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Re: New DS set!!!
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10 Apr 2013, 21:11
9. If x is an integer, is x^2>2x? (1) x is a prime number. (2) x^2 is a multiple of 9. The question stem asks whether x(x2)>0 > x>2 OR x<0. From F.S 1, for x=2, we have a NO. For x=3, we have a YES. Insufficient. From F.S 2, for x=0, we have a NO. For x=3, we have a YES. Insufficient. Taking both together, we know that x =3 and get a YES. Sufficient. C.
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Re: New DS set!!!
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10 Apr 2013, 21:20
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 From F.S 1, for {1,0,1} the product is 0. For {2,3,4} the product is 24. Insufficient. From F.S 2, for {1,0,1} the product is 0. For {3,2,1} the product is6. Insufficient. If both are taken together, the fact that atleast one integer is positive makes it mandatory to choose integers starting from 1.If we choose 2, the last integer would be a 0, which is not positive. Thus, the only sets we could have are {1,0,1} or {0,1,2} ; in both cases the product is 0. Sufficient. C.
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Re: New DS set!!!
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10 Apr 2013, 21:27
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 twodigit prime number (2) x=qy+9, for some positive integer q From F.S 1, for y=23 and x=24, Remainder is 1, for y=23 and x=25, the remainder is 2. Insufficient. From F.S 2, we know that x = qy+9. Now we also know for the remainder(R)> 0<=R<Divisor > 0<=9<y. Thus, depending upon the value of y, we could have a different remainder.Insufficient. When both taken together, we know that 9 is definately less than the two digit prime number. Thus the remainder is 9.Sufficient. C.
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Re: New DS set!!!
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10 Apr 2013, 22:33
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 We know that (ra+rb)*6 = 1 unit of work> (ra+rb) = 1/6. From F.S 1, (1/ra+1/rb)/2 = 25/2 > (1/ra+1/rb) = 25. Need the value of 1/ra. Now we have two equations, multiplying them together , we have : 1+ra/rb+rb/ra+1 = 25/6 > ra/rb+rb/ra = 13/6 = 13/2.3 = (4+9)/2.3 = 2/3+3/2. Thus, ra/rb could be 2/3 or 3/2. Insufficient. From F.S 2, we have ra(t+5) = rb*t = 1. Thus, ra = 1/(t+5) and rb = 1/t. Putting this in the initial equation, we get > 1/(t+5) + 1/t = 1/6 = 5/30 = 2/30+3/30 = 1/15+1/10. Thus, t = 10 days and we can calculate the value of ra.Sufficient. B.
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Re: New DS set!!!
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11 Apr 2013, 00:59
10. What is the value of the median 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. From F.S 1, the given condition is only possible when all the elements in the given set are all equal, where all the elements are equal to the average. Insufficient. From F.S 2, for set {1,2,3} the average = range = 2 and the median = 2. For set {2,4,6} the average = range = 4 and the median = 4. Insufficient. Taking both together, all the elements have to be equal . The range will be zero, hence the average has to be zero. Thus the only set possible is when all the elements are zero.Sufficient. C.
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Re: New DS set!!!
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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!!!
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11 Apr 2013, 10:07
QUESTION 1 (1) We might have a product equal to 0: 0.1.2 = 0 but we might also have a product equal to 6: 1.2.3 = 6 INSUFFICIENT
(2) We might have a product equal to 6: 3.(2).(1) = 6 but we might also have a product equal to 0: 0.1.2 = 0 INSUFFICIENT
(1)+(2) If the integers are k, k1, k2, and we know (stmt2) that: k+ k1 + k2 < 6 => k < 3, we can only have: {0,1,2}or {1,0,1}, and both have a product of ZERO SUFFICIENT
ANS: C



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Re: New DS set!!!
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11 Apr 2013, 10:08
QUESTION 2 (1) Clearly INSUFFICIENT. Take x = 13, y = 11, for which the division gives remainder 2. Take x=12, y=11, for which the division gives remainder 1. (2) x = q.y + 9. We'll know that the remainder of the division is 9, IF y > 9. Take the case x=15,y=2, q=3 (which gives a remainder 1). Now, take the case x= 29, y=10, q= 9 (which gives a remainder 9). INSUFFICIENT
(1) + (2): Given, from stmt1, that y is larger than 9, then x=q.y+9 will always give remainder 9. SUFFICIENT
ANS: C



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Re: New DS set!!!
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11 Apr 2013, 10:08
QUESTION 3 (1) Clearly INSUFFICIENT. If ABC is isosceles, the median BD is also the height of the triangle relative to AC. We can increase the value of AC, as much as we want while also keeping BD equal to 12 cm
(2) AC^2 = AB^2 + BC^2 means B is a right angle (reverse property of the Pythagorean theorem). Which means, that AC is a diameter to the circle to which ABC is inscribed. This means that the the median BD is also equal to AD and to CD, all equal to the radius of the circle. AC = 24 cm SUFFICIENT
ANS: B



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Re: New DS set!!!
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11 Apr 2013, 10:15
QUESTION 4 Given that, Va + Vb = 1/6, essentially we need to know Va.
(1) (1/Va + 1/Vb)/2 = 12.5 => (Va + Vb)/(Va.Vb) = 25 => Va.Vb = 25/6 With Va + Vb = 1/6, and Va.Vb = 25/6, we can find Va and Vb, but we can't tell which is Va. INSUFFICIENT
(2) 1/Va = 1/Vb + 5, along with the information Va + Vb = 1/6, we can find out the solution. Knowing that Va > Vb, also gives us the exact values of Va and Vb. SUFFICIENT
(Observation: although there's no need to solve for the equation, if you're curious, the oslution is: A> 15 days; B> 10 days)
ANS: B



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Re: New DS set!!!
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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!!!
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11 Apr 2013, 10:28
QUESTION 6 x=3k , y=4k, z =8k and Average = (Sum of Ages)/15k.
(1) Average = 600/(15k) < 40? We need to know the value of k. If k=1, then the average = 40. If k=2, then the average is 20. INSUFFICIENT
(2) sum of ages in X = 120k; sum of ages in Y = 80 k; sum of ages in Z = 400k . Therefore Average = (600k)/15k = 40. SUFFICIENT
ANS: B



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Re: New DS set!!!
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Updated on: 11 Apr 2013, 10:33
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 Answer should be CS1 : possible cases are 1 0 1 and 0 1 2 and 1 2 3 and 2 3 4…… out of which 1 0 1 and 0 1 2 would yield product of zero and rest of the sequences would yield product as a certain positive integer. Multiple answers so this statement is not sufficient. S2 : possible cases are 0 1 2 and 1 0 1 and 0 1 2 and 1 2 3 ……… out of which 0 1 2 and 1 0 1 and 0 1 2 would yield product of zero and rest of the sequences would product as a certain negative integer. Multiple answers so this statement is not sufficient. S1 + S2 : 1 0 1 and 0 1 2 and 0 1 2 Each of these sequences yields product as zero so combining these statement together are sufficient to answer the question.
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Originally posted by Narenn on 11 Apr 2013, 10:30.
Last edited by Narenn on 11 Apr 2013, 10:33, edited 1 time in total.



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Re: New DS set!!!
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Updated on: 11 Apr 2013, 10:33
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 twodigit prime number (2) x=qy+9, for some positive integer q Answer should be CS1 : Y could be 11, 13, 17, 23, ……. and no further information about X. hence not sufficient S2 : If y is <9 then there will be multiple remainder. If y>9 then there will be only one remainder i.e.9 Not sufficient S1 + S2 : Y is two digit Prime number and Y is such that x=qy+9 for some Q so X/Y will certainly leave Remainder as 9. Sufficient.
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Originally posted by Narenn on 11 Apr 2013, 10:31.
Last edited by Narenn on 11 Apr 2013, 10:33, edited 1 time in total.



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Re: New DS set!!!
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Updated on: 11 Apr 2013, 10:32
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 Answer should be B (Doubtful about sufficiency/insufficiency of Stat 1) S1 : ABC is an isosceles triangle hence Angle A = Angle C and AD/DC = AB/BC. Further the median BD will act as an altitude. However since we don’t have any information about Angle B, we can not calculate the AC. Not Sufficient. S2 : This statement says Triangle ABC is right triangle with sides BA, BC, Hypotenuse AC and right angled at Angle B . As per the rule, in right triangle Median towards hypotenuse = ½ Hypotenuse. So BD=1/2AC 12 = ½ AC. AC = 24 Sufficient
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Originally posted by Narenn on 11 Apr 2013, 10:31.
Last edited by Narenn on 11 Apr 2013, 10:32, edited 1 time in total.



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Re: New DS set!!!
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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!!!
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11 Apr 2013, 10:34
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. Answer should be BLet the number of employees In 3 companies be 3x, 4x, and 8x. let the average age of employees in three companies be a, b, c respectively Question : is ((3ax + 4bx + 8cx)/15x) < 40 S1 : Total age = 3ax + 4bx + 8cx = 600 > (600/15x)<40> Here we can only derive x>0 So not sufficient. S2 : a=40, b=20, c=50 Average age = {(40 X 3x) + (20 X 4x) + (50 X 8x)}/3x + 4x + 8x > 600x/15x = 40 Sufficient
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Re: New DS set!!!
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11 Apr 2013, 10:35
9. If x is an integer, is x^2>2x? (1) x is a prime number. (2) x^2 is a multiple of 9. Answer should be CX^2>2x > x^22x >0 > x(x2)>0 that means either x and x2 are positive i.e. x>0 and x>2 or x and x2 are negative i.e. x<0 and x<2 Taking extreme values of both intervals it can be inferred that x>2 or x<0 that mean x should not be 0, 1, or 2 S1 : x = prime. So x can be 2,3,5,7,11,…….. x can be 2 Hence Not sufficient. S2 : x^2 has 9 as a factor. That means x should have 3 as a factor or x should be 0 So X can be 0, 3, 6, 9, 12, ……. Not Sufficient S1 + S2 : x is prime and x contain 3 so X must be 3. Sufficient
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Re: New DS set!!!
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11 Apr 2013, 10:50
QUESTION 8 Let r + b = 10, where r and b are the number of red marbles and blue marbles, respectively. The probability that both will be red is: C(r,2)/C(10,2) = r.(r1)/90. For it to be greater than 3/5, we need to have: r.(r1)/90 > 3/5 => r.(r1) > 54 => r can be 9 or 8.
(1) Probability = C(b,2)/C(10,2) = b.(b1)/(10.9) < 1/10. Therefore, b.(b1) < 9. Which means, b can be 2 (r = 8) or 3 (r=7). In the first, the answer to the stem question is YES, and in the second NO. INSUFFICIENT
(2) r > 6. If r = 7, the answer to the stem is NO. If r=8, the answer to the stem is YES. INSUFFICIENT
(1) + (2), No new information: r can be 7 or r can be 8. In the first case, the answer is NO. In the second, YES. INSUFFICIENT
ANS: E



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Re: New DS set!!!
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11 Apr 2013, 10:56
QUESTION 9 (1) For x = 2 (PRIME), we have x^2 = 4 = 2.x, but for x = 3 (PRIME), we have x^2 = 9 > 2.x = 6. INSUFFICIENT
(2) x^2 = 9.k. If x = 0 (and we consider that 0 is a multiple of 9), than x^2 = 2.x = 0. But, if x =3, than x^2=9>2.x=6. INSUFFICIENT
(1) + (2) Note that the functions y=x^2 and y=2x do not cross for x>2 (the parabolic function grows at a faster rate than the linear function for x>2). Since x^2 = 9.k, and x is prime, then x > 2, which means that x^2>2.x in this case. SUFFICIENT
ANS: C







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