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# 12 Easy Pieces (or not?)

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Math Expert
Joined: 02 Sep 2009
Posts: 44654
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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Posts: 44654
Re: 12 Easy Pieces (or not?) [#permalink]

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25 Jan 2012, 04:45
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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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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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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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Joined: 02 Sep 2009
Posts: 44654
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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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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Posts: 44654
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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03 Oct 2012, 01:30
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.

Bunuel..how ans A?? i got D..

i took 90 instead of 18..

90*2/9=20...20 were observe , 20*3/4= 15 were no negative..
so from remaining 70...we should bring 30 negative so ratio would be 2 to 1..between negative and non-negative ..

70*3/7=30...so i got 3/7 ans..

where m wrong??

Thanks bunuel for posting all these question.. i gonna know that where i m in math now..
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Re: 12 Easy Pieces (or not?) [#permalink]

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03 Oct 2012, 03:01
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sanjoo 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.

Bunuel..how ans A?? i got D..

i took 90 instead of 18..

90*2/9=20...20 were observe , 20*3/4= 15 were no negative..
so from remaining 70...we should bring 30 negative so ratio would be 2 to 1..between negative and non-negative ..

70*3/7=30...so i got 3/7 ans..

where m wrong??

Thanks bunuel for posting all these question.. i gonna know that where i m in math now..

The ratio of negative numbers to non-negative numbers to be 2 to 1 there should be total of 90*2/3=60 negative numbers and 30 non-negative.

"2/9 of the numbers in a data set A were observed" --> 20 observed and 90-20=70 numbers left to observe;
"3/4 of those numbers were non-negative" --> 15 non-negative and 5 negative;
In not yet observed part there should be 60-5=55 negative numbers. Thus 55/70=11/14 of the remaining numbers in set A must be negative.

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

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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.

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

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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.

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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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.

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]

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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.

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]

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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.

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]

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02 May 2013, 03:13
Expert's post
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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.

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]

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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.

Sorry, this may be a silly thing to ask, I don't understand the problem.

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]

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07 Jul 2013, 13:10
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.

Sorry, this may be a silly thing to ask, I don't understand the problem.

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]

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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]

Hi bunel
can you explain how we get max,min possible values from -9 to 9
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Posts: 44654
Re: 12 Easy Pieces (or not?) [#permalink]

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17 Sep 2013, 03:53
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]

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]

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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.

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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Posts: 44654
Re: 12 Easy Pieces (or not?) [#permalink]

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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.

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))

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] 18 Sep 2013, 13:11

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