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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#p12119022. 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#p12119033. 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#p12119044. 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#p12119065. 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#p12119076. 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#p12119087. 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#p12119098. 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#p12119109. 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#p121191110. 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#p1211912Kudos points for each correct solution!!!
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I just wanna say: thanks Bunuel !
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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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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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Q8) Whether the probably of drawing 2 red marbles from a jar of 10 marbles is > 3/5 From Stmt 1, the probability that both marbles selected will be blue is < 1/10 and since there are only red and blue marbles in the jarThe probability of selecting both marbles red = 1 - LT 1/10 ==> GT 9/10 - Sufficient From Stmt 2, the number of red marbles in the jar is at least 60 % is >= 6 of 10 marbles in the jarThe probability of selecting both marbles red >= 6/10 * 5/9 >= 0.333 which can be less than 3/5 when there are 6 red marbles and more than 3/5 when there are 9 red marbles - Not Sufficient Hence answer is A
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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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Q10) What is the value of the media of set A?From Stmt 1, No number in set A < Avg, of set AThis implies all the numbers in the set are equal to the average otherwise the Avg of the set A will be more if any one number is more than the Avg. Hence the median of set A in which all the numbers are same as the Avg. is none other than the Avg. value. - Sufficient.
From Stmt 2, Avg. of set A is = Amax - Amin of Set A.This only implies the lower and upper limit of Set A and they could be even or odd number of values in the set. This will result in various values for median within Amin and Amin. - Not Sufficient. Answer is A
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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 SufficientIf 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 SufficientIf 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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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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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
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Q6) Is Avg. age of employees of three companies X, Y and Z < 40 years?Using the unknown multiplier approach, the number of employees in companies X, Y, and Z are 3x, 4x, and 8x respectively From Stmt 1, the total age of all the employees in 3 companies is 600 - Not SufficientIf x = 1, the total employees in companies X, Y and Z is 3 + 4 + 8 = 15, so the avg. age of employees is 600 / 15 = 40 years If x = 2, the total employees in companies X, Y and Z is 6 + 8 + 16 = 30, so the avg. age of employees is 600 / 30 = 20 years From Stmt 2, avg age of X is 40, avg age of Y is 20, and avg age of z is 50 - SufficientIf x = 1, the total employees in companies X, Y and Z is 3 + 4 + 8 = 15, so the avg. age of employees is (40*3+20*4+50*8) / 15 = 40 years If x = 2, the total employees in companies X, Y and Z is 6 + 8 + 16 = 30, so the avg. age of employees is (40*6+20*8+50*16) / 30 = 40 years Answer is B
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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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Q7) Average (arithmetic mean) temperature in city A in March < the average (arithmetic mean) temperature in city B in March?From Stmt 1, average temperate for City A and B cannot be determined from median - Not Sufficientmean = median IFF the temperature in City A and City B in March follows a sequence of evenly spaced values. From Stmt 2, Avg Temp A / Avg Temp B = 3 / 4 - SufficientUsing unknown multiplier approach Avg Temp in A is 3x and Avg Temp in B is 4x so no matter what the value of x the Avg. temp in B > Avg. temp in A Answer is - B
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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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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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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]
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10 Apr 2013, 13:29
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
My answer C. Solution in the attachment
Attachments
DS.png [ 24.17 KiB | Viewed 22814 times ]
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Re: New DS set!!! [#permalink]
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10 Apr 2013, 13:31
My Answers:
1. C
2. B
3. C
4. B
5. B
6. B
7. B
8. B
9. B
10. E
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Re: New DS set!!! [#permalink]
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10 Apr 2013, 13:45
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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]
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10 Apr 2013, 13:57
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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, 15:15, edited 1 time in total.
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