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# 12 Easy Pieces (or not?)

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

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10 Oct 2014, 03:56
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=prime.

For this question, I solved it by this approach. P is prime number so P has to end with 1,3,7 or 9. There for the last digit of 3 nos has to be 1,3, 7. So to maximize P try 67, 71 and 73.

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

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19 Oct 2014, 12:38
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 Bunuel,

I have a query (might be silly).

The question does not asks the probability of getting at least 2 socks of same color. I thought it asked probability of getting precisely 2 socks of same color and the rest 2 socks having the 2 other color.
i.e. I assumed the only eligible combinations could be
2 white, 1 black, 1 grey
or 1 white, 2 black, 1 grey
or 1 white, 1 black, 2 grey

but i excluded the cases such as
3 white, 1 black etc.

Please let me know, if I am missing something.

Thanks,
Sriniwas

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

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15 Jan 2015, 11:32
Concerning the 3rd problem, with the distance 1.5 hours before they meet, I have one question:

The combined rate of 90 means that the distance is shrinking anyway, so this is why we don't need to subtract the 1.5 from the time?

Because I only randomly ended up with the correct response, because 135 appeared in my calculations, which were like this:

90(T-1.5) = 40T. To explain here, I used 40T as the distance, by subtracting 25T (Fs distance) from 65T (As distance).
90T - 135 = 40T
50T = 135
T=2.7.

Then of course, I couldn't find my result after substituting for T and I chose 135, only because it appeared in my calculations.

Could you explain, if possible, what went wrong in my thinking?!

Also, feel free to delete the post as the response is wrong, if you feel it is confusing people!

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

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15 Jan 2015, 12:12
About number 7. I didn't solve it, but after reading Bunuel's explanation I got it (after some effort) so I thought I could share my explanation of Bunuel's explanation, in case it helps someone:

So, we can draw 10 slots, just to visualize it:
__.__.__.__.__.__.__.__.__.__

We know we have 5 red marbles, so they will take up half of the slots. We can start using our red marbles in these 2 ways:
1) Starting from the 1st slot and placing one red marble in every other slot, ending with the last red marble in the second to last slot.
2) The second way is to start from the 2nd slot and end in the last slot.

Our slots now look like this:
R.__.R.__.R.__.R.__.R.__

We now see that for the rest we don't mind where they are being placed, since no matter where each of the rest of the marbles is, no 2 marbles od the same colour will be next to each other, and the last marble will definitely be different than the first.

So, we have 2B, 2G and 1Y still left. This means than we can fill in these slots in:
5! / 2!2!1! = 5! / 2!2! = 30 ways.

In total then it is 2*30 = 60 ways.

I have a deep repulsion towards permutations and combinations, but I am making my piece with that, since at least I am starting to grasp probabilities, a topic that was never my cup of tea!

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

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15 Jan 2015, 12:41
About 9, I solved it by ordering the marbles in such as way that when the last one is red, this marble matched one of the answer choices:

BWRGBYP|BWRGBYP|BWRGBYP|BWRGBYP|BWRGBYP|BWR, this become 7*5+3= 35+3=38, ANS C.

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

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15 Jan 2015, 13:29
Question 11,

I guess it should have stated that we can only unse each one of these 3 nunnbers only once, right?
Because, I used the same approach as Bunuel, but without knowing that you can only use these nubers once the process an be very long... and I am not even sure if it would end up in the same result - i guess not.

Is there sth in the way that the question is phrased that should have warned me that these numbers were only to be used once?

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

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16 Jan 2015, 02:45
pacifist85 wrote:
Question 11,

I guess it should have stated that we can only unse each one of these 3 nunnbers only once, right?
Because, I used the same approach as Bunuel, but without knowing that you can only use these nubers once the process an be very long... and I am not even sure if it would end up in the same result - i guess not.

Is there sth in the way that the question is phrased that should have warned me that these numbers were only to be used once?

We are given a set: {1, 3, 6, 7, 7, 7}. I think it should be clear that 1, 3, and 6 should be used once, and 7 thrice.
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Re: 12 Easy Pieces (or not?) [#permalink]

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17 Jun 2015, 12:31
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.

Here I feel
ymax cant be 9 but 8
y min cant be -7 but -6
and same goes for xmax cant be 5 but 4
xmin cant be -3 but -2

however it seems the answer will not change.

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

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17 Jun 2015, 12:34
anurag356 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.

Here I feel
ymax cant be 9 but 8
y min cant be -7 but -6
and same goes for xmax cant be 5 but 4
xmin cant be -3 but -2

however it seems the answer will not change.

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

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17 Jun 2015, 12:38
Bunuel wrote:
anurag356 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.

Here I feel
ymax cant be 9 but 8
y min cant be -7 but -6
and same goes for xmax cant be 5 but 4
xmin cant be -3 but -2

however it seems the answer will not change.

No wonder you are best ....... I never realised there are more pages.
Thank you

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

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22 Nov 2015, 16:29
For question #8, the solution written here is that , Worst case scenario would be if the first two chips we pick will be of the different colors

Why can't we get 2 different colored chip in the first two chips that we pick?

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

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23 Nov 2015, 12:01
marvas8581 wrote:
For question #8, the solution written here is that , Worst case scenario would be if the first two chips we pick will be of the different colors

Why can't we get 2 different colored chip in the first two chips that we pick?

Your question is not clear at all. Please elaborate what you mean.
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There are 5 pairs of white, 3 pairs of black and 2 pairs of grey socks [#permalink]

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08 Dec 2015, 12:25
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

I totally understand that we don't need to do the calculations here because the answer is obvious. However, I would like to know how we do calculate the answer to get to the official answer.

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

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08 Dec 2015, 13:07
Peltina 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

I totally understand that we don't need to do the calculations here because the answer is obvious. However, I would like to know how we do calculate the answer to get to the official answer.

Search for a question before you post. This has already been discussed at 12-easy-pieces-or-not-126366.html#p1033919

Topics merged.

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

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08 Dec 2015, 20:32
Peltina 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

I totally understand that we don't need to do the calculations here because the answer is obvious. However, I would like to know how we do calculate the answer to get to the official answer.

Hi,

in such questions, we look at the worst scenario..
here it would be getting different coloured socks , every time we pick socks..

but there are only three different coloured socks, so as a worst case , the first three picked up are different colours..
the fourth which you pick up has to be one of these three colours, thus ensuring that we have two socks of same colour..
hence prob is 1, that is it is sure to have two socks of atleast one colour..
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Re: 12 Easy Pieces (or not?) [#permalink]

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09 Dec 2015, 14:34
chetan2u wrote:
Hi,

in such questions, we look at the worst scenario..
here it would be getting different coloured socks , every time we pick socks..

but there are only three different coloured socks, so as a worst case , the first three picked up are different colours..
the fourth which you pick up has to be one of these three colours, thus ensuring that we have two socks of same colour..
hence prob is 1, that is it is sure to have two socks of atleast one colour..

Thank you! I was looking more for the formula that we should use to get to the answer P=1 in case we fail to notice that the answer is obvious and needs no calculation.

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12 Easy Pieces (or not?) [#permalink]

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09 Dec 2015, 22:39
1
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Expert's post
Peltina wrote:
chetan2u wrote:
Hi,

in such questions, we look at the worst scenario..
here it would be getting different coloured socks , every time we pick socks..

but there are only three different coloured socks, so as a worst case , the first three picked up are different colours..
the fourth which you pick up has to be one of these three colours, thus ensuring that we have two socks of same colour..
hence prob is 1, that is it is sure to have two socks of atleast one colour..

5 pairs of white, 3 pairs of black and 2 pairs of grey socks
Thank you! I was looking more for the formula that we should use to get to the answer P=1 in case we fail to notice that the answer is obvious and needs no calculation.

Hi,
there cant be direct formulas for most of the questions but they are just a step to correct answer..
so if you want the answer the formula way, then it would be something like this...

there are three different coloured socks..
let us first find the scenario where we do not get two of one kind...
1)the first can be picked up, say white, prob=5/10..
2)the second can be picked up, say black, prob=3/9..
3)the third can be picked up, the remaining grey, prob=2/8..
4)the fourth picking up is 0 as no other colour is left prob=0/7..
now these four can be arranged in 4!/2! ways..
prob of two colours of one kind = 1- 5/10 *3/9 * 2/8* 0/7 * 4!/2!=1-0=1
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Re: 12 Easy Pieces (or not?) [#permalink]

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21 Oct 2016, 04:09
Peltina wrote:
chetan2u wrote:
Hi,

in such questions, we look at the worst scenario..
here it would be getting different coloured socks , every time we pick socks..

but there are only three different coloured socks, so as a worst case , the first three picked up are different colours..
the fourth which you pick up has to be one of these three colours, thus ensuring that we have two socks of same colour..
hence prob is 1, that is it is sure to have two socks of atleast one colour..

Thank you! I was looking more for the formula that we should use to get to the answer P=1 in case we fail to notice that the answer is obvious and needs no calculation.

Same, is it possible to answer this question with Combinations? if yes, how? I know it is not the best way to do it. But I failed to notice the trick and I went for the Combination solution and i can't arrive to P=1.

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

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27 Mar 2017, 00:32
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.

Hi Bunuel, I tried to refer other posts for this answer but I did not understand. Can you please elaborate this? how M=7k+3..

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

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27 Mar 2017, 00:48
RMD007 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.

Hi Bunuel, I tried to refer other posts for this answer but I did not understand. Can you please elaborate this? how M=7k+3..

The row begins with blue marble and ends with red marble. What cases can we have?

{blue, white, red} = 7*0 + 3
{blue, white, red, green, black, yellow, pink} {blue, white, red} = 7*1 + 3
{blue, white, red, green, black, yellow, pink} {blue, white, red, green, black, yellow, pink} {blue, white, red} = 7*2 + 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} = 7*3 + 3
...

As you can see in any case the number of marbles must be a multiple of 7 plus 3.

Hope it's clear.
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Kudos [?]: 128668 [0], given: 12181

Re: 12 Easy Pieces (or not?)   [#permalink] 27 Mar 2017, 00:48

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