12 Easy Pieces (or not?)

Hi,
there cant be direct formulas for most of the questions but they are just a step to correct answer..
so if you want the answer the formula way, then it would be something like this...

there are three different coloured socks..
let us first find the scenario where we do not get two of one kind...
1)the first can be picked up, say white, prob=5/10..
2)the second can be picked up, say black, prob=3/9..
3)the third can be picked up, the remaining grey, prob=2/8..
4)the fourth picking up is 0 as no other colour is left prob=0/7..
now these four can be arranged in 4!/2! ways..
prob of two colours of one kind = 1- 5/10 *3/9 * 2/8* 0/7 * 4!/2!=1-0=1
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Hi Bunuel, I tried to refer other posts for this answer but I did not understand. Can you please elaborate this? how M=7k+3..

The row begins with blue marble and ends with red marble. What cases can we have?

{blue, white, red} = 7*0 + 3
{blue, white, red, green, black, yellow, pink} {blue, white, red} = 7*1 + 3
{blue, white, red, green, black, yellow, pink} {blue, white, red, green, black, yellow, pink} {blue, white, red} = 7*2 + 3
{blue, white, red, green, black, yellow, pink} {blue, white, red, green, black, yellow, pink} {blue, white, red, green, black, yellow, pink} {blue, white, red} = 7*3 + 3
...

As you can see in any case the number of marbles must be a multiple of 7 plus 3.

Hope it's clear.
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Is it not the other way around where -1/100 is the least number and -1/18 is the max negative number. I would have picked A as the answer.

No.

-1/18 = ~-0.06 and -1/100 = -0.01

-0.06 < -0.01



As you can see -1/100 is to the right of -1/18.


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You can only apply subtraction when the signs of inequalities are in the opposite directions:

If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from).
Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

You can only add inequalities when their signs are in the same direction:

If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\).
Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).

For more check Manipulating Inequalities.
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Hi Bunuel,

I am unable to understand how did we get 0.001 for 1/10^(n+1).
From what i understand 1/10^(n+1) is of form 1/10^n x 10^1.
Would you please explain?

Thanks

1/10^1 = 0.1
1/10^2 = 0.01
1/10^3 = 0.001
...

Thus, both 1/10^(n+1) and 1/10^n when expressed as decimals are of a type 0.001 (some number of zeros before 1). So, not that both are equal to 0.001 but both will be 0. followed by some number of zeros before 1.
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How do you pick a smart number? Can you show how you strategize about this, since we need to make sure its divisible by18 but also the subsequent calculations wont result in fractional values that dont make sense.

We need a number 2/9 of which is an integer, so the number should be a multiple of 9. We also want (x*2/9)*3/4 = x*1/6 to be in integer, so the number also should be a multiple of 2 (and 3 but since we already established that the number should be a multiple of 9, then we can ignore that here). So, the number should be a multiple of 9*2 = 18.

Number plugging:

How to Do Math on the GMAT Without Actually Doing Math
The Power of Estimation for GMAT Quant
How to Plug in Numbers on GMAT Math Questions
Number Sense for the GMAT
Can You Use a Calculator on the GMAT?
Why Approximate?
GMAT Math Strategies — Estimation, Rounding and other Shortcuts
The 4 Math Strategies Everyone Must Master, Part 1 (1. Test Cases and 2. Choose Smart Numbers.)
The 4 Math Strategies Everyone Must Master, part 2 (3. Work Backwards and 4. Estimate)
Intelligent Guessing on GMAT
How to Avoid Tedious Calculations on the Quantitative Section of the GMAT
GMAT Tip of the Week: No Calculator? No Problem.
The Importance of Sorting Answer Choices on the GMAT

For more check Ultimate GMAT Quantitative Megathread

Hope it helps.
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Are these really sub-600 questions...? If so, these must be the areas I am weakest in...

No, those are quite tricky 700+ questions. The name "12 Easy Pieces (or not?)" is meant to be ironic, juxtaposing the simplicity suggested by 'easy pieces' with the underlying complexity of the questions.
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*Breathing a sigh of relief*

Lol. Thank you, Bunuel.

Doing Charles' study plan, on week 1, and got mixed up (and humbled) with the tags.

Bunuel , KarishmaB, Can you please suggest an easier way to figure out that n = 2 in this case ; Can I infer that n = 2 because there are two zero es in front of 737 ; hence n= 2 will satisfy the inequality ?  Hence , .001 < .00737 < .01 ; How do you arrive at this conclusion all on a sudden that n = 2 ? Please help.

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Use the Estimation concept. 

\(0.00737 = \frac{737}{10^5}\)

737 is less than 1000 but greater than 100. 

\(\frac{100}{10^5}<\frac{737}{10^5} < \frac{1000}{10^5}\)

\(\frac{1}{10^3}<\frac{737}{10^5} < \frac{1}{10^2}\)

So n = 2

Answer (B)

Check this video on Estimations: 
https://youtu.be/4Wy7BrQrjkM
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Thank you  KarishmaB for the wonderful explanation

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