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# 12 Easy Pieces (or not?)

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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 $$x_{max}-x_{min}=9-(-9)=18$$.

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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.
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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.

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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.

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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.

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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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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.

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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

There is no need for algebraic manipulation to answer this question.

$$\frac{1}{10^{n+1} }$$ is 10 times smaller than $$\frac{1}{10^n}$$. When expressed as decimals, both have the form 0.001 (with a varying number of zeros preceding the 1). For the given expression to be valid, we should have: $$0.001 < 0.00737 < 0.01$$. This implies that $$n=2$$ (since $$\frac{1}{10^n}=0.01$$, then $$n=2$$).

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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.

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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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2 ) For x to be integer, maximum value of x^2 should be a perfect square i.e 81 and minimum value of x^2 16

so maximim value ox x^2 = 81 , hence x = 9
similarly minimum value of x^2 = 16 hence x = 4
maximim value ox x^2 - minimum value of x^2 = 9-4 =5

3 ) Since Fanny and alexander are travelling towards each other they travel for the same amount of time say 't' and sum of the distance traveled by them is equal to 360

hence 65t+25t = 360
90t = 360
t = 4 hrs

hence when they meet each would have traveled for 4 hrs. 1.5 hrs before they meet, each would travel for 2.5 hrs.

distance traveled by Fanny in 2.5 hrs= 25* 5/2 = 125/2 miles
distance traveled by Alex in 2.5 hrs= 65* 5/2 = 325/2 miles

therfore 1.5 hrs before they meet, distance traveled by them is
125/2+325/2 = 450/2 = 225 miles

Hence they will be 360-225 = 135 miles apart

4) we can find the values of y-x at the boundaries of the given inequality

x=-3 y=-7 y-x = -7-(-3) = -4
x=5 y=-7 y-x = -7-5 = -12
x=-3 y=9 y-x = 9-(-3) = 12
x=5 y=9 y-x = 9-5 =4

minimum value of y-x is greater than -12 and maximim value less than 12

7) Lets say set A has 36 numbers

2/9*36 = 8 numbers were 'observed' and remaining 36-8 = 28 numbers 'not observed'

3/4*8 = 6 of the 'observed' numbers were non negative.
Hence 8-6 =2 of the 'observed' numbers were negative

Ratio of Negative numbers/Non-negative numbers = 2/1
So out of 36 numbers 2/3 rd should be negative i.e 2/3*36 = 24 negative numbers

Total negative numbers in set A= 24

2 of the observed numbers are negative , hence from the remaining 'not observed' numbers 22 should be negative

fraction of the remaining numbers in set A that must be negative = 22/28 = 11/14

8) there are 15 black chips and 5 white chips lets say first chip picked is black. second one is white..

hence third pick (either black or whote)would gurantee that 2 chips are of same color

9) the pattern is blue, white, red, green, black, yellow, pink (total 7 colors)

Since the row begins with blue marble and ends with red marble the number of marbles would repeat for every 7th marble from red

3 10 17 24 31 38 45

Hence D

12) To find minimum value of x^2*y , both x^2 and y should be minimum

minimum value of y is -1/2
since x^2 is positive , minimum value of x^2 would be when x is -1/5

hence x^2 = (-1/5)^2*-1/4 = 1/25*-1/2 = -1/50

Originally posted by anuu on 22 Jan 2012, 15:25.
Last edited by anuu on 22 Jan 2012, 15:38, edited 1 time in total.
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Re: 12 Easy Pieces (or not?) [#permalink]
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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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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]
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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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]
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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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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