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New DS set!!!

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New DS set!!! [#permalink] New post   10 Apr 2013, 08:10
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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


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


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


4. Machine A and B working together at their constant rates can complete a certain task in 6 days. In how many days, working alone, can machine A complete the task?

(1) The average (arithmetic mean) of the respective times A and B would each take to 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 machine B to complete the task.


5. Is the standard deviation of list \(A=\{ 3-2x, \ 3-x, \ 3, \ 3+x, \ 3+2x\}\) greater than the standard deviation of list \(B=\{ 3-2x, \ 3-x, \ 3, \ 3+x, \ 3+2x, \ y \}\)?

(1) The standard deviation of list A is positive.
(2) \(y=3\).


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 years.
(2) The average age of employees in X, Y, and Z, is 40, 20, and 50 years, respectively.


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

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


8. All marbles in a jar containing 10 marbles are either red or blue. If two marbles are drawn from the jar, at random and without replacement, is the probability that both marbles are red greater than \(\frac{3}{5}\)?

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


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

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


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.

The solutions can be found below on this page.

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Re: New DS set!!! [#permalink] New post   14 Apr 2013, 09:00
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2. If \(x\) and \(y\) are both positive integers and \(x \gt y\), what the remainder when \(x\) is divided by \(y\)?

Given that \(x\) and \(y\) are positive integers, there exist unique integers \(q\) and \(r\), called the quotient and remainder, respectively, such that \(x = yq + r\) and \(0 \leq r < y\).

(1) \(y\) is a two-digit prime number.

This statement is clearly insufficient, as we have no information about \(x\).

(2) \(x=qy+9\), for some positive integer \(q\).

It might be tempting to consider this statement sufficient and conclude that \(r = 9\), given that the equation resembles \(x = yq + r\). However, we don't know whether \(y > 9\): the remainder must be less than the divisor.

For example:

If \(x = 10\) and \(y = 1\), then \(10 = 1*1 + 9\), but the remainder upon dividing 10 by 1 is zero.

If \(x = 11\) and \(y = 2\), then \(11 = 1*2 + 9\), but the remainder upon dividing 11 by 2 is one.

Not sufficient.

(1)+(2) From (2), we know that \(x = qy + 9\), and from (1), we know that \(y\) is greater than 9 (since it's a two-digit number). Thus, we have a direct formula for the remainder, as given above. Sufficient.


Answer: C
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Re: New DS set!!! [#permalink] New post   14 Apr 2013, 09:02
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5. Is the standard deviation of list \(A=\{ 3-2x, \ 3-x, \ 3, \ 3+x, \ 3+2x\}\) greater than the standard deviation of list \(B=\{ 3-2x, \ 3-x, \ 3, \ 3+x, \ 3+2x, \ y \}\)?

First, note that the mean of list A is \(\frac{(3-2x) +(3-x)+3+(3+x)+(3+2x)}{5}=3\). Additionally, it's important to note that list B is simply list A with the addition of the element \(y\).

(1) The standard deviation of list A is positive.

We know that the standard deviation of any list is greater than or equal to zero. The standard deviation of a list is zero only when the list consists of identical elements. So, this statement implies that list A does NOT consist of identical elements or that \(x\) does not equal zero. However, this statement is not sufficient to answer the question.

(2) \(y=3\).

We know that the mean of list A is 3. If \(x \neq 0\) (e.g., if \(x=1\)), then \(A=\{1, \ 2, \ 3, \ 4, \ 5\}\) and \(B = \{1, \ 2, \ 3, \ 3, \ 4, \ 5\}\). The standard deviation of B would be smaller than the standard deviation of A, since the elements of B are less widespread than the elements of list A. However, if \(x=0\), then \(A=\{3, \ 3, \ 3, \ 3, \ 3\}\) and \(B=\{3, \ 3, \ 3, \ 3, \ 3, \ 3\}\), so both will have a standard deviation of zero. This statement alone is not sufficient.

(1)+(2) We know that the mean of list A is 3 and it does not consist of identical elements. Since \(y=3\), then list B is list A with the addition of the element 3. The addition of a new element equal to the mean in list B makes list B less widespread than list A, resulting in a decrease in the standard deviation. Consequently, \(SD(A) \gt SD(B)\). Combining both statements provides sufficient information to answer the question.


Answer: C
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Re: New DS set!!! [#permalink] New post   14 Apr 2013, 09:03
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8. All marbles in a jar containing 10 marbles are either red or blue. If two marbles are drawn from the jar, at random and without replacement, is the probability that both marbles are red greater than \(\frac{3}{5}\)?

The question asks whether \(P(R \ and \ R)=\frac{R}{10}*\frac{R-1}{9} \gt \frac{3}{5}\), where R is the number of red marbles in the jar.

Is \(R(R-1) \gt 54\)?

By plugging in values for \(R\), we find that, for the above inequality to be true, \(R\) must be greater than 7. Hence, the question asks whether \(R > 7\), so whether the number of red marbles is 8, 9, or 10..

(1) The probability that both marbles selected will be blue is less than \(\frac{1}{10}\).

This implies that \(\frac{B}{10}*\frac{B-1}{9} \lt \frac{1}{10}\), where B is the number of blue marbles in the jar. Simplifying this inequality, we get \(B(B-1) < 9\), By plugging in values for \(B\), we find that \(B\) must be less than 4. Therefore, the number of red marbles in the jar can be 7, 8, 9, or 10: \(R > 6\). Not sufficient.

(2) At least 60% of the marbles in the jar are red. This means that the number of red marbles is greater than or equal to 6: \(R \geq 6\). Not sufficient.

(1)+(2) Combining the two statements, we know that the number of red marbles in the jar is between 6 and 10: \(R \gt 6\). If \(R\) is 6 or 7, answer is NO but if \(R\) is 8, 9, or 10, the answer is YES. Not sufficient.


Answer: E
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Re: New DS set!!! [#permalink] New post   14 Apr 2013, 09:03
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7. Was the average (arithmetic mean) temperature in degrees Celsius in city A in March less than the average (arithmetic mean) temperature in degrees Celsius in city B in March?

(1) The median temperature in degrees Celsius in City A in March was lower than the median temperature in degrees Celsius in city B.

This information is clearly insufficient.

(2) The ratio of the average temperatures in degrees Celsius in City A and City B in March was 3 to 4, respectively.

Since temperatures can be negative, this statement is also insufficient. Consider \(T(A)=3\) and \(T(B)=4\) as well as \(T(A)=-3\) and \(T(B)=-4\).

(1)+(2) Combining both statements does not provide any additional useful information. Therefore, it is still insufficient.


Answer: E
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Re: New DS set!!! [#permalink] New post   14 Apr 2013, 09:01
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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. Clearly insufficient.

(2) \(AC^2 = AB^2 + BC^2\).

This statement implies that ABC is a right triangle and AC is its hypotenuse. Important property: median from right angle is half of the hypotenuse, hence BD = 12 = AC/2, from which we have that AC = 24. Sufficient.


Answer: B
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Re: New DS set!!! [#permalink] New post   14 Apr 2013, 09:01
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4. Machine A and B working together at their constant rates can complete a certain task in 6 days. In how many days, working alone, can machine A complete the task?

Let A and B be the times needed for machines A and B to complete the task working alone, respectively. Thus, we have \(\frac{1}{A}+\frac{1}{B}=\frac{1}{6}\).

(1) The average (arithmetic mean) of the respective times A and B would each take to complete the task working alone is 12.5 days.

This implies that \(A+B=2*12.5=25\). However, since we do not know which machine is faster (we cannot differentiate between A and B), even if we substitute B with \(25-A\) into \(\frac{1}{A}+\frac{1}{B}=\frac{1}{6}\) to get \(\frac{1}{A} + \frac{1}{25-A} = \frac{1}{6}\) and solve, we must get two different answers for A and B: \(A \lt B\) and \(A \gt B\). Not sufficient.

(2) It would take machine A 5 more days to complete the task alone than it would take machine B to complete the task.

This implies that \(A=B+5\). Substituting this into the equation, we get \(\frac{1}{A}+\frac{1}{A-5}=\frac{1}{6}\). We can solve for \(A\) to get \(A=2\) or \(A=15\). However, \(A=2\) cannot be true since it would make \(B\) negative, so \(A = 15\) is the only valid solution. Sufficient.


Answer: B
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Re: New DS set!!! [#permalink] New post   14 Apr 2013, 09:04
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10. What is the median of data set A?

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

Since no number is less than the average, it follows that no number is more than the average, suggesting that all elements in the set are identical: A = {x, x, x, ...}. This implies that the average equals the median. However, the actual value of \(x\) remains unknown, so this statement is insufficient.

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

This statement is also insufficient on its own. For instance, if A = {0, 0, 0, 0}, the median is 0, but if A={1, 2, 2, 3}, the median is 2.

(1)+(2) From (1), we know that the set contains identical elements. The range of any such set is 0. Therefore, from (2), we get that (the average) = (the range) = 0. And since from (1), we have established that (the average) = (the median), it follows that (the median) = 0. Sufficient.


Answer: C
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Re: New DS set!!! [#permalink] New post   14 Apr 2013, 08:59
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SOLUTION:

1. What is the product of three consecutive integers?

(1) At least one of the integers is positive.

We can have three cases:

(i) All three integers are positive. In this case, the product will obviously be positive.

(ii) Two of the integers are positive: {0, 1, 2}. In this case, the product will be zero.

(iii) Only one of the integers is positive: {-1, 0, 1}. In this case, the product will be zero.

Not sufficient.

(2)The sum of the integers is less than 6.

Clearly insufficient; consider {-1, 0, 1} and {-3, -2, -1}.

(1)+(2) The second statement implies that we cannot have case (i) from (1), since the least sum of three positive consecutive integers is 1+2+3=6. Thus, we have either case (ii) or case (iii). Therefore, the product of the integers is zero. Sufficient.


Answer: C
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Re: New DS set!!! [#permalink] New post   14 Apr 2013, 09:02
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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 question asks whether the average age, which is equal to \(\frac{(total \ age)}{(number \ of \ employees)} < 40\). This is equivalent to asking whether \(\frac{(total \ age)}{3x+4x+8x} < 40\), or in other words: is \((total \ age) < 600x\)?

(1) The total age of all the employees in these companies is 600 years.

The question simplifies to: is \(600 < 600x\)? Or equivalently, is \(1 < x\)?

We do not know that: if \(x=1\), then the answer is NO, but if \(x > 1\), then the answer is YES. Not sufficient.

(2) The average age of employees in X, Y, and Z, is 40, 20, and 50 years, respectively.

The above implies that \((total \ age)=40*3x+20*4x+50*8x=600x\), so the answer to the question is NO. Sufficient.


Answer: B
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Re: New DS set!!! [#permalink] New post   14 Apr 2013, 09:04
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9. Is \(x(x-2) > 0\)?

\(x(x-2) > 0\) is true when \(x < 0\) or \(x > 2\). Essentially, if \(x\) is not 0, 1, or 2, we have a YES answer to the question.

(1) \(x\) is a prime number.

If \(x=2\), then the answer is NO, but if \(x\) is some other prime, then the answer is YES. Not sufficient.

(2) \(x^2\) is a multiple of 9.

If \(x=0\), then the answer is NO, but if \(x=3\), then the answer is YES. Not sufficient.

(1)+(2) Since from (1), \(x\) is a prime, and from (2), \(x^2\) is a multiple of 9, then \(x\) can only be 3. Therefore, the answer to the question is YES. Sufficient.


Answer: C
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Re: New DS set!!! [#permalink] New post   10 Apr 2013, 08:20
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I just wanna say: thanks Bunuel !
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Re: New DS set!!! [#permalink] New post   Updated on: 10 Apr 2013, 09:57
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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

Originally posted by Zarrolou on 10 Apr 2013, 09:12.
Last edited by Zarrolou on 10 Apr 2013, 09:57, edited 1 time in total.
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Re: New DS set!!! [#permalink] New post   10 Apr 2013, 10:11
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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   11 Apr 2013, 10:07
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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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Re: New DS set!!! [#permalink] New post   11 Apr 2013, 10:08
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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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Re: New DS set!!! [#permalink] New post   11 Apr 2013, 10:08
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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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Re: New DS set!!! [#permalink] New post   11 Apr 2013, 10:50
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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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Re: New DS set!!! [#permalink] New post   11 Apr 2013, 11:06
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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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Re: New DS set!!! [#permalink] New post   11 Apr 2013, 14:52
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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

Answer is B
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Re: New DS set!!! [#permalink]
11 Apr 2013, 14:52
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