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12 Easy Pieces (or not?)

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Bunuel
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Re: 12 Easy Pieces (or not?) [#permalink] New post  16 Jan 2015, 02:45
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pacifist85 wrote:
Question 11,

I guess it should have stated that we can only unse each one of these 3 nunnbers only once, right?
Because, I used the same approach as Bunuel, but without knowing that you can only use these nubers once the process an be very long... and I am not even sure if it would end up in the same result - i guess not.

Is there sth in the way that the question is phrased that should have warned me that these numbers were only to be used once?


We are given a set: {1, 3, 6, 7, 7, 7}. I think it should be clear that 1, 3, and 6 should be used once, and 7 thrice.
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Re: 12 Easy Pieces (or not?) [#permalink] New post  08 Dec 2015, 12:25
There are 5 pairs of white, 3 pairs of black and 2 pairs of grey socks in a drawer. If four socks are picked at random what is the probability of getting two socks of the same color?
A. 1/5
B. 2/5
C. 3/4
D. 4/5
E. 1


I totally understand that we don't need to do the calculations here because the answer is obvious. However, I would like to know how we do calculate the answer to get to the official answer.
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chetan2u
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Re: 12 Easy Pieces (or not?) [#permalink] New post  08 Dec 2015, 20:32
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Peltina wrote:
There are 5 pairs of white, 3 pairs of black and 2 pairs of grey socks in a drawer. If four socks are picked at random what is the probability of getting two socks of the same color?
A. 1/5
B. 2/5
C. 3/4
D. 4/5
E. 1


I totally understand that we don't need to do the calculations here because the answer is obvious. However, I would like to know how we do calculate the answer to get to the official answer.


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..
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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  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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Bunuel
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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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Bunuel
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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  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  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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