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12 Easy Pieces (or not?)

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Peltina
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Re: 12 Easy Pieces (or not?) [#permalink] New post   09 Dec 2015, 14:34
chetan2u wrote:
Hi,

in such questions, we look at the worst scenario..
here it would be getting different coloured socks , every time we pick socks..

but there are only three different coloured socks, so as a worst case , the first three picked up are different colours..
the fourth which you pick up has to be one of these three colours, thus ensuring that we have two socks of same colour..
hence prob is 1, that is it is sure to have two socks of atleast one colour..



Thank you! I was looking more for the formula that we should use to get to the answer P=1 in case we fail to notice that the answer is obvious and needs no calculation.
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Re: 12 Easy Pieces (or not?) [#permalink] New post   09 Dec 2015, 22:39
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Peltina wrote:
chetan2u wrote:
Hi,

in such questions, we look at the worst scenario..
here it would be getting different coloured socks , every time we pick socks..

but there are only three different coloured socks, so as a worst case , the first three picked up are different colours..
the fourth which you pick up has to be one of these three colours, thus ensuring that we have two socks of same colour..
hence prob is 1, that is it is sure to have two socks of atleast one colour..


5 pairs of white, 3 pairs of black and 2 pairs of grey socks
Thank you! I was looking more for the formula that we should use to get to the answer P=1 in case we fail to notice that the answer is obvious and needs no calculation.


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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Re: 12 Easy Pieces (or not?) [#permalink] New post   27 Mar 2017, 00:32
Bunuel wrote:
9. Julie is putting M marbles in a row in a repeating pattern: blue, white, red, green, black, yellow, pink. If the row begins with blue marble and ends with red marble, then which of the following could be the value of M?
A. 22
B. 30
C. 38
D. 46
E. 54

There are total of 7 different color marbles in a pattern. Now, as the row begins with blue marble and ends with red marble (so ends with 3rd marble in a pattern) then M=7k+3. The only answer choice which is multiple of 7 plus 3 is 38=35+3.

Answer: C.


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..
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Re: 12 Easy Pieces (or not?) [#permalink] New post   27 Mar 2017, 00:48
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RMD007 wrote:
Bunuel wrote:
9. Julie is putting M marbles in a row in a repeating pattern: blue, white, red, green, black, yellow, pink. If the row begins with blue marble and ends with red marble, then which of the following could be the value of M?
A. 22
B. 30
C. 38
D. 46
E. 54

There are total of 7 different color marbles in a pattern. Now, as the row begins with blue marble and ends with red marble (so ends with 3rd marble in a pattern) then M=7k+3. The only answer choice which is multiple of 7 plus 3 is 38=35+3.

Answer: C.


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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Re: 12 Easy Pieces (or not?) [#permalink] New post   29 Mar 2017, 13:59
Bunuel wrote:
12. If \({-\frac{1}{3}}\leq{x}\leq{-\frac{1}{5}}\) and \({-\frac{1}{2}}\leq{y}\leq{-\frac{1}{4}}\), what is the least value of \(x^2*y\) possible?
A. -1/100
B. -1/50
C. -1/36
D. -1/18
E. -1/6

To get the least value of \(x^2*y\), which obviously will be negative, try to maximize absolute value of \(x^2*y\), as more is the absolute value of a negative number "more" negative it is (the smallest it is).

To maximize \(|x^2*y|\) pick largest absolute values possible for \(x\) and \(y\): \((-\frac{1}{3})^2*(-\frac{1}{2})=-\frac{1}{18}\). Notice that: -1/18<-1/36<-1/50<-1/100, so -1/100 is the largest number and -1/18 is the smallest number (we cannot obtain -1/6 from x^2*y or else it would be the correct answer).

Answer: D.


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.
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Re: 12 Easy Pieces (or not?) [#permalink] New post   29 Mar 2017, 21:38
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vs224 wrote:
Bunuel wrote:
12. If \({-\frac{1}{3}}\leq{x}\leq{-\frac{1}{5}}\) and \({-\frac{1}{2}}\leq{y}\leq{-\frac{1}{4}}\), what is the least value of \(x^2*y\) possible?
A. -1/100
B. -1/50
C. -1/36
D. -1/18
E. -1/6

To get the least value of \(x^2*y\), which obviously will be negative, try to maximize absolute value of \(x^2*y\), as more is the absolute value of a negative number "more" negative it is (the smallest it is).

To maximize \(|x^2*y|\) pick largest absolute values possible for \(x\) and \(y\): \((-\frac{1}{3})^2*(-\frac{1}{2})=-\frac{1}{18}\). Notice that: -1/18<-1/36<-1/50<-1/100, so -1/100 is the largest number and -1/18 is the smallest number (we cannot obtain -1/6 from x^2*y or else it would be the correct answer).

Answer: D.


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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Re: 12 Easy Pieces (or not?) [#permalink] New post   04 Aug 2019, 13:58
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Shef08 wrote:
Bunuel wrote:
4. If -3<x<5 and -7<y<9, which of the following represent the range of all possible values of y-x?
A. -4<y-x<4
B. -2<y-x<4
C. -12<y-x<4
D. -12<y-x<12
E. 4<y-x<12

To get max value of y-x take max value of y and min value of x: 9-(-3)=12;
To get min value of y-x take min value of y and max value of x: -7-(5)=-12;

Hence, the range of all possible values of y-x is -12<y-x<12.

Answer: D.




Why is it not A? Can’t we simply subtract x from y? Then the answer will be A?

Posted from my mobile device



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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Re: 12 Easy Pieces (or not?) [#permalink] New post   28 Aug 2019, 06:37
Bunuel wrote:
10. If \(n\) is an integer and \(\frac{1}{10^{n+1}}<0.00737<\frac{1}{10^n}\), then what is the value of n?
A. 1
B. 2
C. 3
D. 4
E. 5

Also no need for algebraic manipulation. 1/10^(n+1) is 10 times less than 1/10^n, and both when expressed as decimals are of a type 0.001 (some number of zeros before 1) --> so the given expression to hold true we should have: 0.001<0.00737<0.01, which means that n=2 (1/10^n=0.01 --> n=2).

Answer: B.


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
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Re: 12 Easy Pieces (or not?) [#permalink] New post   28 Aug 2019, 06:53
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swatjazz wrote:
Bunuel wrote:
10. If \(n\) is an integer and \(\frac{1}{10^{n+1}}<0.00737<\frac{1}{10^n}\), then what is the value of n?
A. 1
B. 2
C. 3
D. 4
E. 5

Also no need for algebraic manipulation. 1/10^(n+1) is 10 times less than 1/10^n, and both when expressed as decimals are of a type 0.001 (some number of zeros before 1) --> so the given expression to hold true we should have: 0.001<0.00737<0.01, which means that n=2 (1/10^n=0.01 --> n=2).

Answer: B.


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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Re: 12 Easy Pieces (or not?) [#permalink] New post   29 Dec 2021, 17:11
Bunuel wrote:
7. After 2/9 of the numbers in a data set A were observed, it turned out that 3/4 of those numbers were non-negative. What fraction of the remaining numbers in set A must be negative so that the total ratio of negative numbers to non-negative numbers be 2 to 1?
A. 11/14
B. 13/18
C. 4/7
D. 3/7
E. 3/14

If choose variable for set A there will be too many fractions to manipulate with, so pick some smart #: let set A contain 18 numbers.

"2/9 of the numbers in a data set A were observed" --> 4 observed and 18-4=14 numbers left to observe;
"3/4 of those numbers were non-negative" --> 3 non-negative and 1 negative;
Ratio of negative numbers to non-negative numbers to be 2 to 1 there should be total of 18*2/3=12 negative numbers, so in not yet observed part there should be 12-1=11 negative numbers. Thus 11/14 of the remaining numbers in set A must be negative.

Answer: A.



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.
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Re: 12 Easy Pieces (or not?) [#permalink] New post   30 Dec 2021, 01:31
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testtakerstrategy wrote:
Bunuel wrote:
7. After 2/9 of the numbers in a data set A were observed, it turned out that 3/4 of those numbers were non-negative. What fraction of the remaining numbers in set A must be negative so that the total ratio of negative numbers to non-negative numbers be 2 to 1?
A. 11/14
B. 13/18
C. 4/7
D. 3/7
E. 3/14

If choose variable for set A there will be too many fractions to manipulate with, so pick some smart #: let set A contain 18 numbers.

"2/9 of the numbers in a data set A were observed" --> 4 observed and 18-4=14 numbers left to observe;
"3/4 of those numbers were non-negative" --> 3 non-negative and 1 negative;
Ratio of negative numbers to non-negative numbers to be 2 to 1 there should be total of 18*2/3=12 negative numbers, so in not yet observed part there should be 12-1=11 negative numbers. Thus 11/14 of the remaining numbers in set A must be negative.

Answer: A.



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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Re: 12 Easy Pieces (or not?) [#permalink] New post   13 Nov 2023, 18:06
Are these really sub-600 questions...? If so, these must be the areas I am weakest in...
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Re: 12 Easy Pieces (or not?) [#permalink] New post   13 Nov 2023, 23:26
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username10101 wrote:
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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12 Easy Pieces (or not?) [#permalink] New post   15 Nov 2023, 19:39
Bunuel wrote:
username10101 wrote:
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.


*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.
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Re: 12 Easy Pieces (or not?) [#permalink] New post   15 Apr 2024, 15:50
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.
Bunuel wrote:
10. If \(n\) is an integer and \(\frac{1}{10^{n+1}}<0.00737<\frac{1}{10^n}\), then what is the value of n?
A. 1
B. 2
C. 3
D. 4
E. 5

There is no need for algebraic manipulation to answer this question.

\(\frac{1}{10^{n+1} }\) is 10 times smaller than \(\frac{1}{10^n}\). When expressed as decimals, both have the form 0.001 (with a varying number of zeros preceding the 1). For the given expression to be valid, we should have: \(0.001 < 0.00737 < 0.01\). This implies that \(n=2\) (since \(\frac{1}{10^n}=0.01\), then \(n=2\)).

Answer: B.

­
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12 Easy Pieces (or not?) [#permalink] New post   Updated on: 16 Apr 2024, 00:35
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sayan640 wrote:
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.
Bunuel wrote:
10. If \(n\) is an integer and \(\frac{1}{10^{n+1}}<0.00737<\frac{1}{10^n}\), then what is the value of n?
A. 1
B. 2
C. 3
D. 4
E. 5

There is no need for algebraic manipulation to answer this question.

\(\frac{1}{10^{n+1} }\) is 10 times smaller than \(\frac{1}{10^n}\). When expressed as decimals, both have the form 0.001 (with a varying number of zeros preceding the 1). For the given expression to be valid, we should have: \(0.001 < 0.00737 < 0.01\). This implies that \(n=2\) (since \(\frac{1}{10^n}=0.01\), then \(n=2\)).

Answer: B.

­

­
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
 ­

Originally posted by KarishmaB on 15 Apr 2024, 21:06.
Last edited by KarishmaB on 16 Apr 2024, 00:35, edited 1 time in total.
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12 Easy Pieces (or not?) [#permalink] New post   15 Apr 2024, 21:20
Thank you  KarishmaB for the wonderful explanation :)
KarishmaB wrote:
sayan640 wrote:
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.
Bunuel wrote:
10. If \(n\) is an integer and \(\frac{1}{10^{n+1}}<0.00737<\frac{1}{10^n}\), then what is the value of n?
A. 1
B. 2
C. 3
D. 4
E. 5

There is no need for algebraic manipulation to answer this question.

\(\frac{1}{10^{n+1} }\) is 10 times smaller than \(\frac{1}{10^n}\). When expressed as decimals, both have the form 0.001 (with a varying number of zeros preceding the 1). For the given expression to be valid, we should have: \(0.001 < 0.00737 < 0.01\). This implies that \(n=2\) (since \(\frac{1}{10^n}=0.01\), then \(n=2\)).

Answer: B.

­

­
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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Re: 12 Easy Pieces (or not?) [#permalink] New post   22 Apr 2024, 05:07
sayan640 wrote:
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.
Bunuel wrote:
10. If \(n\) is an integer and \(\frac{1}{10^{n+1}}<0.00737<\frac{1}{10^n}\), then what is the value of n?
A. 1
B. 2
C. 3
D. 4
E. 5

There is no need for algebraic manipulation to answer this question.

\(\frac{1}{10^{n+1} }\) is 10 times smaller than \(\frac{1}{10^n}\). When expressed as decimals, both have the form 0.001 (with a varying number of zeros preceding the 1). For the given expression to be valid, we should have: \(0.001 < 0.00737 < 0.01\). This implies that \(n=2\) (since \(\frac{1}{10^n}=0.01\), then \(n=2\)).

Answer: B.

­


If it were me, I'd get rid of the pesky decimal.

.00737*10^5 =737

So multiply all three by 10^5:

10^(4-n)<737<10^(5-n)

737 is less than 1000 and greater than 100, meaning n must be 2 by observation.

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Re: 12 Easy Pieces (or not?) [#permalink]
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