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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, 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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24 Jan 2014, 09:22
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, Could you please explain this question. I didn't get it.

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

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24 Jan 2014, 09:26
Vidhi1 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 Bunuel, Could you please explain this question. I didn't get it.

Check here: 12-easy-pieces-or-not-126366-20.html#p1104551 Also, can you please be more specific when asking a question? Thank you.
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Re: 12 Easy Pieces (or not?) [#permalink]

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24 Jan 2014, 20:02
For #3 I used little to no math at all.

I've found that when I was struggling with the GMAT it was because I was looking into the math TOO HARD. This is a logic base question

*We known that the two will meet at $$Point 0$$ and the distance will be $$0$$
They go at a constant rate and $$1 Hr$$ before they meet they are $$65+25 miles$$ apart
Sooooo at $$1.5 Hr$$ it has to be $$<90 miles$$ from $$Point 0$$

E.

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

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30 Jan 2014, 00:23
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.

Hi Bunuel,

Thanks for this explanation. I tried real hard to solve this ques, when I first attempted, algebraically but couldnt do it... I know its far too easy to solve it this way .. but I always go for algebraic ... please help...

If possible please to provide an algebraic solution for this.... thanks a ton for all the help...

Cheers!

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

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30 Jan 2014, 01:11
rawjetraw 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.

Hi Bunuel,

Thanks for this explanation. I tried real hard to solve this ques, when I first attempted, algebraically but couldnt do it... I know its far too easy to solve it this way .. but I always go for algebraic ... please help...

If possible please to provide an algebraic solution for this.... thanks a ton for all the help...

Cheers!

Algebraic solution is a trap here and a waste of time... But anyway:

Say there are x numbers in set A.

"2/9 of the numbers in a data set A were observed" --> $$\frac{2x}{9}$$ observed and $$x-\frac{2x}{9}=\frac{7x}{9}$$ numbers left to observe;

"3/4 of those numbers were non-negative" --> $$\frac{3}{4}*\frac{2x}{9}=\frac{x}{6}$$ non-negative and $$\frac{1}{4}*\frac{2x}{9}=\frac{x}{18}$$ negative;

Ratio of negative numbers to non-negative numbers to be 2 to 1 there should be total of $$x*\frac{2}{3}$$ negative numbers, so in not yet observed part there should be $$\frac{2x}{3}-\frac{x}{18}=\frac{11x}{18}$$ negative numbers. Thus $$\frac{(\frac{11x}{18})}{(\frac{7x}{9})}=\frac{11}{14}$$ of the remaining numbers in set A must be negative.

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

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30 Jan 2014, 01:45
Bunuel wrote:

Algebraic solution is a trap here and a waste of time... But anyway:

Say there are x numbers in set A.

"2/9 of the numbers in a data set A were observed" --> $$\frac{2x}{9}$$ observed and $$x-\frac{2x}{9}=\frac{7x}{9}$$ numbers left to observe;

"3/4 of those numbers were non-negative" --> $$\frac{3}{4}*\frac{2x}{9}=\frac{x}{6}$$ non-negative and $$\frac{1}{4}*\frac{2x}{9}=\frac{x}{18}$$ negative;

Ratio of negative numbers to non-negative numbers to be 2 to 1 there should be total of $$x*\frac{2}{3}$$ negative numbers, so in not yet observed part there should be $$\frac{2x}{3}-\frac{x}{18}=\frac{11x}{18}$$ negative numbers. Thus $$\frac{(\frac{11x}{18})}{(\frac{7x}{9})}=\frac{11}{14}$$ of the remaining numbers in set A must be negative.

Hope it helps.

Whoa! You are so awesome. Crystal clear now, mate!

I just find that I am good at algebraically solving a problem .. i am much more faster that way than thinking of a possible number ... what if i choose a wrong number n lead to more difficult a path to solution and so on n so forth .. but this time I just couldnt do it..

I just kept messing at this part... and got godknows what answers ..

"3/4 of those numbers were non-negative" --> $$\frac{3}{4}*\frac{2x}{9}=\frac{x}{6}$$ non-negative and $$\frac{1}{4}*\frac{2x}{9}=\frac{x}{18}$$ negative;

i kept doing this .. since $$\frac{x}{6}$$ non-negative ... then we have $$\frac{5x}{6}$$ negative .. just went on solving in that direction .. which is clearly wrong ... i tried to model my algebraic solution exactly the way you've done with choosing-a-number method .. and would get stumped at this place all the time ..
now i get it ..

Thanks a ton for the prompt response.. Cheers!

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

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28 Apr 2014, 09:54
honey86 wrote:
Bunuel,

how I approached this questions was -
combined rate = 90 (65+25), and total distance to cover = 360 which will be completed in 3 hours.
Therefore, (total distance) - (distance covered in 2.5 hours at the speed of 90) will give us our answer i.e. 135

However, your solution seems quick but I did not completely understand the logic.
Distance traveled at 90 for 1.5 hours - shouldn't this distance be lesser than what is required ? (I know that right answer is obtained through this, but I am still trying to understand the logic behind your solution. Please help me understand this).

Could not understand the red part. Please elaborate.
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Re: 12 Easy Pieces (or not?) [#permalink]

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28 Apr 2014, 10:27
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

Distance traveled at 90 for 1.5 hours - shouldn't this distance be lesser than what is required ?

i.e. distance traveled at relative speed of 90 mph and time duration of 1.5 hours.
In your solution, you calculated the distance traveled at rate 90 mph, during the time interval of 1.5 hours.
This is the distance that is covered by F & A(approaching towards each other), but we need to calculate the distance that is left between them 1.5 hours before they met.

I am trying to understand how are these two distances equal.

one correction -
combined rate = 90 (65+25), and total distance to cover = 360 which will be completed in 4 hours.

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

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28 Apr 2014, 10:29
honey86 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

Distance traveled at 90 for 1.5 hours - shouldn't this distance be lesser than what is required ?

i.e. distance traveled at relative speed of 90 mph and time duration of 1.5 hours.
In your solution, you calculated the distance traveled at rate 90 mph, during the time interval of 1.5 hours.
This is the distance that is covered by F & A(approaching towards each other), but we need to calculate the distance that is left between them 1.5 hours before they met.

I am trying to understand how are these two distances equal.

one correction -
combined rate = 90 (65+25), and total distance to cover = 360 which will be completed in 4 hours.

The distance 1.5 hours before they meet is the distance they cover in 1.5 hours, isn't it?
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Re: 12 Easy Pieces (or not?) [#permalink]

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05 May 2014, 05:40
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.

Dear Bunnel

The qs asks us the probability when we pick up 4 socks individually i.e. no pair... so there is a probability that we pick up 4 blacks as there are 10 individual black socks.... i didnt understand the logic given in the answer.
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Re: 12 Easy Pieces (or not?) [#permalink]

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05 May 2014, 05:49
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.

why arent we taking 77 as one of the number?
77+76+73 = 226... which is not a prime
76+73+71 = 220 ... not a prime

then how do we directly assume 67...??? no where does it say the numbers are distinct.
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Re: 12 Easy Pieces (or not?) [#permalink]

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05 May 2014, 05:58
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.

in this how r we assuming 9 as the largest value of y??? it is not given that y <= 9... doesnt it mean that teh largest value can be 8? and same for x the minimum can be -2??
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Re: 12 Easy Pieces (or not?) [#permalink]

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05 May 2014, 05:59
nandinigaur 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.

Dear Bunnel

The qs asks us the probability when we pick up 4 socks individually i.e. no pair... so there is a probability that we pick up 4 blacks as there are 10 individual black socks.... i didnt understand the logic given in the answer.

There are 3 pairs of black socks not 5. Anyway if you pick 4 black socks won't you have two pairs of the same color there?
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Re: 12 Easy Pieces (or not?) [#permalink]

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05 May 2014, 06:01
nandinigaur wrote:
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.

why arent we taking 77 as one of the number?
77+76+73 = 226... which is not a prime
76+73+71 = 220 ... not a prime

then how do we directly assume 67...??? no where does it say the numbers are distinct.

In 77+76+73 you used four 7's while we have only 3.
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Re: 12 Easy Pieces (or not?) [#permalink]

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05 May 2014, 06:03
nandinigaur 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.

in this how r we assuming 9 as the largest value of y??? it is not given that y <= 9... doesnt it mean that teh largest value can be 8? and same for x the minimum can be -2??

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

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05 May 2014, 06:07
because they are in parentheses means they are a set and no number can repeat? as in we can have only 3 7s, one 3, one 6 and one 1?
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Re: 12 Easy Pieces (or not?) [#permalink]

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05 May 2014, 06:09
nandinigaur wrote:
because they are in parentheses means they are a set and no number can repeat? as in we can have only 3 7s, one 3, one 6 and one 1?

Yes, we have one 1, one 3, one 6, and three 7's to use for three 2-digit numbers.
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Re: 12 Easy Pieces (or not?) [#permalink]

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24 Jun 2014, 04:23
Q11: You can use the Answer choices to get to the answer quickly.

The smallest possible sum you can get is 13+67+77 = 157
Right of the bat this eliminates A&B 97 & 151 as they are smaller than 157.
Next C: 209 is divisible by 19 So C is not Prime and gets eliminated.
Next E: 219 is divisible by 3 So E is not prime and gets eliminated.

The only choice left is D

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

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18 Sep 2014, 23:54
Bunuel wrote:
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

Worst case scenario would be if the first two chips we pick will be of the different colors. But the next chip must match with either of two, so 3 is the answer.

Bunuel i m not able to get the essence of this one..esp the "least" term here. Also, while attempting such worst case scenario problems what will be deciding factor int he question stem??
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Re: 12 Easy Pieces (or not?) [#permalink]

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19 Sep 2014, 02:24
Vinitkhicha1111 wrote:
Bunuel wrote:
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

Worst case scenario would be if the first two chips we pick will be of the different colors. But the next chip must match with either of two, so 3 is the answer.

Bunuel i m not able to get the essence of this one..esp the "least" term here. Also, while attempting such worst case scenario problems what will be deciding factor int he question stem??

The question asks about the minimum number of chips we should pick to guarantee that we have 2 chips of the same color.
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Re: 12 Easy Pieces (or not?)   [#permalink] 19 Sep 2014, 02:24

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