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New DS set!!! [#permalink] New post 10 Apr 2013, 07:10
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The next set of medium/hard DS questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers.

1. What is the product of three consecutive integers?

(1) At least one of the integers is positive
(2) The sum of the integers is less than 6

Solution: new-ds-set-150653-60.html#p1211902


2. If x and y are both positive integers and x>y, what the remainder when x is divided by y?

(1) y is a two-digit prime number
(2) x=qy+9, for some positive integer q

Solution: new-ds-set-150653-60.html#p1211903


3. The length of the median BD in triangle ABC is 12 centimeters, what is the length of side AC?

(1) ABC is an isosceles triangle
(2) AC^2 = AB^2 + BC^2

Solution: new-ds-set-150653-60.html#p1211904


4. Two machines, A and B, each working at a constant rate, can complete a certain task working together in 6 days. In how many days, working alone, can machine A complete the task?

(1) The average time A and B can complete the task working alone is 12.5 days.
(2) It would take machine A 5 more days to complete the task alone than it would take for machine B to complete the task

Solution: new-ds-set-150653-60.html#p1211906


5. Set A={3-2x, 3-x, 3, 3+x, 3+2x}, where x is an integer. Is the standard deviation of set A more than the standard deviation of set B={3-2x, 3-x, 3, 3+x, 3+2x, y}

(1) The standard deviation of set A is positive
(2) y=3

Solution: new-ds-set-150653-60.html#p1211907


6. The ratio of the number of employees of three companies X, Y and Z is 3:4:8, respectively. Is the average age of all employees in these companies less than 40 years?

(1) The total age of all the employees in these companies is 600
(2) The average age of employees in X, Y, and Z, is 40, 20, and 50, respectively.

Solution: new-ds-set-150653-80.html#p1211908


7. Was the average (arithmetic mean) temperature in city A in March less than the average (arithmetic mean) temperature in city B in March?

(1) The median temperature in City A in March was less than the median temperature in city B
(2) The ratio of the average temperatures in A and B in March was 3 to 4, respectively

Solution: new-ds-set-150653-80.html#p1211909


8. Two marbles are drawn from a jar with 10 marbles. If all marbles are either red of blue, is the probability that both marbles selected will be red greater than 3/5?

(1) The probability that both marbles selected will be blue is less than 1/10
(2) At least 60% of the marbles in the jar are red

Solution: new-ds-set-150653-80.html#p1211910


9. If x is an integer, is x^2>2x?

(1) x is a prime number.
(2) x^2 is a multiple of 9.

Solution: new-ds-set-150653-80.html#p1211911


10. What is the value of the media of set A?

(1) No number in set A is less than the average (arithmetic mean) of set A.
(2) The average (arithmetic mean) of set A is equal to the range of set A.

Solution: new-ds-set-150653-80.html#p1211912


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8. Two marbles are drawn from a jar with 10 marbles. If all marbles are either red of blue, is the probability that both marbles selected will be red greater than 3/5?

The question: is P(R and R)=R/10*(R-1)/9>3/5?
Is R(R-1)>54?
Is R>7? (By number plugging) So, the question asks whether the number of red marbles is more than 7 (8, 9, or 10).

(1) The probability that both marbles selected will be blue is less than 1/10. This implies that B/10*(B-1)/9<1/10. So, we have that B(B-1)<9, thus B<4, so the number of red marbles in the jar is 7, 8, 9, or 10. Not sufficient.

(2) At least 60% of the marbles in the jar are red. This implies that the number of red marbles is 6 or more. Not sufficient.

(1)+(2) From above we have that R>6. Not sufficient.

Answer: E.
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7. Was the average (arithmetic mean) temperature in city A in March less than the average (arithmetic mean) temperature in city B in March?

(1) The median temperature in City A in March was less than the median temperature in city B. Clearly insufficient.

(2) The ratio of the average temperatures in A and B in March was 3 to 4, respectively. Temperatures can be negative, thus this statement is also not sufficient. Consider T(A)=3 and T(B)=4 AND T(A)=-3 and T(B)=-4.

(1)+(2) We have no additional useful info. Not sufficient.

Answer: E.
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Re: New DS set!!! [#permalink] New post 10 Apr 2013, 08:12
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9. If x is an integer, is x^2>2x?
x^2-2x>0
x(x-2)>0
The question asks is x in one of those intervals?
x<0 , x>2

(1) x is a prime number.
Sufficient All prime numbers are greater than 2 so x\geq{2} so we are in the right interval.
What a stupid mistake!Not suff, x can be 2
(2) x^2 is a multiple of 9.
Not sufficient. If x=0, x^2 =0 is a multiple of 9 but 0 isn't in the intervals, if x=3 x^2=9 is a multiple of nine and 3 is in our intervals.

C
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Re: New DS set!!! [#permalink] New post 11 Apr 2013, 15:37
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See Img for my solutions.These are very good question...
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Re: New DS set!!! [#permalink] New post 11 Apr 2013, 20:52
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9. If x is an integer, is x^2>2x?
(1) x is a prime number.
(2) x^2 is a multiple of 9.


1) If x = 2, x^2 = 2x. For any other integer, x^2 > 2x. Not sufficient.
2) As x^2 is a multiple of 9, x is not 2. So, x^2 > 2x. Sufficient.

Together: As x is prime number, x cannot be 0. As x^2 is multiple of 9, x can be 3, 6, 9, 12, …..
So, x^2 > 2x as x must be > 2. Sufficient.

Answer is C.
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Re: New DS set!!! [#permalink] New post 12 Apr 2013, 02:48
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1. (1) Insufficient. It could be {0,1,2} and the product is 0, or it could be {1,2,3} and their product is 6.
(2) Insufficient. It could be {0,1,2} and the product is 0, or it could be {-3,-2,-1} and the product is -6

(1)+(2) Sufficient. Suppose the least number is n, so the second is (n+1) and the third is (n+2). Their sum must be less than 6: n+(n+1)+(n+2)<6 or 3n+3<6 or n<1. By first statement at least one of the integers must be positive, it means that the largest is positive: n+2>0 or n>-2. So n=-1 or n=0. Therefore, there are two possible sets {-1, 0, 1} or {0, 1, 2}. The product anyway is 0.

The correct answer is C.

2. (1) Insufficient. For x=11 and y=10 the remainder when x is divided by y is 1, for x=12 and y=2 the remainder when x is divied by y is 2.
(2) Insufficient. The main point here is the remainder must be less than the divisor. We don't know is y>9 or y<=9. For y=1 the remainder when x is divided by y is 0, for y=2, x=2q+9 the remainder is 1 when x is divided by 2.

(1)+(2) Sufficient. If y is a two-digit number the when qy is divided by y there is remainder 0, and when 9 is divided by y the remainder is 9, since 9<y. So, x when divided by y gives the remainder 9.

The correct answer C
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5. Set A={3-2x, 3-x, 3, 3+x, 3+2x}, where x is an integer. Is the standard deviation of set A more than the standard deviation of set B={3-2x, 3-x, 3, 3+x, 3+2x, y}

(1) The standard deviation of set A is positive. We know that the standard deviation of any set is more than or equal to zero. The standard deviation of a set is zero only when the set consists of identical elements. So, this statement implies that set A does NOT consists of identical elements or that x does not equal to zero. Still this statement is not sufficient to answer the question.

(2) y=3. The mean of set A is 3. Now, if x\neq{0} for example if x=1, then the standard deviation of B would be smaller that the standard deviation A, since the elements of B would be less widespread than the element of set A. But if x=0, then A={3, 3, 3, 3, 3} and B={3, 3, 3, 3, 3, 3}, so both will have the standard deviation of zero. Bot sufficient.

(1)+(2) Since from (1) x\neq{0}, then adding a new element which equals to the mean will shrink the standard deviation, thus SD(A)>SD(B). Sufficient.

Answer: C.
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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10. What is the value of the media of set A?

(1) No number in set A is less than the average (arithmetic mean) of set A.

Since no number is less than the average, then no number is more than the average, which implies that the list contains identical elements: A={x, x, x, ...}. From this it follows that (the average)=(the median). But we don't know the value of x, thus this statement is NOT sufficient.

(2) The average (arithmetic mean) of set A is equal to the range of set A.

Not sufficient: if A={0, 0, 0, 0}, then (the median)=0, but if A={1, 2, 2, 3}, then (the median)=2.

(1)+(2) From (1) we have that the list contains identical elements. The range of all such sets is 0. Therefore, from (2) we have that (the average)=(the range)=0 and since from (1) we also know that (the average)=(the median), then (the median)=0. Sufficient.

Answer: C..
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NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: New DS set!!! [#permalink] New post 10 Apr 2013, 07:33
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1. What is the product of three consecutive integers?

(1) At least one of the integers is positive
Clearly Not sufficient. {0, 1, 2} product=0; {1, 2, 3} product = 6
(2) The sum of the integers is less than 6
Clearly Not sufficient. {0, 1, 2} product=0; {-1, -2, -3} product = 6

(1)+(2) From 2 we know that a+(a+1)+(a+2)<6 so a<1
The three integers are a,a+1,a+2, and at least one is positive (knowing that a is an integer less than 1) : 1)a>0 this cannot be: there is no integer between 0 and 1;
2)a+1>0 a>-1 with a<1 there is one solution a=0 and the product is 0 {0,1,2};3) a+2>0 a>-2 here we have 2 solutions a=-1 and the numbers are {-1,0,1}=>prod=0 or a=0 and the product is 0.
IMO C
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Re: New DS set!!! [#permalink] New post 10 Apr 2013, 07:49
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2. If x and y are both positive integers and x>y, what the remainder when x is divided by y?

(1) y is a two-digit prime number
Not sufficient. ie 22/11 R=0, 23/11 R=1
(2) x=qy+9, for some positive integer q
Not sufficient. ie x= 29 = 20+9 = 4*5+9 or 5*4+9 I don't know if 4 or 5 is y and this changes the R.

(1)+(2) If y is a two-digit prime number x=qy+9 for any q says that the R of x/y is 9. x/y=q+9/y
IMO C
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Re: New DS set!!! [#permalink] New post 10 Apr 2013, 08:33
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Q9) If x is an integer, is x^2>2x?

From Stmt 1) x is 2, 3, 5, 7, 11, ... all primes - Not Sufficient

If x = 2 results in x^2 = 2x and if x = 3, 4, .. only then x^2 > 2x

From Stmt 2) x is -3, -6, 0, 6, 9, ... - Not Sufficient

If x = 0 results in x^2 = 2x and if x = -3, 4, .. only then x^2 > 2x

From Stmt 1 and Stmt 2, the number that can meet the requirement is 3

Answer is C
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Re: New DS set!!! [#permalink] New post 10 Apr 2013, 08:54
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6. The ratio of the number of employees of three companies X, Y and Z is 3:4:8, respectively. Is the average age of all employees in these companies less than 40 years?

(1) The total age of all the employees in these companies is 600
Not sufficient. ie totEmployees= 15 (3+4+8) => average age = 600/15=40 or totEmployees = 30 (2*(3+4+8))=> average age = 600/30=20. In the first case the answer is NO, in the second is YES
(2) The average age of employees in X, Y, and Z, is 40, 20, and 50, respectively.
\frac{(40*3+20*4+50*8)}{15}=\frac{120+80+400}{15}=\frac{600}{15}=40 If you weight the average age with the ratios you find out that the average age is 40, which is NOT less than 40. Sufficient
IMO B
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Re: New DS set!!! [#permalink] New post 10 Apr 2013, 09:08
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Bunuel wrote:
The next set of medium/hard DS questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers.

1. What is the product of three consecutive integers?

(1) At least one of the integers is positive -Can not determine the values
(2) The sum of the integers is less than 6- Can not determine the values

(1)+(2)-> at least one positive and <6 so can't be 1,2,3 -> so 0 must be part of the sequence .


IMO: C


Bunuel wrote:
2. If x and y are both positive integers and x>y, what the remainder when x is divided by y?

(1) y is a two-digit prime number - Can not determine the values
(2) x=qy+9, for some positive integer q - Can not determine the values
1+2=>y>9 , as y is two digit ...so remainder 9



IMO: C

Bunuel wrote:
3. The length of the median BD in triangle ABC is 12 centimeters, what is the length of side AC?

(1) ABC is an isosceles triangle No idea about which sides are same
(2) AC^2 = AB^2 + BC^2 D is mid point of AC , SO AD= DC=BD = 12 => AC

IMO:B

Bunuel wrote:
4. Two machines, A and B, each working at a constant rate, can complete a certain task working together in 6 days. In how many days, working alone, can machine A complete the task?

(1) The average time A and B can complete the task working alone is 12.5 days.say A can complete in x , then b in = 12.5*2 - x
hence \frac{1}{x}+\frac{1}{25-x}=\frac{1}{6}=> x= 15 or 10

(2) It would take machine A 5 more days to complete the task alone than it would take for machine B to complete the task
\frac{1}{x}+\frac{1}{x+5}=\frac{1}{6}=> x= 10
So A 15 days
[/color]

IMO: B

Bunuel wrote:
5. Set A={3-2x, 3-x, 3, 3+x, 3+2x}, where x is an integer. Is the standard deviation of set A more than the standard deviation of set B={3-2x, 3-x, 3, 3+x, 3+2x, y}

(1) The standard deviation of set A is positiveNot sufficient
(2) y=3So we added 3 to the existing series , where 3 was mean=> SD will decrease

IMO: B


Bunuel wrote:
6. The ratio of the number of employees of three companies X, Y and Z is 3:4:8, respectively. Is the average age of all employees in these companies less than 40 years?

(1) The total age of all the employees in these companies is 600Can 't do anything
(2) The average age of employees in X, Y, and Z, is 40, 20, and 50, respectively.can determine
40*3x + 20*4x + 50 *8x = total age , now dividing by 15x we can get avg age


IMO: B


Bunuel wrote:
7. Was the average (arithmetic mean) temperature in city A in March less than the average (arithmetic mean) temperature in city B in March?

(1) The median temperature in City A in March was less than the median temperature in city BNo idea about avg
(2) The ratio of the average temperatures in A and B in March was 3 to 4, respectivelyit means Avg(A) < Avg(B)

IMO: B

Bunuel wrote:
8. Two marbles are drawn from a jar with 10 marbles. If all marbles are either red of blue, is the probability that both marbles selected will be red greater than 3/5?

(1) The probability that both marbles selected will be blue is less than 1/10
say there are b blue and 10-b red=> \frac{(bC2)}{(10C2) < 0.1=> b(b-1) < 9 Not sufficient}
(2) At least 60% of the marbles in the jar are red
So Red GE 6 and Blue LE 4 => for 6 red p(selecting 2 red out of 6) = 0.33 < \frac{3}{5} but if all are red P(red) = 1 > \frac{3}{5}
so insufficient



Now adding 1+2 => blue can be 4 ,3,2,1,0 both blue should be less than 0.1 => so b can't be 4 it can only 3,2
So red balls can be 7 or 8-> no definite ans

IMO: E


Bunuel wrote:
9. If x is an integer, is x^2>2x?

(1) x is a prime number.not sufficient can be 2 or 3..
(2) x^2 is a multiple of 9. =>X is a multiple of 3 when we do squaring we multiply with a multiple of 3 and whem we do 2X we multiply with 2
so x^2>2x is always true except x= 0


adding 1 +2 => x can't be zero
IMO: C

Bunuel wrote:
10. What is the value of the media of set A?

(1) No number in set A is less than the average (arithmetic mean) of set A. so all the numbers must be same =>median is any number of the set, sufficient to determine
(2) The average (arithmetic mean) of set A is equal to the range of set A.not sufficient

IMO: A


Please let me know how many i got correct

Last edited by focus2k13 on 11 Apr 2013, 19:00, edited 1 time in total.
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Re: New DS set!!! [#permalink] New post 10 Apr 2013, 09:11
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5. Set A={3-2x, 3-x, 3, 3+x, 3+2x}, where x is an integer. Is the standard deviation of set A more than the standard deviation of set B={3-2x, 3-x, 3, 3+x, 3+2x, y}

(1) The standard deviation of set A is positive
Not sufficient.ie with x=1
A={1,2, 3, 4, 5} B={1, 2, 3, 4, 5, y} All depends on y.
(2) y=3
Not sufficient.ie if x=0
A={3,3,3,3,3} B={3,3,3,3,3,3} STD of B is = STD of A
if x=1
A={1,2, 3, 4, 5} B={1, 2, 3, 4, 5, 3} STD of B is < STD of A

(1)+(2) From 1 we know that x\neq{0} and from 2 that y=3. Sufficient. ie:x=1 : A={1,2, 3, 4, 5} B={1, 2, 3, 4, 5, 3} STD of B is < STD of A
x=1000 : A={-1997,-997, 3, 1003, 2003} B={-1997,-997, 3, 1003, 2003, 3} STD of B is < STD of A
A and B share 4 elements in common that are different from 3, but because B has one more 3 than X it will have a STD lesser than A
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Re: New DS set!!! [#permalink] New post 10 Apr 2013, 09:46
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8. Two marbles are drawn from a jar with 10 marbles. If all marbles are either red of blue, is the probability that both marbles selected will be red greater than 3/5?

(1) The probability that both marbles selected will be blue is less than 1/10
\frac{B}{10}*\frac{B-1}{9}<\frac{1}{10}
B^2-B-9<0 b<3,5xxx so b can be (0,1,2,3) If b=0 the answer is YES; if b=3 the answer is NO (\frac{7}{10}*\frac{6}{9}=\frac{42}{90}<\frac{3}{5})
Not sufficient.
(2) At least 60% of the marbles in the jar are red
R\geq{6}. Clearly not sufficient.

(1)+(2) Since together they add no new info IMO E
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Re: New DS set!!! [#permalink] New post 10 Apr 2013, 11:59
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1. What is the product of three consecutive integers?

(1) At least one of the integers is positive
(2) The sum of the integers is less than 6

My answer C

Stmt 1: Three consecutive integers can be any among (-1,0,1) or (1,2,3) or (99,100,101) etc...no sufficient.
Stmt 2: Three consecutive integers can be any among (-100,-99,-98) or (-1,0,1) or (-16,-15,-14) etc...no sufficient.

combining both statements we have only 2 sets (-1,0,1) or (0,1,2). and in ether case product is 0. hence C
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Re: New DS set!!! [#permalink] New post 10 Apr 2013, 12:08
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2. If x and y are both positive integers and x>y, what the remainder when x is divided by y?

(1) y is a two-digit prime number
(2) x=qy+9, for some positive integer q


My answer C.

Stmt 1: Y can be among 11,13,37,61 etc.. and so can X be any number satisfying X>Y.Hence reminder when x is divided by y cannot be uniquely determined.

Stmt 2 : x=qy+9. Say Q=2 and Y=3 .=>X=15 and X/Y leave no reminder.
Say Q=3 and Y=4 => X=21 and X/Y leaves a reminder of 1.. Hence insufficient.

combining : we know that Y>9..hence the reminder when X=qy+9/y will always leave a reminder of 9
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Re: New DS set!!! [#permalink] New post 10 Apr 2013, 12:45
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4. Two machines, A and B, each working at a constant rate, can complete a certain task working together in 6 days. In how many days, working alone, can machine A complete the task?

(1) The average time A and B can complete the task working alone is 12.5 days.
(2) It would take machine A 5 more days to complete the task alone than it would take for machine B to complete the task

My answer B


Time taken by A alone is a and by B is b.

Stmt 1: (a+b)/2= 12.5 => a+b=25. together they complete the task in 6 days => (1/a)+(1/b)=1/6 =>(ab)/(a+b) =6
a+b=25 => ab = 150. So (a,b) can be (10,15) or (15,10). insufficient.

Stmt 2: b=a-5.=> (1/a)+(1/a-5) = 1/6. on simplification we get a quadratic a = a^2-17a+30=0 solving , a=15 or 2.
a cant be 2 ,as a=2=> b=-3 which on this earth is not possible. Hence a=15. sufficient.
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Re: New DS set!!! [#permalink] New post 10 Apr 2013, 12:57
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Q1. Let 3 integers be (x-1), x and (x+1)

St1: At least 1 integer is positive.
Case 1 (1 Integers positive): 1 -- 2 -- 3 (Product = 6)
Case 2 (2 Integers positive): 0 -- 1 -- 2 (Product = 0)
Case 3 (All 3 Integers positive): -1 -- 0 -- 1 (Product = 0)
Hence Not Sufficient

St2: sum of 3 integers < 6
x - 1 + x + x + 1 < 6
=> x < 2
if integers are 0, 1, 2 (product = 0)
if integers are -6, -2, -1 (product = -12)
Insufficient

Together: atleast 1 integer is negative and sum < 6
Case1: if 2 integers >0, 0, 1, 2 (product = 0)
Case1: if 1 integer is positive, -1, 0, 1 (product = 0)
both case will include a 0 hence the product will be 0
Together sufficient.
Ans C

Last edited by srcc25anu on 10 Apr 2013, 14:15, edited 1 time in total.
Re: New DS set!!!   [#permalink] 10 Apr 2013, 12:57
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