Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

The next set of medium/hard DS number properties questions. I'll post OA's with detailed explanations on Friday. Please, post your solutions along with the answers.

1. If x is an integer, what is the value of x?

(1) |23x| is a prime number (2) \(2\sqrt{x^2}\) is a prime number.

Hello, for question 9, could you pls explain how the maximum value of [a]+[b] =-2? I dont get it...I assumed a=-1 so [a] = -1, and b=-2, so [b]=-2...that means ab=2...but [a]+[b]= -1-2=-3...how do I get a -2? (as a maximum value if both are negative). thank you

6. Set S consists of more than two integers. Are all the integers in set S negative?

(1) The product of any three integers in the set is negative. If the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}. If the set consists of more than 3 terms, then the set can only have negative numbers. Not sufficient.

(2) The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. Thus the set consists of only negative or only positive integers. Not sufficient.

(1)+(2) Since the second statement rules out {negative, positive, positive} case which we had from (1), then we have that the set must have only negative integers. Sufficient.

Answer: C.

a doubt t on explanation for statement 1 if a set S={A(+ve) , B(-ve), C(+ve) , D (+ve)} and we take product ACD->it will be positive ..so how can you say that if set contains more than 3 element ,any product would be negative only

statement 2: agree that the smallest and largest no must be of same sign ..but based on this info how can we comment on the sign of the remaining elements.. pls help me understand what am i missing

(1) \(|23x|\) is a prime number. From this statement it follows that x=1 or x=-1. Not sufficient.

(2) \(2\sqrt{x^2}\) is a prime number. The same here: x=1 or x=-1. Not sufficient.

(1)+(2) x could be 1 or -1. Not sufficient.

Answer: E.

a doubt on the statement 2 \sqrt{x*x}=-1 0r 1 but since its given that 2*(sqrtX^2)=prime; x can only take positive value as -2 would not be a prime no..

thus according to me statement two should be sufficient

(1) \(|23x|\) is a prime number. From this statement it follows that x=1 or x=-1. Not sufficient.

(2) \(2\sqrt{x^2}\) is a prime number. The same here: x=1 or x=-1. Not sufficient.

(1)+(2) x could be 1 or -1. Not sufficient.

Answer: E.

a doubt on the statement 2 \sqrt{x*x}=-1 0r 1 but since its given that 2*(sqrtX^2)=prime; x can only take positive value as -2 would not be a prime no..

thus according to me statement two should be sufficient

\(2\sqrt{x^2}\) won't give negative value for any x, because \(\sqrt{non-negative \ number}\geq{0}\): the square root function cannot give negative result.

If x=1 or x=-1, then \(2\sqrt{x^2}=2\sqrt{1}=2=prime\).

6. Set S consists of more than two integers. Are all the integers in set S negative?

(1) The product of any three integers in the set is negative. If the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}. If the set consists of more than 3 terms, then the set can only have negative numbers. Not sufficient.

(2) The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. Thus the set consists of only negative or only positive integers. Not sufficient.

(1)+(2) Since the second statement rules out {negative, positive, positive} case which we had from (1), then we have that the set must have only negative integers. Sufficient.

Answer: C.

a doubt t on explanation for statement 1 if a set S={A(+ve) , B(-ve), C(+ve) , D (+ve)} and we take product ACD->it will be positive ..so how can you say that if set contains more than 3 element ,any product would be negative only

statement 2: agree that the smallest and largest no must be of same sign ..but based on this info how can we comment on the sign of the remaining elements.. pls help me understand what am i missing

You misinterpreted the statement. It says: the product of ANY three integers in the set is negative. So, the set cannot be {positive, positive, positive, negative} because the product of the first 3 terms there is NOT negative. Basically if the set has 4 or more terms, and not all of them are negative, then this statement will be violated, because we'll be able to pick two negative terms and one non-negative, which will not give negative product.

6. Set S consists of more than two integers. Are all the integers in set S negative?

(1) The product of any three integers in the set is negative. If the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}. If the set consists of more than 3 terms, then the set can only have negative numbers. Not sufficient.

(2) The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. Thus the set consists of only negative or only positive integers. Not sufficient.

(1)+(2) Since the second statement rules out {negative, positive, positive} case which we had from (1), then we have that the set must have only negative integers. Sufficient.

Answer: C.

a doubt t on explanation for statement 1 if a set S={A(+ve) , B(-ve), C(+ve) , D (+ve)} and we take product ACD->it will be positive ..so how can you say that if set contains more than 3 element ,any product would be negative only

statement 2: agree that the smallest and largest no must be of same sign ..but based on this info how can we comment on the sign of the remaining elements.. pls help me understand what am i missing

Hello somyakalra,

For st1: it is given that product of any 3 integers is negative. The word "any" is of prime importance here

The case that you have taken for ACD will not be part of this set since it does not meet the condition in st 1...for a set of 4 integers which should satisfy st1 all 4 nos have to be negative only.You cannot question the given statements in DS questions.

For st 2: Note prime nos are always positive

Consider a set of negative nos -1 to -5

Largest no: -1 , smallest: -5 Product of these no: 5 which is prime. All the members in this set will have to be in this range only of -1 to -5 only

Consider another case meeting condition given in st2

Another group with smallest integer :1 and largest integer:5 Their product is 5.

Again we get 2 cases ( all members negative and all members positive) for st2 and hence it not sufficient

Combining you get Ans C as you are left with only 1 criteria which satisfies st1 and st2 and that all nos in the set are negative

Hope it helps

Posted from my mobile device

Posted from my mobile device

Posted from my mobile device _________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

6. Set S consists of more than two integers. Are all the integers in set S negative?

(1) The product of any three integers in the set is negative. If the set consists of only 3 terms, then the set could be either {negative, negative, negative} or {negative, positive, positive}. If the set consists of more than 3 terms, then the set can only have negative numbers. Not sufficient.

(2) The product of the smallest and largest integers in the set is a prime number. Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign. Thus the set consists of only negative or only positive integers. Not sufficient.

(1)+(2) Since the second statement rules out {negative, positive, positive} case which we had from (1), then we have that the set must have only negative integers. Sufficient.

Answer: C.

a doubt t on explanation for statement 1 if a set S={A(+ve) , B(-ve), C(+ve) , D (+ve)} and we take product ACD->it will be positive ..so how can you say that if set contains more than 3 element ,any product would be negative only

statement 2: agree that the smallest and largest no must be of same sign ..but based on this info how can we comment on the sign of the remaining elements.. pls help me understand what am i missing

As for the second statement.

(2) says: The product of the smallest and largest integers in the set is a prime number.

Since only positive numbers can be primes, then the smallest and largest integers in the set must be of the same sign.

So either: smallest * greatest = negative * negative and in this case as both the smallest and the greatest are negative then ALL integers in the list are negative OR smallest * greatest = positive * positive and in this case as both the smallest and the greatest are positive then ALL integers in the list are positive.

9. If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + [b] = 1 ?

Given that some function [] rounds DOWN a number to the nearest integer. For example [1.5]=1, [2]=2, [-1.5]=-2, ...

(1) ab = 2. First of all this means that a and b are of the same sign.

If both are negative, then the maximum value of [a] + [b] is -2, for any negative a and b. So, this case is out.

If both are positive, then in order [a] + [b] = 1 to hold true, must be true that [a]=0 and [b]=1 (or vise-versa). Which means that \(0\leq{a}<1\) and \(1\leq{b}<2\) (or vise-versa). But in this case ab cannot be equal to 2. So, this case is also out.

We have that the answer to the question is NO. Sufficient.

(2) 0 < a < b < 2. If a=1/2 and b=1, then [a] + [b] = 0 + 1 = 1 but if a=1/4 and b=1/2, then [a] + [b] = 0 + 0 = 0. Not sufficient.

Answer: A.

Bunuel how we do get the above highlighted part ? How about when a and b are equal to -4 and -0.5

9. If [x] denotes the greatest integer less than or equal to x for any number x, is [a] + [b] = 1 ?

Given that some function [] rounds DOWN a number to the nearest integer. For example [1.5]=1, [2]=2, [-1.5]=-2, ...

(1) ab = 2. First of all this means that a and b are of the same sign.

If both are negative, then the maximum value of [a] + [b] is -2, for any negative a and b. So, this case is out.

If both are positive, then in order [a] + [b] = 1 to hold true, must be true that [a]=0 and [b]=1 (or vise-versa). Which means that \(0\leq{a}<1\) and \(1\leq{b}<2\) (or vise-versa). But in this case ab cannot be equal to 2. So, this case is also out.

We have that the answer to the question is NO. Sufficient.

(2) 0 < a < b < 2. If a=1/2 and b=1, then [a] + [b] = 0 + 1 = 1 but if a=1/4 and b=1/2, then [a] + [b] = 0 + 0 = 0. Not sufficient.

Answer: A.

Bunuel how we do get the above highlighted part ? How about when a and b are equal to -4 and -0.5

Can you please suggest in question number 9 why can't we have values a = 0.1 and b = 20 so a*b will be 2 and equation won't be satisfied

The point is that [a] + [b] = 1 won't be true for any a and b if ab = 2, so in all cases we have a NO answer to the question (not only for your case). _________________

6. Set S consists of more than two integers. Are all the numbers in set S negative?

(1) The product of any three integers in the list is negative (2) The product of the smallest and largest integers in the list is a prime number.

I took your posibilities here for this problem as follows. S(smallest), L(largest) -S,-,-L +S,-,+L

Now here, (1) & (2) satisfies as prod of 3 will be -ve and prod of S&L will be +ve.. So we have 2 solutions even after combining the 2 equations. So E should be the answer. Please clarify..

6. Set S consists of more than two integers. Are all the numbers in set S negative?

(1) The product of any three integers in the list is negative (2) The product of the smallest and largest integers in the list is a prime number.

I took your posibilities here for this problem as follows. S(smallest), L(largest) -S,-,-L +S,-,+L

Now here, (1) & (2) satisfies as prod of 3 will be -ve and prod of S&L will be +ve.. So we have 2 solutions even after combining the 2 equations. So E should be the answer. Please clarify..

Thanks & Regards, Lokesh.

How can positive number in your second set be the smallest one in the set if you have a negative number there? _________________

6. Set S consists of more than two integers. Are all the numbers in set S negative?

(1) The product of any three integers in the list is negative (2) The product of the smallest and largest integers in the list is a prime number.

I took your posibilities here for this problem as follows. S(smallest), L(largest) -S,-,-L +S,-,+L

Now here, (1) & (2) satisfies as prod of 3 will be -ve and prod of S&L will be +ve.. So we have 2 solutions even after combining the 2 equations. So E should be the answer. Please clarify..

Thanks & Regards, Lokesh.

How can positive number in your second set be the smallest one in the set if you have a negative number there?

(1) \(|23x|\) is a prime number. From this statement it follows that x=1 or x=-1. Not sufficient.

(2) \(2\sqrt{x^2}\) is a prime number. The same here: x=1 or x=-1. Not sufficient.

(1)+(2) x could be 1 or -1. Not sufficient.

Answer: E.

Hi Bunuel sorry to ask a silly doubt,but prime numbers are always positive so can't we discard -1 ? Thanks Anupama

(2) \(2\sqrt{x^2}\) is a prime number, not x. For \(2\sqrt{x^2}\) to a prime number, x must be 1 or -1. So, we have two possible values of x, which makes the statement insufficient. _________________

8. Set A consist of 10 terms, each of which is a reciprocal of a prime number, is the median of the set less than 1/5?

(1) Reciprocal of the median is a prime number. If all the terms equal 1/2, then the median=1/2 and the answer is NO but if all the terms equal 1/7, then the median=1/7 and the answer is YES. Not sufficient.

(2) The product of any two terms of the set is a terminating decimal. This statement implies that the set must consists of 1/2 or/and 1/5. Thus the median could be 1/2, 1/5 or (1/5+1/2)/2=7/20. None of the possible values is less than 1/5. Sufficient.

Answer: B.

Theory: Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are non-negative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^2\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\).

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example \(\frac{x}{2^n5^m}\), (where x, n and m are integers) will always be the terminating decimal.

We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \(\frac{6}{15}\) has 3 as prime in denominator and we need to know if it can be reduced.

I believe you meant to write "1/5 + 1/2 = 7/10", not 7/20 and that "250 = 2 * 5^3" not 2 * 5^2? just want to be clear. thank you.

8. Set A consist of 10 terms, each of which is a reciprocal of a prime number, is the median of the set less than 1/5?

(1) Reciprocal of the median is a prime number. If all the terms equal 1/2, then the median=1/2 and the answer is NO but if all the terms equal 1/7, then the median=1/7 and the answer is YES. Not sufficient.

(2) The product of any two terms of the set is a terminating decimal. This statement implies that the set must consists of 1/2 or/and 1/5. Thus the median could be 1/2, 1/5 or (1/5+1/2)/2=7/20. None of the possible values is less than 1/5. Sufficient.

Answer: B.

Theory: Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are non-negative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^2\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\).

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example \(\frac{x}{2^n5^m}\), (where x, n and m are integers) will always be the terminating decimal.

We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \(\frac{6}{15}\) has 3 as prime in denominator and we need to know if it can be reduced.

I believe you meant to write "1/5 + 1/2 = 7/10", not 7/20 and that "250 = 2 * 5^3" not 2 * 5^2? just want to be clear. thank you.

It's (1/5 + 1/2)/2 = 7/20, so no typo there. You are right about 250 = 2 * 5^3, though. Edited. Thank you. _________________

Excellent posts dLo saw your blog too..!! Man .. you have got some writing skills. And Just to make an argument = You had such an amazing resume ; i am glad...

So Much $$$ Business school costs a lot. This is obvious, whether you are a full-ride scholarship student or are paying fully out-of-pocket. Aside from the (constantly rising)...

I barely remember taking decent rest in the last 60 hours. It’s been relentless with submissions, birthday celebration, exams, vacating the flat, meeting people before leaving and of...

Rishabh from Gyan one services, India had a one to one interview with me where I shared my experience at IMD till now. http://www.gyanone.com/blog/life-at-imd-interview-with-imd-mba/ ...