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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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18 Sep 2013, 12:11
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]

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21 Sep 2013, 07:47
Bunuel Can you please explain ques no. 12 ? how is the least value -1/18 ? why is it not -1/100 ? I am not understanding it.
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Re: 12 Easy Pieces (or not?) [#permalink]

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21 Sep 2013, 07:49
mamacta wrote:
Bunuel Can you please explain ques no. 12 ? how is the least value -1/18 ? why is it not -1/100 ? I am not understanding it.

Because -1/18<-1/100.
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Re: 12 Easy Pieces (or not?) [#permalink]

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21 Sep 2013, 09:08
Ok got it ... don't know what i was thinking before .. thanx a lot .. sorry for asking a dumb question.
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Re: 12 Easy Pieces (or not?) [#permalink]

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01 Oct 2013, 04:07
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 was wondering... if the question asked: if 2 socks are picked randomly, what is the probability of getting 2 socks of the same color, am I correct in going $$\frac{10}{20}*\frac{9}{19} + \frac{6}{20}*\frac{5}{19}+\frac{4}{20}*\frac{3}{19}$$?
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Re: 12 Easy Pieces (or not?) [#permalink]

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01 Oct 2013, 12:08
Bunuel wrote:
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?

2 should be D, -9 is the smallest since -9^2=81, and 9 is the largest since 9^2=81 as well, both of those fall between 9 and 99.
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Re: 12 Easy Pieces (or not?) [#permalink]

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01 Oct 2013, 21:28
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

This is how i solved the ques:

We have to deal with 3 two digit numbers whose sum is as given in the ques. Considering the unit's digit in the various answer choices we have 9,7,1

Next look from the given set of integers (1, 3, 6, 7, 7, 7) , by a way of chosing any 3 values we get a possiblity of only "1" (--3+--7+--1)as a the unit digit.
Hence our options are narowed to B/D i.e 151 , 211

Since we need to find the largest possible value check for option 211 which can be writen as 67+73+71=211

Also can be thought about in this way : If i add only unit digits of --3+--7+--1 i get the sum with 1 as the unit digit and a carry of 1 to the tens digit . So we have to see if there are any 3 numbers which can add to a sum of 20 , so that i can add the carry forward of "1" to make the number 21.
A quick scan leaves with the possibility of 6,7,7 to make up for the 10's digit....

Hope this technique of looking at problems helps...
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Re: 12 Easy Pieces (or not?) [#permalink]

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08 Oct 2013, 06:12
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 $$x_{max}-x_{min}=9-(-9)=18$$.

X(max) should be 9.99 and X(min) should be -9.99

\sqrt{99} = 9.99

19.98 will be the correct answer.
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Re: 12 Easy Pieces (or not?) [#permalink]

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08 Oct 2013, 06:14
honchos 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 $$x_{max}-x_{min}=9-(-9)=18$$.

X(max) should be 9.99 and X(min) should be -9.99

\sqrt{99} = 9.99

Notice that we are told that x is an integer.
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Re: 12 Easy Pieces (or not?) [#permalink]

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08 Oct 2013, 06:16
Bunuel wrote:
honchos 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 $$x_{max}-x_{min}=9-(-9)=18$$.

X(max) should be 9.99 and X(min) should be -9.99

\sqrt{99} = 9.99

Notice that we are told that x is an integer.

Sorry boss, these silly mistakes will kill me in exam.

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

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23 Oct 2013, 12:22
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.

Hey Bunuel,
Can you please advice how you chose 18 as the smart number?
Thanks,
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Re: 12 Easy Pieces (or not?) [#permalink]

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24 Oct 2013, 01:22
prsnt11 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.

Hey Bunuel,
Can you please advice how you chose 18 as the smart number?
Thanks,

Choose a number which is a multiple of 9 (since we are told that "2/9 of the numbers in a data set..."). Try to choose a multiple so that you get another number ("3/4 of those numbers...") an integer too.
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Re: 12 Easy Pieces (or not?) [#permalink]

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06 Nov 2013, 20:01
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.

Hi Bunnel,
Isn't this is the probability of getting atleast 1 pair of socks? This will include cases where we have 2 pairs or all socks of same color.
Question seems to be asking the probability of exactly 1 pair.
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Re: 12 Easy Pieces (or not?) [#permalink]

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07 Nov 2013, 01:29
cumulonimbus 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.

Hi Bunnel,
Isn't this is the probability of getting atleast 1 pair of socks? This will include cases where we have 2 pairs or all socks of same color.
Question seems to be asking the probability of exactly 1 pair.

In that case it would be asking about the probability of EXACTLY two socks of the same color.
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Re: 12 Easy Pieces (or not?) [#permalink]

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26 Nov 2013, 06:04
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]

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26 Nov 2013, 07:11
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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
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]

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26 Nov 2013, 08:03
Thanks, you are right, I didn't read the question well.
If it were for "...2 chips of different colours", then the answer would be 16. Right?
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Re: 12 Easy Pieces (or not?) [#permalink]

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26 Nov 2013, 08:05
Pmar2012 wrote:
Thanks, you are right, I didn't read the question well.
If it were for "...2 chips of different colours", then the answer would be 16. Right?

Yes, that's correct.
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Re: 12 Easy Pieces (or not?) [#permalink]

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21 Jan 2014, 12:33
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 colours. How many different arrangements are possible?
A. 30
B. 60
C. 120
D. 240
E. 480

Seems tough and complicated but if we read the stem carefully we find that the only way both conditions to be met for 5 red marbles, which are half of total marbles, they can be arranged only in two ways: R*R*R*R*R* or *R*R*R*R*R.

Here comes the next good news, in these cases BOTH conditions are met for all other marbles as well: no two adjacent marbles will be of the same color and the first and the last marbles will be of different colors.

Now, it's easy: 2 blue, 2 green and 1 yellow can be arranged in 5 empty slots in 5!/(2!*2!)=30 ways (permutation of 5 letters BBGGY out of which 2 B's and 2 G' are identical). Finally as there are two cases (R*R*R*R*R* and *R*R*R*R*R. ) then total # of arrangement is 30*2=60.

what about R B G R Y R G R B R ?
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Re: 12 Easy Pieces (or not?) [#permalink]

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23 Jan 2014, 02:53
mariofelix 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 colours. How many different arrangements are possible?
A. 30
B. 60
C. 120
D. 240
E. 480

Seems tough and complicated but if we read the stem carefully we find that the only way both conditions to be met for 5 red marbles, which are half of total marbles, they can be arranged only in two ways: R*R*R*R*R* or *R*R*R*R*R.

Here comes the next good news, in these cases BOTH conditions are met for all other marbles as well: no two adjacent marbles will be of the same color and the first and the last marbles will be of different colors.

Now, it's easy: 2 blue, 2 green and 1 yellow can be arranged in 5 empty slots in 5!/(2!*2!)=30 ways (permutation of 5 letters BBGGY out of which 2 B's and 2 G' are identical). Finally as there are two cases (R*R*R*R*R* and *R*R*R*R*R. ) then total # of arrangement is 30*2=60.

what about R B G R Y R G R B R ?

This arrangement violates the condition that the first and the last marbles must be of different colours.
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