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# 12 Easy Pieces (or not?)

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Math Expert
Joined: 02 Sep 2009
Posts: 41886

Kudos [?]: 128668 [37], given: 12181

12 Easy Pieces (or not?) [#permalink]

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21 Jan 2012, 06:10
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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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Kudos [?]: 128668 [37], given: 12181

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Joined: 02 Sep 2009
Posts: 41886

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

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25 Jan 2012, 04:45
3
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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=prime.

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Posts: 41886

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

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25 Jan 2012, 05:10
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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).

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Posts: 41886

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

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25 Jan 2012, 05:39
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, 22:00
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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Posts: 41886

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

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26 Jan 2012, 02:50
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, 06:09
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, 06:11
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, 23:47
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, 07:51
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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Kudos [?]: 57 [0], given: 23

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Joined: 02 Sep 2009
Posts: 41886

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

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06 Apr 2012, 08:01
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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

# of arrangements of 2 blue, 2 green and 1 yellow marbles (BBGGY) in 5 slots is 5!/(2!*2!*1!)=30 not 5!, since 2 B's and 2 G's are identical.

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:

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

For example number of permutation of the letters of the word "gmatclub" is $$8!$$ as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is $$\frac{6!}{2!2!}$$, as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be $$\frac{9!}{4!3!2!}$$.

So, for our case # of permutations of 5 letters BBGGYout of which 2 B's and 2 G's are identical is $$\frac{5!}{2!*2!}$$.

Hope it's clear.
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Re: 12 Easy Pieces (or not?) [#permalink]

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06 Apr 2012, 09:09
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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16 May 2012, 00:02
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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16 May 2012, 00:24
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, 02:22
good work pal keep them comming....

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

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17 May 2012, 02:23
i wonder if any one got all 12 right....

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

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17 May 2012, 02:56
the answers given are not in order...

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

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14 Jul 2012, 15:56
Bunuel wrote:
SOLUTIONS:

Notice that most of the problems have short, easy and elegant solutions, since you've noticed a trick/trap hidden in the questions.

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

No formula is need to answer this one. The trick here is that we have only 3 different color socks but we pick 4 socks, which ensures that in ANY case we'll have at least one pair of the same color (if 3 socks we pick are of the different color, then the 4th sock must match with either of previously picked one). P=1.

But what if all three chosen pair of socks were of the white color? I think it's possible as there are 5 pairs of white socks. Sorry I don't really understand how the probability is 1 here.

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

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15 Jul 2012, 05:09
dianamao wrote:
Bunuel wrote:
SOLUTIONS:

Notice that most of the problems have short, easy and elegant solutions, since you've noticed a trick/trap hidden in the questions.

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

No formula is need to answer this one. The trick here is that we have only 3 different color socks but we pick 4 socks, which ensures that in ANY case we'll have at least one pair of the same color (if 3 socks we pick are of the different color, then the 4th sock must match with either of previously picked one). P=1.

But what if all three chosen pair of socks were of the white color? I think it's possible as there are 5 pairs of white socks. Sorry I don't really understand how the probability is 1 here.

First of all, we are not choosing 4 PAIRS of socks, we are choosing 4 socks. Next, I think you didn't understand the question properly: the question asks "what is the probability of getting two socks of the same color?"

Now, ask yourself: can we choose 4 socks, so that not to have two socks of the same color? Can we choose 4 socks of different colors? Since there are only 3 colors, then the answer is NO, hence the probability of getting two socks of the same color is 100% or 1.

Hope it's clear.
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Re: 12 Easy Pieces (or not?) [#permalink]

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08 Aug 2012, 06:54
Bunuel wrote:
SOLUTIONS:

Notice that most of the problems have short, easy and elegant solutions, since you've noticed a trick/trap hidden in the questions.

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

No formula is need to answer this one. The trick here is that we have only 3 different color socks but we pick 4 socks, which ensures that in ANY case we'll have at least one pair of the same color (if 3 socks we pick are of the different color, then the 4th sock must match with either of previously picked one). P=1.

Bunuel,
I agree with your explanation, but I thought of solving this using the algebraic method, just to test my understanding.

Here's what I did long method) :

10 = number of blue socks (5*2)
6 = number of black socks (3*2)
4 = number of grey socks (2*2)

10C2 {6C1 * 2C1 + 6C2 + 2C2} + 6C2*(2C2 + 10C1*2C1) + 2C2*6C1*10C1
--------------------------------------------------------------------
20C4

{Quick explanation - First parenthesis => Choose two blue socks, and then I could choose two grey, or two black or 1 black and 1 grey}
Similarly second parenthesis => Choose two black socks, and then I could choose two grey, or 1 blue and 1 grey; I have already chosen two black and two blue in the first parenthesis;
Third parenthesis => Choose two grey, 1 black and 1 blue; I have already chosen two grey+two blue AND two grey+two black)}

45*{12+15+1} + 15{1+20} + 60
= ----------------------------
4845

1260 + 315 + 60
= ---------------
4845

is not equal to 1! Why?

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

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16 Aug 2012, 10:48
Bunuel wrote:

Now, ask yourself: can we choose 4 socks, so that not to have two socks of the same color? Can we choose 4 socks of different colors? Since there are only 3 colors, then the answer is NO, hence the probability of getting two socks of the same color is 100% or 1.

Hope it's clear.

Bunuel, these are 700+ questions? Do you know their difficulty level?

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Re: 12 Easy Pieces (or not?)   [#permalink] 16 Aug 2012, 10:48

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