit is 0
3: yes, because the sum of digits S=5+7+2+4=18 can be divided by 3
4: yes, because the integer composed of the last 2 digits is 100 which can be divided by 4
5: yes, because the last digit is a 0
6: yes, because it can be divided by 2 and 3 (see above)
8: yes, because the integer composed of the last 3 digits is 400 which can be divided by 8
9: yes, because the sum of digits S=5+7+2+4=18 can be divided by 9
10: yes, because the last digit is a 0
11: yes, because the alternating sum of digits AS=0-0+4-2+0-7+5=0 can be divided by 11
Can n=21,828,534 be divided by 7 and 13?

7: yes, because the special sum of digits SS=4+3*3+2*5-8-3*2-2*8+1+3*2=0 can be divided by 7
13: yes, because the special sum of digits SS=4-3*3+9*5-8+3*2-9*8+1-3*2=-39 can be divided by 13