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12 Easy Pieces (or not?)

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Muki
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Re: 12 Easy Pieces (or not?) [#permalink] New post   20 Oct 2012, 00:10
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

This is from Bunuel's "12 Easy Pieces (or not?)" collection. I understand and agree with the simple explanation provided by Bunuel. However, I am wondering what should be the correct algebraic approach to find this probability using Combinations.

Please provide explanations, as the answer has already been provided by Bunuel.
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Re: 12 Easy Pieces (or not?) [#permalink] New post   20 Oct 2012, 02:11
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Muki 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

This is from Bunuel's "12 Easy Pieces (or not?)" collection. I understand and agree with the simple explanation provided by Bunuel. However, I am wondering what should be the correct algebraic approach to find this probability using Combinations.

Please provide explanations, as the answer has already been provided by Bunuel.


Here is one:

There are \(3^4=81\) possibilities to choose the colors of the 4 socks. Here order counts - 3 choices for the first sock, 3 for the second,...We have all the time 3 color choices, because we have at least 4 socks of the same color for each color.

Let's count the number of possibilities to choose at least twice the same color:
2 socks of the same color and 1 of each of the other two colors - \(3\cdot\frac{4!}{2!}=36\) - 3 choices for the color with 2 socks, then 4! permutations of the 4 socks, divide by 2! because 2 socks are of the same color
2 colors, 2 socks of each color - \(\frac{3\cdot{2}}{2}\cdot{\frac{4!}{2!2!}}=18\) - choose 2 colors out of 3, then 4! permutations ... divide...
3 socks of the same color, 1 of a different color - \(3\cdot{2}\cdot{\frac{4!}{3!}}=24\) - 3 choices for the first color (with 3 socks), 2 choices for the other sock, 4! ... divide ...
finally, all 4 socks of the same color - 3 possibilities

\(\frac{36+18+24+3}{81}=\frac{81}{81}=1\)

Is this worth doing?

If you think of the complementary event - no two socks of the same color:
1st sock - 3 color choices
2nd sock - 2 color choices
3rd sock - 1 color choice
4th sock - 0 choice, we don't have a fourth color
So, number of choices for 4 socks of different colors is 0. Doesn't matter how many for the total number of possible choices, 0 divided by anything not zero is still 0!!! So the requested probability is 1 - 0 = 1.
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Re: 12 Easy Pieces (or not?) [#permalink] New post   22 Oct 2012, 03:40
Bunuel wrote:
5. The angles in a triangle are x, 3x, and 5x degrees. If a, b and c are the lengths of the sides opposite to angles x, 3x, and 5x respectively, then which of the following must be true?
I. c>a+b
II. c^2>a^2+b^2
III. c/a/b=10/6/2

A. I only
B. II only
C. III only
D. I and III only
E. II and III only

According to the relationship of the sides of a triangle: the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. Thus I and III can never be true: one side (c) can not be larger than the sum of the other two sides (a and b). Note that III is basically the same as I: if c=10, a=6 and b=2 then c>a+b, which can never be true. Thus even not considering the angles, we can say that only answer choice B (II only) is left.

Answer: B.

Now, if interested why II is true: as the angles in a triangle are x, 3x, and 5x degrees then x+3x+5x=180 --> x=20, 3x=60, and 5x=100. Next, if angle opposite c were 90 degrees, then according to Pythagoras theorem c^2=a^+b^2, but since the angel opposite c is more than 90 degrees (100) then c is larger, hence c^2>a^+b^2.


Let the side opposite to 5x = 10 , 3x = 9 and x = 8
C^2 > a^2 + b^2
wont hold true
pls explain
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Re: 12 Easy Pieces (or not?) [#permalink] New post   22 Oct 2012, 05:21
Archit143 wrote:
Bunuel wrote:
5. The angles in a triangle are x, 3x, and 5x degrees. If a, b and c are the lengths of the sides opposite to angles x, 3x, and 5x respectively, then which of the following must be true?
I. c>a+b
II. c^2>a^2+b^2
III. c/a/b=10/6/2

A. I only
B. II only
C. III only
D. I and III only
E. II and III only

According to the relationship of the sides of a triangle: the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. Thus I and III can never be true: one side (c) can not be larger than the sum of the other two sides (a and b). Note that III is basically the same as I: if c=10, a=6 and b=2 then c>a+b, which can never be true. Thus even not considering the angles, we can say that only answer choice B (II only) is left.

Answer: B.

Now, if interested why II is true: as the angles in a triangle are x, 3x, and 5x degrees then x+3x+5x=180 --> x=20, 3x=60, and 5x=100. Next, if angle opposite c were 90 degrees, then according to Pythagoras theorem c^2=a^+b^2, but since the angel opposite c is more than 90 degrees (100) then c is larger, hence c^2>a^+b^2.


Let the side opposite to 5x = 10 , 3x = 9 and x = 8
C^2 > a^2 + b^2
wont hold true
pls explain


The angles are uniquely determined: 20, 60 and 100.
Then, the sides cannot be anything you wish. All the triangles with angles 20, 60, and 180 are similar, which means, once you have fixed one side, the other two are uniquely determined.
For example, if you consider that the side opposing the 20 angle is 10, then the side opposing the 60 angle should be approximately 25.3209 , it cannot be 9.
You need trigonometry (which is out of the scope of the GMAT) to determine the sides. But definitely, they cannot be 10, 9 and 8.
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Re: 12 Easy Pieces (or not?) [#permalink] New post   01 May 2013, 23:23
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.


Those values of x and y you had considered would be right if there was an <= symbol at all the places where there are inequality signs. But, provided there is no = symbol along with < and >, then won't the range be -10 to +10? Please explain!
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Re: 12 Easy Pieces (or not?) [#permalink] New post   02 May 2013, 03:13
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sharmila79 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.


Those values of x and y you had considered would be right if there was an <= symbol at all the places where there are inequality signs. But, provided there is no = symbol along with < and >, then won't the range be -10 to +10? Please explain!


If y=8.9 and x=-2.9, then y-x=11.8.
If y=-6.9 and x=4.9, then y-x=-11.8.

So, your range (-10 , 10) is clearly wrong.

Consider the following approach, we have -3<x<5 and -7<y<9,

Add y<9 and -3<x --> y-3<9+x --> y-x<12;
Add -7<y and x<5 --> -7+x<y+5 --> -12<y-x;

So, we have that -12<y-x<12.

Hope it's clear.
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Re: 12 Easy Pieces (or not?) [#permalink] New post   07 Jul 2013, 12:59
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.


Sorry, this may be a silly thing to ask, I don't understand the problem.
What exactly is it asking?

Can someone please explain?

I get that it is asking how many marbles does Julie have. Here is what I am understanding:

Basically there are marbles with seven different colors. Out of which blue, white and red always stay in that order since these three form a pattern. The rest of the four marbles of different colors can be in any order. But the 8th marble will always be blue followed by white and followed by white.
Is this thinking/approach correct??

If yes, where does 38 come from and if it is not correct, please tell me what's wrong.

Thanks
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Re: 12 Easy Pieces (or not?) [#permalink] New post   07 Jul 2013, 13:10
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jjack0310 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.


Sorry, this may be a silly thing to ask, I don't understand the problem.
What exactly is it asking?

Can someone please explain?

I get that it is asking how many marbles does Julie have. Here is what I am understanding:

Basically there are marbles with seven different colors. Out of which blue, white and red always stay in that order since these three form a pattern. The rest of the four marbles of different colors can be in any order. But the 8th marble will always be blue followed by white and followed by white.
Is this thinking/approach correct??

If yes, where does 38 come from and if it is not correct, please tell me what's wrong.

Thanks


No, that's not correct.

The question asks to determine how many marbles Julie has.

The pattern is always the same {blue, white, red, green, black, yellow, pink}{blue, white, red, green, black, yellow, pink}{blue, white, red, green, black, yellow, pink}...

At some point Julie does not have enough marbles to end the pattern and the row ends with a red marble: {blue, white, red}.

For example, it could happen if she had 7+3=10 marbles:
{blue, white, red, green, black, yellow, pink}{blue, white, red}

Or 7*2+3=17 marbles:
{blue, white, red, green, black, yellow, pink}{blue, white, red, green, black, yellow, pink}{blue, white, red}

Or: 7*3+3=24 marbles:
{blue, white, red, green, black, yellow, pink}{blue, white, red, green, black, yellow, pink}{blue, white, red, green, black, yellow, pink}{blue, white, red}

As you can see the number of marbles is always a multiple of 7 plus 3.

The only answer choice which is multiple of 7 plus 3 is 38 = 7*5+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, green, black, yellow, pink}{blue, white, red, green, black, yellow, pink}{blue, white, red}.

Hope it's clear.
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Re: 12 Easy Pieces (or not?) [#permalink] New post   17 Sep 2013, 03:13
Bunuel wrote:
2. If x is an integer and 9<x^2<99, then what is the value of maximum possible value of x minus minimum possible value of x?
A. 5
B. 6
C. 7
D. 18
E. 20

Also tricky. Notice that \(x\) can take positive, as well as negative values to satisfy \(9<x^2<99\), hence \(x\) can be: -9, -8, -7, -6, -4, 4, 5, 6, 7, 8, or 9. We asked to find the value of \(x_{max}-x_{min}\), ans since \(x_{max}=9\) and \(x_{min}=-9\) then \(x_{max}-x_{min}=9-(-9)=18\).
[/square_root]
Answer: D.




Hi bunel
can you explain how we get max,min possible values from -9 to 9
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Re: 12 Easy Pieces (or not?) [#permalink] New post   17 Sep 2013, 03:53
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sivapavan wrote:
Bunuel wrote:
2. If x is an integer and 9<x^2<99, then what is the value of maximum possible value of x minus minimum possible value of x?
A. 5
B. 6
C. 7
D. 18
E. 20

Also tricky. Notice that \(x\) can take positive, as well as negative values to satisfy \(9<x^2<99\), hence \(x\) can be: -9, -8, -7, -6, -4, 4, 5, 6, 7, 8, or 9. We asked to find the value of \(x_{max}-x_{min}\), ans since \(x_{max}=9\) and \(x_{min}=-9\) then \(x_{max}-x_{min}=9-(-9)=18\).
[/square_root]
Answer: D.




Hi bunel
can you explain how we get max,min possible values from -9 to 9


Sure.

Since x is an integer and 9<x^2<99, then the least value of x is -9 --> (-9)^2<99 (x cannot be -10 because 10^2=100>99).
The same way, the max value of x is 9 --> 9^2<99 (x cannot be 10 because 10^2=100>99).

Does this make sense?
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Re: 12 Easy Pieces (or not?) [#permalink] New post   18 Sep 2013, 13:08
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.




This answer is not 100% right, because there is not the sign <= but only <.
therefore (assuming that X and Y are integers the answer is: (-6-(-4))<Y-X<(8-(-2))
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Re: 12 Easy Pieces (or not?) [#permalink] New post   18 Sep 2013, 13:11
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einnocenti 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.




This answer is not 100% right, because there is not the sign <= but only <.
therefore (assuming that X and Y are integers the answer is: (-6-(-4))<Y-X<(8-(-2))


The answer IS 100% correct.

First of all we are not told that x and y are integers.

Next, consider the following approach, we have -3<x<5 and -7<y<9,

Add y<9 and -3<x --> y-3<9+x --> y-x<12;
Add -7<y and x<5 --> -7+x<y+5 --> -12<y-x;

So, we have that -12<y-x<12.

Hope it's clear.
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Re: 12 Easy Pieces (or not?) [#permalink] New post   26 Nov 2013, 07:04
Hi Bunuel,

Thank you for the very high quality questions.
Can you please explain your thinking behind the solution in question 8?

My confusion stems from the word "guarantee". Ideally the least number of chosen chips, that could result in 2 different colours, is indeed 3. However, the probability of this event is certainly not 1.
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Re: 12 Easy Pieces (or not?) [#permalink] New post   26 Nov 2013, 08:11
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Pmar2012 wrote:
Hi Bunuel,

Thank you for the very high quality questions.
Can you please explain your thinking behind the solution in question 8?

My confusion stems from the word "guarantee". Ideally the least number of chosen chips, that could result in 2 different colours, is indeed 3. However, the probability of this event is certainly not 1.


We are not interested in the probability in question 8:
8. There are 15 black chips and 5 white chips in a jar. What is the least number of chips we should pick to guarantee that we have 2 chips of the same color?

Similar questions to practice:
in-his-pocket-a-boy-has-3-red-marbles-4-blue-marbles-and-85216.html
of-the-science-books-in-a-certain-supply-room-50-are-on-131100.html
in-a-deck-of-52-cards-each-card-is-one-of-4-different-color-83183.html
a-box-contains-10-red-pills-5-blue-pills-12-yellow-56779.html
each-of-the-integers-from-0-to-9-inclusive-is-written-on-130562.html
a-student-is-asked-to-pick-marbles-from-a-bag-that-contains-72390.html
of-the-science-books-in-a-certain-supply-room-50-are-on-131100.html
if-a-librarian-randomly-removes-science-books-from-a-library-93861.html
m10-q24-ps-69233.html#p1237169

Hope this helps.
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Re: 12 Easy Pieces (or not?) [#permalink] New post   30 Jan 2014, 00:23
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.


Hi Bunuel,

Thanks for this explanation. I tried real hard to solve this ques, when I first attempted, algebraically but couldnt do it... I know its far too easy to solve it this way .. but I always go for algebraic ... please help...

If possible please to provide an algebraic solution for this.... thanks a ton for all the help...

Cheers! :o
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Re: 12 Easy Pieces (or not?) [#permalink] New post   30 Jan 2014, 01:11
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rawjetraw 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.


Hi Bunuel,

Thanks for this explanation. I tried real hard to solve this ques, when I first attempted, algebraically but couldnt do it... I know its far too easy to solve it this way .. but I always go for algebraic ... please help...

If possible please to provide an algebraic solution for this.... thanks a ton for all the help...

Cheers! :o


Algebraic solution is a trap here and a waste of time... But anyway:

Say there are x numbers in set A.

"2/9 of the numbers in a data set A were observed" --> \(\frac{2x}{9}\) observed and \(x-\frac{2x}{9}=\frac{7x}{9}\) numbers left to observe;

"3/4 of those numbers were non-negative" --> \(\frac{3}{4}*\frac{2x}{9}=\frac{x}{6}\) non-negative and \(\frac{1}{4}*\frac{2x}{9}=\frac{x}{18}\) negative;

Ratio of negative numbers to non-negative numbers to be 2 to 1 there should be total of \(x*\frac{2}{3}\) negative numbers, so in not yet observed part there should be \(\frac{2x}{3}-\frac{x}{18}=\frac{11x}{18}\) negative numbers. Thus \(\frac{(\frac{11x}{18})}{(\frac{7x}{9})}=\frac{11}{14}\) of the remaining numbers in set A must be negative.

Hope it helps.
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Re: 12 Easy Pieces (or not?) [#permalink] New post   15 Jan 2015, 13:29
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?
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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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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]
08 Dec 2015, 20:32
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