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# New DS set!!!

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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

Kudos points for each correct solution!!!
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10 Apr 2013, 20:27
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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

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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10 Apr 2013, 21:33
1
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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

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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10 Apr 2013, 23:59
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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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11 Apr 2013, 04:22
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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11 Apr 2013, 09:07
1
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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, k-1, k-2, and we know (stmt2) that: k+ k-1 + k-2 < 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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11 Apr 2013, 09:08
1
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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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11 Apr 2013, 09:08
1
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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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11 Apr 2013, 09:15
1
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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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11 Apr 2013, 09:23
QUESTION 5

A={3 -2x, 3-x , 3, 3+x, 3+2x} -> mean = 3, deviations from the mean: 2x, x, 0, x, 2x
B={3- 2x, 3-x , 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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11 Apr 2013, 09:28
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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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11 Apr 2013, 09:30
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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

S1 :- 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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11 Apr 2013, 09:31
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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

S1 :- 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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11 Apr 2013, 09:31
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Expert's post
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

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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11 Apr 2013, 09:32
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

The 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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11 Apr 2013, 09:34
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Expert's post
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.

Let 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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11 Apr 2013, 09:35
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Expert's post
9. If x is an integer, is x^2>2x?

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

X^2>2x -------> x^2-2x >0 -----> x(x-2)>0 that means either x and x-2 are positive i.e. x>0 and x>2 or x and x-2 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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11 Apr 2013, 09:50
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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.(r-1)/90.
For it to be greater than 3/5, we need to have: r.(r-1)/90 > 3/5 => r.(r-1) > 54 => r can be 9 or 8.

(1) Probability = C(b,2)/C(10,2) = b.(b-1)/(10.9) < 1/10. Therefore, b.(b-1) < 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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11 Apr 2013, 09:56
1
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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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11 Apr 2013, 10:06
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QUESTION 10
(1) Sum (x1+x2+x3+..xn)/ n >= xk. For any Xk, we have, x1+x2+...+xn >= n.xk. Therefore, x1=x2=...=xn. In this case, the median is equal to the mean, which is
equal to any number in the set. But we do not know which number that is. INSUFFICIENT

(2) Take {1,2,3} with range and average both equal to 2. In this case, median =2. Now, take {3,6,9}, with average and range equal to 6. In this case, median = 6.
INSUFFICIENT

(1) + (2) All numbers have to be equal, but the range is zero. Therefore, all items are equal to zero. In this case, the median
is ZERO. SUFFICIENT

ANS: C
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11 Apr 2013, 13:52
Q5) Is the standard deviation of set A {3-2x, 3-x, 3, 3+x, 3+2x} > the standard deviation of set B={3-2x, 3-x, 3, 3+x, 3+2x, y}

From Stmt 1, y is not known and only SD > 0
- Not Sufficient
If 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 - Sufficient
When y = 3 and for any integer value for x SD is always >= 0 and SD of set A > SD of set B

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