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# 12 Easy Pieces (or not?)

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12 Easy Pieces (or not?) [#permalink]

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21 Jan 2012, 05:10
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After posting some 700+ questions, I've decided to post the problems which are not that hard. Though each question below has a trap or trick so be careful when solving. I'll post OA's with detailed solutions after some discussion. Good luck.

1. 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

Solution: 12-easy-pieces-or-not-126366.html#p1033919

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

Solution: 12-easy-pieces-or-not-126366.html#p1033921

3. Fanny and Alexander are 360 miles apart and are traveling in a straight line toward each other at a constant rate of 25 mph and 65 mph respectively, how far apart will they be exactly 1.5 hours before they meet?
A. 25 miles
B. 65 miles
C. 70 miles
D. 90 miles
E. 135 miles

Solution: 12-easy-pieces-or-not-126366.html#p1033924

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

Solution: 12-easy-pieces-or-not-126366.html#p1033925

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

Solution: 12-easy-pieces-or-not-126366.html#p1033930

6. Anna has 10 marbles: 5 red, 2 blue, 2 green and 1 yellow. She wants to arrange all of them in a row so that no two adjacent marbles are of the same color and the first and the last marbles are of different colors. How many different arrangements are possible?
A. 30
B. 60
C. 120
D. 240
E. 480

Solution: 12-easy-pieces-or-not-126366.html#p1033932

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

Solution: 12-easy-pieces-or-not-126366.html#p1033933

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?
A. 3
B. 5
C. 6
D. 16
E. 19

Solution: 12-easy-pieces-or-not-126366.html#p1033935

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

Solution: 12-easy-pieces-or-not-126366.html#p1033936

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

Solution: 12-easy-pieces-or-not-126366.html#p1033938

11. The numbers {1, 3, 6, 7, 7, 7} are used to form three 2-digit numbers. If the sum of these three numbers is a prime number p, what is the largest possible value of p?
A. 97
B. 151
C. 209
D. 211
E. 219

Solution: 12-easy-pieces-or-not-126366-20.html#p1033939

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

Solution: 12-easy-pieces-or-not-126366-20.html#p1033949

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

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20 Oct 2012, 01:11
1
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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.

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]

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29 Apr 2013, 10:18
1
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Q4.
x > -3 and x < 5
y > -7 and y < 9
y-x < 9 - (-3) = 12
y-x > -7-5 = -12

hence -12 < y-x < 12
Ans D
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Re: 12 Easy Pieces (or not?) [#permalink]

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26 Nov 2013, 07:11
1
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Expert's post
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
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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12 Easy Pieces (or not?) [#permalink]

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09 Dec 2015, 21:39
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Expert's post
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]

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22 Jan 2012, 10:17
Ans
1-E
2-A
3-E
4-D
5-B
6-C
7-A
8-A
9-C
10-D
11-D
12-B
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Re: 12 Easy Pieces (or not?) [#permalink]

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22 Jan 2012, 11:10
vinayaerostar wrote:
Ans
1-E
2-A
3-E
4-D
5-B
6-C
7-A
8-A
9-C
10-D
11-D
12-B

Great! 2/3 of the questions answered correctly. Would you like to provide explanations along with answers?
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Re: 12 Easy Pieces (or not?) [#permalink]

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22 Jan 2012, 14:37
anuu wrote:

9) Hence D

Good job! Kudos given.

4 out of 7 answers are correct (though one of the questions you've answered correctly has easier and more elegant solution).

Dare to try rest of the questions?
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Re: 12 Easy Pieces (or not?) [#permalink]

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23 Jan 2012, 11:07
2-d
3-e
4-d
5-b
6-d
7-d
8-c
9-c
10-b
11-d
12-d

Still working on the 1st one...plz check
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Re: 12 Easy Pieces (or not?) [#permalink]

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25 Jan 2012, 04:39
Just posted solutions. Kudos points given to everyone with correct solutions. Let me know if I missed someone.
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Re: 12 Easy Pieces (or not?) [#permalink]

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25 Jan 2012, 21:00
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 C (III only) is left.

Now, if interested why III 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.

Bunnel: If I am correct you meant to say that II is the correct choice????
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Re: 12 Easy Pieces (or not?) [#permalink]

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26 Jan 2012, 01:50
subhajeet 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.

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.

Bunnel: If I am correct you meant to say that II is the correct choice????

Yes, correct choice is c^2>a^2+b^2, as explained, so B.
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Re: 12 Easy Pieces (or not?) [#permalink]

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26 Jan 2012, 05:09
Got 3 answers wrong Bunnel thanks for the questions
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Re: 12 Easy Pieces (or not?) [#permalink]

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26 Jan 2012, 05:11
subhajeet wrote:
Got 3 answers wrong Bunnel thanks for the questions

Those are quite tricky questions so 9(!) correct answers out of 12 is pretty good result. Well done!
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Re: 12 Easy Pieces (or not?) [#permalink]

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26 Jan 2012, 22:47
Got only 8 correct out of 12 questions.

The most shameful part is that all of my incorrect responses are related to easier problems than those problems that are given correct responses by me. Sometimes, I just dive into doing maths rather than using the wits.

Learning from this series of questions: GIVE YOURSELF A FEW SECONDS OF REFLECTION TIME BEFORE YOU START SOLVING A QUANT PROBLEM. MAYBE, THE SOLUTION DOESN'T EVEN REQUIRE DOING MATH (e.g. Question#1)!

Thank you, Bunuel, as always, for this series of questions.
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Re: 12 Easy Pieces (or not?) [#permalink]

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06 Apr 2012, 06:51
Bunuel wrote:
6. Anna has 10 marbles: 5 red, 2 blue, 2 green and 1 yellow. She wants to arrange all of them in a row so that no two adjacent marbles are of the same color and the first and the last marbles are of different colors. How many different arrangements are possible?
A. 30
B. 60
C. 120
D. 240
E. 480
....
as there are two cases (R*R*R*R*R* and *R*R*R*R*R. ) then total # of arrangement is 30*2=60.

I am little confused ...after this bold part.. since we have 5 slots available to be filled by 5 marbles and we can pick any marbles given our setting of marbles (without disturbing any conditions in the stem ?). now # arrangements in either case is going to be 5! (=120) so total # arrangements = 2*120= 240...

what am I missing here
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Re: 12 Easy Pieces (or not?) [#permalink]

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06 Apr 2012, 08:09
Bunuel wrote:
yogesh1984 wrote:
Bunuel wrote:
6. Anna has 10 marbles: 5 red, 2 blue, 2 green and 1 yellow. She wants to arrange all of them in a row so that no two adjacent marbles are of the same color and the first and the last marbles are of different colors. How many different arrangements are possible?
A. 30
B. 60
C. 120
D. 240
E. 480
....
as there are two cases (R*R*R*R*R* and *R*R*R*R*R. ) then total # of arrangement is 30*2=60.

I am little confused ...after this bold part.. since we have 5 slots available to be filled by 5 marbles and we can pick any marbles given our setting of marbles (without disturbing any conditions in the stem ?). now # arrangements in either case is going to be 5! (=120) so total # arrangements = 2*120= 240...

what am I missing here

THEORY.
Permutations of $$n$$ things of which $$P_1$$ are alike of one kind, $$P_2$$ are alike of second kind, $$P_3$$ are alike of third kind ... $$P_r$$ are alike of $$r_{th}$$ kind such that: $$P_1+P_2+P_3+..+P_r=n$$ is:

$$\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}$$.

Hope it's clear.

How can I do this feels like slapping myself ! anyway seems time to turn over to the basics.

Thanks for the explanation mate
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Re: 12 Easy Pieces (or not?) [#permalink]

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15 May 2012, 23:02
Bunuel a slightly different query, based on the difficulty level how much time would you give to solve these questions. Got Q no. 2 & 6 wrong!
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Re: 12 Easy Pieces (or not?) [#permalink]

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15 May 2012, 23:24
Bunuel wrote:
11. The numbers {1, 3, 6, 7, 7, 7} are used to form three 2-digit numbers. If the sum of these three numbers is a prime number p, what is the largest possible value of p?
A. 97
B. 151
C. 209
D. 211
E. 219

What is the largest possible sum of these three numbers that we can form? Maximize the first digit: 76+73+71=220=even, so not a prime. Let's try next largest sum, switch digits in 76 and we'll get: 67+73+71=211. Question is it a prime number? If you notice 210=2*3*5*7=the product of the first four primes. So, 210+1=211 must be a prime. For example: 2+1=3=prime, 2*3+1=7=prime, 2*3*5+1=31=prime.

Bunuel could you elaborate on the observation you presented. Is it that product of consecutive primes +1 is prime or is it something else?
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Re: 12 Easy Pieces (or not?) [#permalink]

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17 May 2012, 01:22
good work pal keep them comming....
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Re: 12 Easy Pieces (or not?) [#permalink]

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17 May 2012, 01:23
i wonder if any one got all 12 right....
Re: 12 Easy Pieces (or not?)   [#permalink] 17 May 2012, 01:23

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