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New DS set!!! [#permalink]
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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: newdsset15065360.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 twodigit prime number (2) x=qy+9, for some positive integer q Solution: newdsset15065360.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: newdsset15065360.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: newdsset15065360.html#p12119065. Set A={32x, 3x, 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={32x, 3x, 3, 3+x, 3+2x, y}(1) The standard deviation of set A is positive (2) y=3 Solution: newdsset15065360.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: newdsset15065380.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: newdsset15065380.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: newdsset15065380.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: newdsset15065380.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: newdsset15065380.html#p1211912Kudos points for each correct solution!!!
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Re: New DS set!!! [#permalink]
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10 Apr 2013, 14:41
Q7: Is avg temp in A < avg temp in B?
St1: median temp in A < median temp in B case 1: Temp in A for 3 days = 1  1  10 (mean = 4, median = 1) Temp in B for 3 days = 2  2  2 (mean = 2, median = 2) MEAN of A > MEAN of B
Case 2:Temp in A for 3 days = 1  1  2 (mean = 4/3, median = 1) Temp in B for 3 days = 2  2  2 (mean = 2, median = 2) MEAN of A < MEAN of B hence insufficient
St2: Ratio of avg March temp in A : Ratio of avg March temp in B = 3:4 clearly implies Avg Temp of A for March < Avg temp of B for MArch Sufficient
Answer B



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Re: New DS set!!! [#permalink]
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10 Apr 2013, 14:44
5. Set A={32x, 3x, 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={32x, 3x, 3, 3+x, 3+2x, y}
(1) The standard deviation of set A is positive (2) y=3
My answer C
SD of A = 2*x^2. and of B = (10*x^2 +(3y)^2)/6. we need to check if SD(A)SD(B)>0
Stmt 1: from this we know that x^2>=1. But not the value of Y. SD(A)SD(B)= .34*x^2((3y)^2)/6. We dont know the values of X or Y. hence cant decide.
Stmt 2: Y =3. hence SD(B)=1.66*x^2. Hence SD(A)SD(B)=.34*x^2>=0. 0 for x=0. Hence insufficient.
Combining We know X^2>=1 and Y=3 hence SD(A)>SD(B). Sufficient.



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Re: New DS set!!! [#permalink]
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11 Apr 2013, 05:22
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. From F.S 1, if the no of employees is in X,Y,Z is 3k,4k,8k respectively;k is a positive integer; the average age = 600/(3+4+8)k = 40/k, hence a NO for k=1,a YES for k=2. Insufficient. From F.S 2, the average age of all the employees for no of employees in the given ratio= (3k*40+4k*20+8k*50)/15k ; where k is a positive integer > 600/15 = 40,hence a NO irrepective of the value of k. Sufficient. B.
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Re: New DS set!!! [#permalink]
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11 Apr 2013, 10:23
QUESTION 5
A={3 2x, 3x , 3, 3+x, 3+2x} > mean = 3, deviations from the mean: 2x, x, 0, x, 2x B={3 2x, 3x , 3, 3+x, 3+2x, y} > mean = (15/6 +y/6),
(1) SD of A is positive, then x>0. (no other restrictions on x, so let's take for example x = 1). The deviations of set A are 2, 1,1,2. If y is a gigantic number, and x = 1, it is obvious that the deviations from the mean in set B will be much higher than 2, 1,0,1,2. However, if y = 3, the deviations will also be 2,1,1,2. INSUFFICIENT
(2) y = 3, means that the mean for set B = 3, and deviations are: 2x, x,0,0,x,2x. Therefore SD for B and A are the same. SUFFICIENT
ANS: B



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Re: New DS set!!! [#permalink]
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11 Apr 2013, 10:32
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 Answer should be CThe work is 100% Machine A and B, working together, can finish a work in 6 days. So in a day they are completing 100/6 = 16.66% of work. A + B = 16.66 S1 : (100/A) + (100/B) = 25 Not Sufficient. S2 : (100/A) – (100/B) = 5 Not Sufficient. S1 + S2 : Solving two equations we would get A= 10 Sufficient.
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Re: New DS set!!! [#permalink]
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11 Apr 2013, 14:52
Q5) Is the standard deviation of set A {32x, 3x, 3, 3+x, 3+2x} > the standard deviation of set B={32x, 3x, 3, 3+x, 3+2x, y} From Stmt 1, y is not known and only SD > 0  Not SufficientIf y <= 3 SD of set A > SD of set B and for other values SD of set A < SD of set B From Stmt 2, y = 3  SufficientWhen y = 3 and for any integer value for x SD is always >= 0 and SD of set A > SD of set B Answer is B
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Re: New DS set!!! [#permalink]
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11 Apr 2013, 21:51
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
1) This says nothing about temperatures of other days. Not sufficient. 2) This says that the average temperature in city A in March was less than the average temperature in city B in March. Sufficient.
Answer is B.



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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?Given that the ratio of the number of employees is 3x:4x:8x, for some positive multiple x. The questions asks whether (average age)=(total age)/(# of employees)<40, or whether (total age)/(3x+4x+8x)<40, which is the same as: is (total age)<600x? (1) The total age of all the employees in these companies is 600. The question becomes: is 600<600x? Or is 1<x. We don't know that: f x=1, then the answer is NO but if x>1, then the answer is YES. Not sufficient. (2) The average age employees in X, Y, and Z, is 40, 20, and 50, respectively. (total age)=40*3x+20*4x+50*8x=600x, so the answer to the question is NO. Sufficient. Answer: B.
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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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Please help:
1) Is there a fixed time frame to post the answers ? (Background: I tried posting yesterday, 11 AM IST, but the post was locked)
2) How can I quickly know about similar posts which have open questions accompanied with kudos for answers ? Subscribing to a topic didn't work for me. I am a corporate slave.



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Re: New DS set!!! [#permalink]
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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?
Given that the ratio of the number of employees is 3x:4x:8x, for some positive multiple x.
The questions asks whether (average age)=(total age)/(# of employees)<40, or whether (total age)/(3x+4x+8x)<40, which is the same as: is (total age)<600x?
(1) The total age of all the employees in these companies is 600. The question becomes: is 600<600x? Or is 1<x. We don't know that: f x=1, then the answer is NO but if x>1, then the answer is YES. Not sufficient.
(2) The average age employees in X, Y, and Z, is 40, 20, and 50, respectively. (total age)=40*3x+20*4x+50*8x=600x, so the answer to the question is NO. Sufficient.
Answer: B.  Hi Bunuel, Doubt in this Q. Pls help & suggest where am I going Wrong in this Question ...... Q .. 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 Now, it gives the ratio of number of employees i.e.; 3:4:8 & then we can take this as real number of employees like 3 employees in Company X, 4 in Y & 8 in company Z ...therefore, we have 3+4+8=15 , 15 employees in total....& average age of employees in company X must be something, lets say its x for Company X, so the total age for company X will be 3x, similarly, for Company Y the average age is y therefore, total age for company Y will be 4y & for Company Z, the total age will be 8z. & the average of the total ages will be 3x+4y+8z/15 ...... & as the question asks, Is the average age of all employees in these companies less than 40 years ?? i.e.; 3x+4y+8z/15 <40 or 3x+4y+8z<600...... so we have to find out if 3x+4y+8z<600 ..... rephrasing complete. now statement (1) The total age of all the employees in these companies is 600. that means 3x+4y+8z=600 but we have to check if 3x+4y+8z<600. so it is clearly NO. statement (2) The average age of employees in X, Y, and Z, is 40, 20, and 50, respectively therefore, total age = 3*40+4*20+8*50 = 600. as we have to check if 3x+4y+8z<600. so it is also a clear NO. & hence , answer is D. Now please let me know where am I going wrong. Your help will be appreciated. Thanks !!!!
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15 Apr 2013, 00:21
manishuol wrote: 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?
Given that the ratio of the number of employees is 3x:4x:8x, for some positive multiple x.
The questions asks whether (average age)=(total age)/(# of employees)<40, or whether (total age)/(3x+4x+8x)<40, which is the same as: is (total age)<600x?
(1) The total age of all the employees in these companies is 600. The question becomes: is 600<600x? Or is 1<x. We don't know that: f x=1, then the answer is NO but if x>1, then the answer is YES. Not sufficient.
(2) The average age employees in X, Y, and Z, is 40, 20, and 50, respectively. (total age)=40*3x+20*4x+50*8x=600x, so the answer to the question is NO. Sufficient.
Answer: B.  Hi Bunuel, Doubt in this Q. Pls help & suggest where am I going Wrong in this Question ...... Q .. 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 Now, it gives the ratio of number of employees i.e.; 3:4:8 & then we can take this as real number of employees like 3 employees in Company X, 4 in Y & 8 in company Z ...therefore, we have 3+4+8=15 , 15 employees in total....& average age of employees in company X must be something, lets say its x for Company X, so the total age for company X will be 3x, similarly, for Company Y the average age is y therefore, total age for company Y will be 4y & for Company Z, the total age will be 8z. & the average of the total ages will be 3x+4y+8z/15 ...... & as the question asks, Is the average age of all employees in these companies less than 40 years ?? i.e.; 3x+4y+8z/15 <40 or 3x+4y+8z<600...... so we have to find out if 3x+4y+8z<600 ..... rephrasing complete. now statement (1) The total age of all the employees in these companies is 600. that means 3x+4y+8z=600 but we have to check if 3x+4y+8z<600. so it is clearly NO. statement (2) The average age of employees in X, Y, and Z, is 40, 20, and 50, respectively therefore, total age = 3*40+4*20+8*50 = 600. as we have to check if 3x+4y+8z<600. so it is also a clear NO. & hence , answer is D. Now please let me know where am I going wrong. Your help will be appreciated. Thanks !!!! Given that the ratio of the number of employees of three companies X, Y and Z is 3:4:8, respectively, so the number of employees could be: 3, 4, 8; 3* 2=6, 4* 2=8, 8* 2=16; 3* 3=9, 4* 3=12, 8* 3=24; 3* 4=12, 4* 4=16, 8* 4=32; ... Notice that the multiple is the same in each case. Thus the ratio of the number of employees is 3x:4x:8x, for some positive multiple x, not 3x:4y:8z. Hope it's clear.
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15 Apr 2013, 00:51
Bunuel wrote: manishuol wrote: 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?
Given that the ratio of the number of employees is 3x:4x:8x, for some positive multiple x.
The questions asks whether (average age)=(total age)/(# of employees)<40, or whether (total age)/(3x+4x+8x)<40, which is the same as: is (total age)<600x?
(1) The total age of all the employees in these companies is 600. The question becomes: is 600<600x? Or is 1<x. We don't know that: f x=1, then the answer is NO but if x>1, then the answer is YES. Not sufficient.
(2) The average age employees in X, Y, and Z, is 40, 20, and 50, respectively. (total age)=40*3x+20*4x+50*8x=600x, so the answer to the question is NO. Sufficient.
Answer: B.  Hi Bunuel, Doubt in this Q. Pls help & suggest where am I going Wrong in this Question ...... Q .. 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 Now, it gives the ratio of number of employees i.e.; 3:4:8 & then we can take this as real number of employees like 3 employees in Company X, 4 in Y & 8 in company Z ...therefore, we have 3+4+8=15 , 15 employees in total....& average age of employees in company X must be something, lets say its x for Company X, so the total age for company X will be 3x, similarly, for Company Y the average age is y therefore, total age for company Y will be 4y & for Company Z, the total age will be 8z. & the average of the total ages will be 3x+4y+8z/15 ...... & as the question asks, Is the average age of all employees in these companies less than 40 years ?? i.e.; 3x+4y+8z/15 <40 or 3x+4y+8z<600...... so we have to find out if 3x+4y+8z<600 ..... rephrasing complete. now statement (1) The total age of all the employees in these companies is 600. that means 3x+4y+8z=600 but we have to check if 3x+4y+8z<600. so it is clearly NO. statement (2) The average age of employees in X, Y, and Z, is 40, 20, and 50, respectively therefore, total age = 3*40+4*20+8*50 = 600. as we have to check if 3x+4y+8z<600. so it is also a clear NO. & hence , answer is D. Now please let me know where am I going wrong. Your help will be appreciated. Thanks !!!! Given that the ratio of the number of employees of three companies X, Y and Z is 3:4:8, respectively, so the number of employees could be: 3, 4, 8; 3* 2=6, 4* 2=8, 8* 2=16; 3* 3=9, 4* 3=12, 8* 3=24; 3* 4=12, 4* 4=16, 8* 4=32; ... Notice that the multiple is the same in each case. Thus the ratio of the number of employees is 3x:4x:8x, for some positive multiple x, not 3x:4y:8z. Hope it's clear.  HI, Thanks for ur quick reply ...... yes, you are right, the ratio of number of employees is 3x:4x:8x but I didn't mean that the ratio of number of employees is 3x:4y:8z instead I said that 3x+4y+8z can be the total weight of the employees of three companies X,Y, Z if we assume that the average age in Company X is x & average age in Company Y is y & average age in Company Z is z then the total age of employees of company X will be 3x & the total age of employees of company Y will be 4y & the total age of employees of company Z will be 8z & therefore, the total age of all the employees of the three companies will be 3x+4y+8z & then we can calculate the average. Please review & give me some advise on that. Thanks !! in advance !!!!
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manishuol wrote: Bunuel wrote: manishuol wrote: 
Hi Bunuel,
Doubt in this Q. Pls help & suggest where am I going Wrong in this Question ......
Q .. 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
Now, it gives the ratio of number of employees i.e.; 3:4:8 & then we can take this as real number of employees like 3 employees in Company X, 4 in Y & 8 in company Z ...therefore, we have 3+4+8=15 , 15 employees in total....& average age of employees in company X must be something, lets say its x for Company X, so the total age for company X will be 3x, similarly, for Company Y the average age is y therefore, total age for company Y will be 4y & for Company Z, the total age will be 8z. & the average of the total ages will be 3x+4y+8z/15 ...... & as the question asks, Is the average age of all employees in these companies less than 40 years ?? i.e.; 3x+4y+8z/15 <40 or 3x+4y+8z<600...... so we have to find out if 3x+4y+8z<600 ..... rephrasing complete.
now statement (1) The total age of all the employees in these companies is 600. that means 3x+4y+8z=600 but we have to check if 3x+4y+8z<600. so it is clearly NO.
statement (2) The average age of employees in X, Y, and Z, is 40, 20, and 50, respectively therefore, total age = 3*40+4*20+8*50 = 600. as we have to check if 3x+4y+8z<600. so it is also a clear NO. & hence , answer is D.
Now please let me know where am I going wrong. Your help will be appreciated. Thanks !!!! Given that the ratio of the number of employees of three companies X, Y and Z is 3:4:8, respectively, so the number of employees could be: 3, 4, 8; 3* 2=6, 4* 2=8, 8* 2=16; 3* 3=9, 4* 3=12, 8* 3=24; 3* 4=12, 4* 4=16, 8* 4=32; ... Notice that the multiple is the same in each case. Thus the ratio of the number of employees is 3x:4x:8x, for some positive multiple x, not 3x:4y:8z. Hope it's clear.  HI, Thanks for ur quick reply ...... yes, you are right, the ratio of number of employees is 3x:4x:8x but I didn't mean that the ratio of number of employees is 3x:4y:8z instead I said that 3x+4y+8z can be the total weight of the employees of three companies X,Y, Z if we assume that the average age in Company X is x & average age in Company Y is y & average age in Company Z is z then the total age of employees of company X will be 3x & the total age of employees of company Y will be 4y & the total age of employees of company Z will be 8z & therefore, the total age of all the employees of the three companies will be 3x+4y+8z & then we can calculate the average. Please review & give me some advise on that. Thanks !! in advance !!!! Can you please tell me which step in the solution ( newdsset15065380.html#p1211908) you don't understand. This way I think it would be easier to explain.
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15 Apr 2013, 01:23
okay ..sir .. as you said in the first line of ur solution to this Q ... Given that the ratio of the number of employees is 3x:4x:8x, for some positive multiple x.,, I think you have taken x as an integer so that it is flexible to say tat this ratio can be 3:4:8 or 3*2 : 4*2 : 8*2 or 6:8:16 or 9:12:24... I would like to know why can't we change this ratio of number employees to the ratio of weights of employees of the three companies...Can we change the Ratio of number of Employees to the Ratio of weights of Employees of three companies .... Pls Advise , Thanks in Advance !!
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