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12 Easy Pieces (or not?)

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12 Easy Pieces (or not?) [#permalink] New post 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] New post 26 Sep 2012, 05:30
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

Here according to the solution, the max value for y is 9 but Y's value is between -7 and 9 so max value of y is 8 right?
Please help.
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Re: 12 Easy Pieces (or not?) [#permalink] New post 03 Oct 2012, 00:30
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.


Answer: A.


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] New post 03 Oct 2012, 02:01
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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.


Answer: A.


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] New post 03 Oct 2012, 03:31
ohkey bunuel....!! now i got that..its clear now :)) . Thank you MATH GURU :)..
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Re: 12 Easy Pieces (or not?) [#permalink] New post 03 Oct 2012, 07:12
and here it is +1 for u bunuell !!!
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Re: 12 Easy Pieces (or not?) [#permalink] New post 04 Oct 2012, 10:07
difficult problems very intelligently solved,
kudos.
hats off.
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There are 5 pairs of white, 3 pairs of black and 2 pairs of [#permalink] New post 19 Oct 2012, 23:10
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: There are 5 pairs of white, 3 pairs of black and 2 pairs of [#permalink] New post 20 Oct 2012, 01:11
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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 20 Oct 2012, 20:13
Thanks a lot Eva. This helps.
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Re: 12 Easy Pieces (or not?) [#permalink] New post 22 Oct 2012, 02:40
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, 04:21
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 08 Apr 2013, 08:41
Bunuel wrote:
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

Make it simple! The question is: how far apart will they be exactly 1.5 hours before they meet? As Fanny and Alexander's combined rate is 25+65 mph then 1.5 hours before they meet they'll be (25+65)*1.5=135 miles apart.

Answer: E.

Bunuel,

Shouldn't this be 360 - 135? Because, when Fanny and Alexander meet, they would have traveled 360 miles and 135 miles is the distance traveled in 1.5 hours and not the distance apart.
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Re: 12 Easy Pieces (or not?) [#permalink] New post 09 Apr 2013, 02:41
Expert's post
nt2010 wrote:
Bunuel wrote:
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

Make it simple! The question is: how far apart will they be exactly 1.5 hours before they meet? As Fanny and Alexander's combined rate is 25+65 mph then 1.5 hours before they meet they'll be (25+65)*1.5=135 miles apart.

Answer: E.

Bunuel,

Shouldn't this be 360 - 135? Because, when Fanny and Alexander meet, they would have traveled 360 miles and 135 miles is the distance traveled in 1.5 hours and not the distance apart.


The question asks about the distance between them 1.5 hours before they meet, which is 135 miles.

Hope it's clear.
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Re: 12 Easy Pieces (or not?) [#permalink] New post 29 Apr 2013, 07:57
Bunuel wrote:

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.


Hi Experts,

I am having trouble understanding the above quoted part.
I understand that -side opposite to biggest angle is longest and opposite is also true.
Also, I know that acc to Pythagoras theorem c^2=a^+b^2,
However, I am failed to understand the reasoning behind:
Quote:
since the angle opposite c is more than 90 degrees (100) then c is larger, hence c^2>a^+b^2.


At 90 degree- Biggest Side = c; Other two sides are - a and b
At 100 degree- Biggest Side is = (C+delta c); and other two sides are -(a-delta a)and (b-delta b)
where, delta a - change in value of a
hence, Equation becomes-

(C+delta c)^2 >=< (a-delta a)^2and (b-delta b)^2
Now, how to be sure about Inequality sign? Since lengths of triangle varies per trigonometry rules..

Please help.
Sorry, if i went too ahead.

Regards,
H
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Re: 12 Easy Pieces (or not?) [#permalink] New post 29 Apr 2013, 10:10
Q2. 9 < x^2 < 99
x^2 = {16,25,36,49,64,81}
Max X = 9
Min X= -9
Max - Min = 9 - (-9) = 18
Hence D
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Re: 12 Easy Pieces (or not?) [#permalink] New post 29 Apr 2013, 10:15
Q3.
Relative distance covered per hour = 25mph + 65 mph = 90 mph
they meet in 360 / 90 = 4 hours
1.5 hrs before they meet = 4 - 1.5 = 2.5 hours
In 2.5 hours: Fanny travelled 2.5 * 25 = 62.5 miles and Alex travelled 2.5 * 65 = 162.5 miles
Distance between them = 360 - (62.5 + 162.5) = 135 miles

Ans E
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Re: 12 Easy Pieces (or not?) [#permalink] New post 29 Apr 2013, 10:18
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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] New post 29 Apr 2013, 10:29
Q7: Lets assume total numbers = 36 (I took an LCM of 9 and 4 so that I have integers to work with)
Of 8 numbers (2/9 of 36), 6 are Non-Negative or 2 are negative
Remaining numbers = 36 - 8 = 28

let negative numbers out of these 28 be X and Non-negative numbers be 28-x
(x + 2) / (28 - x + 6) = 2/1
or x = 22

Hence 22 numbers in remaining 28 must be negative.
22 / 28 = 11/14
Ans A
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Re: 12 Easy Pieces (or not?) [#permalink] New post 29 Apr 2013, 10:32
Q8:
Out of any 3 chips, 2 will be of the same color
it could either be BWB OR WBW
ans A
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Re: 12 Easy Pieces (or not?) [#permalink] New post 29 Apr 2013, 10:37
Q9.
BLUE W R G BLACK Y P
sequence starts with Blue and ends with Red
it is an arithmetic progression series with first a = 3 and d = 7
so any of the options that satify a + INTEGER * 7 will be correct.

lets check the options.
A. 22. 3 + 7d = 22 or 7d = 19 NOT AN INTEGER
B. 30. 3 + 7d = 30 or 7d = 27 NOT AN INTEGER
C. 38. 3 + 7d = 38 or 7d = 35 INTEGER (Correct Answer)
D. 46. 3 + 7d = 46 or 7d = 43 NOT AN INTEGER
E. 54. 3 + 7d = 54 or 7d = 51 NOT AN INTEGER

Ans C
Re: 12 Easy Pieces (or not?)   [#permalink] 29 Apr 2013, 10:37
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