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 350,000 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:

Re: Good set of DS 3 [#permalink]
17 Oct 2009, 19:43

Expert's post

GMAT TIGER wrote:

Bunuel wrote:

Yes and about the ZIP trap:

GMAT likes to act in the zone -1<=x<=1. So I always ask myself:

Did I assumed, with no ground for it, that variable can not be Zero? Check 0! Did I assumed, with no ground for it, that variable is an Integer? Check fractions! Did I assumed, with no ground for it, that variable is Positive? Check negative values!

I called it ZIP trap. Helps me a lot especially with number property problems.

Thats cool.

You can say PINZF (or better) trap as well:

P = positive I = integer N = negative Z = zero F = fraction

Sure you can call whatever suits you, no copyright on that term, for me ZIP sounds good.))) _________________

Re: Good set of DS 3 [#permalink]
25 Oct 2009, 08:38

GMAT TIGER wrote:

Bunuel wrote:

1. The sum of n consecutive positive integers is 45. What is the value of n? (1) n is even (2) n < 9

1. n could be 2 or 6 or 10 a + a+1 = 45 a = 22 n = 2

a + a+1 + a+2 + a+3 + a+4 + a+5 = 45 a = 5 n = 6

2. n could be 2, 3, 5 or 6

1&2: n could be 2 or 6. E.

Bunuel wrote:

10. The sum of n consecutive positive integers is 45. What is the value of n? (1) n is even (2) n >= 9

1. n could be 2 or 6 or 10

n = 2: a + a+1 = 45 a = 22

n = 6: a + a+1 + a+2 + a+3 + a+4 + a+5 = 45 a = 5

2. n could be 9 or 10 or 14 or 15 or 18 & so on...

1&2: n could be 10 or 14 or 18. E.

How can we find out 10, 14, 15 & 18??? I mean total is 45, how did tiger wrote 14, 15, 18 mark straight away?? I think its not possible!!! Pls help!! _________________

Re: Good set of DS 3 [#permalink]
25 Oct 2009, 19:21

Bunuel wrote:

ANSWERS:

10. The sum of n consecutive positive integers is 45. What is the value of n? (1) n is odd (2) n >= 9

Look at the Q 1 we changed even to odd and n<9 to n>=9

(1) not sufficient see Q1. (2) As we have consecutive positive integers max for n is 9: 1+2+3+...+9=45. (If n>9=10 first term must be zero. and we are given that all terms are positive) So only case n=9. Sufficient.

Answer: B.

sum of n integers = (n*(n+1))/2

(n*(n+1))/2 = 45 which yields n = 9 could be the only answer

Stmt 1 and 2 (D) are both sufficient.

Does GMAT assume that we would not indulge in any formula? Just curious

Re: Good set of DS 3 [#permalink]
25 Oct 2009, 19:53

Expert's post

goldgoldandgold wrote:

Bunuel wrote:

ANSWERS:

10. The sum of n consecutive positive integers is 45. What is the value of n? (1) n is odd (2) n >= 9

Look at the Q 1 we changed even to odd and n<9 to n>=9

(1) not sufficient see Q1. (2) As we have consecutive positive integers max for n is 9: 1+2+3+...+9=45. (If n>9=10 first term must be zero. and we are given that all terms are positive) So only case n=9. Sufficient.

Answer: B.

sum of n integers = (n*(n+1))/2

(n*(n+1))/2 = 45 which yields n = 9 could be the only answer

Stmt 1 and 2 (D) are both sufficient.

Does GMAT assume that we would not indulge in any formula? Just curious

This is not correct.

First of all the formula you referring n(1+n)/2 is the formula for counting the sum of n FIRST integers (meaning that starting from 1). Question stem does not give us that information: "The sum of n consecutive positive integers is 45." You can not assume anything.

The formula for counting the sum of consecutive integers (not necessarily first n integers), which is in fact sum of AP (arithmetic progression) is: Sn=n*(a1+an)/2 or Sn=n*(2a1+d(n-1))/2, you can substitute d=1, as the numbers are consecutive.

If you further substitute a1 (the first term of the progression) with 1 you'll get exactly the formula you wrote: Sn=n(2+n-1)/2=n(n+1)/2

Second, you can find the n consecutive odd positive integers (n=odd) to total 45: n=3 --> 14, 15, 16 n=5 --> 7, 8, 9, 10, 11.

Re: Good set of DS 3 [#permalink]
26 Oct 2009, 00:12

Quote:

3. Multiplication of the two digit numbers wx and cx, where w,x and c are unique non-zero digits, the product is a three digit number. What is w+c-x? (1) The three digits of the product are all the same and different from w c and x. (2) x and w+c are odd numbers.

(1) wx+cx=aaa (111, 222, ... 999=37*k) --> As x is the units digit in both numbers, a can be 1,4,6 or 9 (2,3,7 out because x^2 can not end with 2,3, or 7. >>What about 8?5 is out because in that case x also should be 5 and we know that x and a are distinct numbers). 1 is also out because 111=37*3 and we need 2 two digit numbers. 444=37*12 no good we need units digit to be the same. 666=37*18 no good we need units digit to be the same. 999=37*27 is the only possibility all digits are distinct except the unit digits of multiples. Sufficient (2) x and w+c are odd numbers. Number of choices: 13 and 23 or 19 and 29 and w+c-x is the different even number.

Answer: A.

Could somebody please elaborate the solution..i don't understand:( _________________

Re: Good set of DS 3 [#permalink]
27 Oct 2009, 21:19

tejal777 wrote:

Quote:

3. Multiplication of the two digit numbers wx and cx, where w,x and c are unique non-zero digits, the product is a three digit number. What is w+c-x?

(1) The three digits of the product are all the same and different from w c and x. (2) x and w+c are odd numbers.

(1) wx+cx=aaa (111, 222, ... 999=37*k) --> As x is the units digit in both numbers, a can be 1,4,6 or 9 (2,3,7 out because x^2 can not end with 2,3, or 7. >>What about 8?5 is out because in that case x also should be 5 and we know that x and a are distinct numbers). 1 is also out because 111=37*3 and we need 2 two digit numbers. 444=37*12 no good we need units digit to be the same. 666=37*18 no good we need units digit to be the same. 999=37*27 is the only possibility all digits are distinct except the unit digits of multiples. Sufficient

(2) x and w+c are odd numbers. Number of choices: 13 and 23 or 19 and 29 and w+c-x is the different even number.

Answer: A.

Could somebody please elaborate the solution..i don't understand:(

8 is also not possible. ........................................................................

1. wx * cx = aaa (aaa can only be 111 (for 9), 444 (for 2), 666 (for 8), and 999 (for 3 and 7)). 222, 333, 555, 777, and 888 are not possible as noneo f the unit difgits of the 3 digit numbers is square of an integer.

wx * cx = 111 = 3x37. Not possible as there is only one 2 digit number. Needs 2 two digit numbers. wx * cx = 444 = 2x3x37 = 6x37. Not possible as there is only one 2 digit number. Needs 2 two digit numbers. wx * cx = 666 = 2x3x3x37 = 18x37. Not possible as the unit digit of these 2 digit numbers are not the same. wx * cx = 999 = 3x3x3x37 = 27x37. Possible as the unit digit of these 2 digit numbers are same.

Re: Good set of DS 3 [#permalink]
07 Nov 2009, 23:46

GMAT TIGER wrote:

tejal777 wrote:

Quote:

3. Multiplication of the two digit numbers wx and cx, where w,x and c are unique non-zero digits, the product is a three digit number. What is w+c-x?

(1) The three digits of the product are all the same and different from w c and x. (2) x and w+c are odd numbers.

(1) wx+cx=aaa (111, 222, ... 999=37*k) --> As x is the units digit in both numbers, a can be 1,4,6 or 9 (2,3,7 out because x^2 can not end with 2,3, or 7. >>What about 8?5 is out because in that case x also should be 5 and we know that x and a are distinct numbers). 1 is also out because 111=37*3 and we need 2 two digit numbers. 444=37*12 no good we need units digit to be the same. 666=37*18 no good we need units digit to be the same. 999=37*27 is the only possibility all digits are distinct except the unit digits of multiples. Sufficient

(2) x and w+c are odd numbers. Number of choices: 13 and 23 or 19 and 29 and w+c-x is the different even number.

Answer: A.

Could somebody please elaborate the solution..i don't understand:(

8 is also not possible. ........................................................................

1. wx * cx = aaa (aaa can only be 111 (for 9), 444 (for 2), 666 (for 8), and 999 (for 3 and 7)). 222, 333, 555, 777, and 888 are not possible as noneo f the unit difgits of the 3 digit numbers is square of an integer.

wx * cx = 111 = 3x37. Not possible as there is only one 2 digit number. Needs 2 two digit numbers. wx * cx = 444 = 2x3x37 = 6x37. Not possible as there is only one 2 digit number. Needs 2 two digit numbers. wx * cx = 666 = 2x3x3x37 = 18x37. Not possible as the unit digit of these 2 digit numbers are not the same. wx * cx = 999 = 3x3x3x37 = 27x37. Possible as the unit digit of these 2 digit numbers are same.

2. is self explanatory..

I am sorry.. but I still dont understand how did you arrive at that 37..

Re: Good set of DS 3 [#permalink]
26 Dec 2009, 08:04

2

This post received KUDOS

Expert's post

jan4dday wrote:

please explain 2nd Q.

i want an example where XYZ can be prime using STATEMENT 1 ALONE

First note prime numbers are only positive. (Also note that \(x\), \(y\) and \(z\) are integers)

Q: \(xyz=p\), is \(p\) prime?

(1) \(x=-y\) --> \(p=-x^2z\). Let's check when this expression gives a prime number:

Well first of all \(p\) to be prime \(z\) MUST be negative, as \(p\) MUST be positive to be a prime.

Next if \(x>|1|\), (eg \(|2|\), \(|3|\), ...) OR equals to zero, \(p\) won't be prime. So \(x\) must be equal to \(|1|\).

But it's not enough. We'll have \(p=-x^2z=-z\), so \(p\) to be a prime number \(z\) must be equal to \(-prime\).

You are asking how using statement (1) \(p\) could be a prime: according to above, when \(|x|=1\) and \(z=-p\). eg.: \(x=-1\) --> \(y=1\) --> z\(=-7\) --> \(p=(-1)*1*(-7)=7\), which is prime.

Statement (1) may or may not give the prime number for \(xyz\). Not sufficient.

(2) \(z=1\) --> \(p=xy\). Again for \(p\) to be a prime number \(xy\) must be \(>0\) (both positive or both negative). Then if \(x=|prime|\) and \(y=|1|\), OR \(y=|prime|\) and \(x=|1|\), so that \(xy>0\), then \(xy\) is a prime number. For any other values or combinations of \(x\) and \(y\), \(p\) won't be a prime. Not sufficient.

(1)+(2) \(p=xyz=-x^2\) (as \(x=-y\) and \(z=1\)). \(-x^2\) is never positive, hence \(p\) is not a prime. Sufficient.

Re: Good set of DS 3 [#permalink]
02 Jan 2010, 08:07

Great Questions and Great Tips Bunuel!!

+1 Kudos! for this

Cheers! JT _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

Re: Good set of DS 3 [#permalink]
02 Jan 2010, 08:20

Bunuel wrote:

5. If a and b are integers, and a not= b, is |a|b > 0? (1) |a^b| > 0 (2) |a|^b is a non-zero integer

This is tricky |a|b > 0 to hold true: a#0 and b>0.

(1) |a^b|>0 only says that a#0, because only way |a^b| not to be positive is when a=0. Not sufficient. NOTE having absolute value of variable |a|, doesn't mean it's positive. It's not negative --> |a|>=0

(2) |a|^b is a non-zero integer. What is the difference between (1) and (2)? Well this is the tricky part: (2) says that a#0 and plus to this gives us two possibilities as it states that it's integer: A. -1>a>1 (|a|>1), on this case b can be any positive integer: because if b is negative |a|^b can not be integer. OR B. |a|=1 (a=-1 or 1) and b can be any integer, positive or negative. So (2) also gives us two options for b. Not sufficient.

(1)+(2) nothing new: a#0 and two options for b depending on a. Not sufficient.

Answer: E.

Bunuel... apologize... but am I little lost with this explanation. I am not able to understand what exactly you mean by a#0 - Does this mean a = 0? _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

Re: Good set of DS 3 [#permalink]
02 Jan 2010, 08:24

Expert's post

jeeteshsingh wrote:

Bunuel wrote:

5. If a and b are integers, and a not= b, is |a|b > 0? (1) |a^b| > 0 (2) |a|^b is a non-zero integer

This is tricky |a|b > 0 to hold true: a#0 and b>0.

(1) |a^b|>0 only says that a#0, because only way |a^b| not to be positive is when a=0. Not sufficient. NOTE having absolute value of variable |a|, doesn't mean it's positive. It's not negative --> |a|>=0

(2) |a|^b is a non-zero integer. What is the difference between (1) and (2)? Well this is the tricky part: (2) says that a#0 and plus to this gives us two possibilities as it states that it's integer: A. -1>a>1 (|a|>1), on this case b can be any positive integer: because if b is negative |a|^b can not be integer. OR B. |a|=1 (a=-1 or 1) and b can be any integer, positive or negative. So (2) also gives us two options for b. Not sufficient.

(1)+(2) nothing new: a#0 and two options for b depending on a. Not sufficient.

Answer: E.

Bunuel... apologize... but am I little lost with this explanation. I am not able to understand what exactly you mean by a#0 - Does this mean a = 0?

\(a\) does not equal to \(b\). Sorry for confusion. _________________

Re: Good set of DS 3 [#permalink]
02 Jan 2010, 09:28

Hi Bunuel.....

Would appreciate if you could explain this in detail as I am confused...

Thanks! JT _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

Re: Good set of DS 3 [#permalink]
02 Jan 2010, 10:58

Expert's post

jeeteshsingh wrote:

Hi Bunuel.....

Would appreciate if you could explain this in detail as I am confused...

Thanks! JT

If a and b are integers, and a not= b, is |a|b > 0? (1) |a^b| > 0 (2) |a|^b is a non-zero integer.

\(|a|b > 0\) is true when \(b>0\) and \(a\) does not equal to zero.

(1) \(|a^b| > 0\) --> \(a\) does not equal to zero, but we don't know about \(b\), it can be any value, positive or negative. Not sufficient.

(2) \(|a|^b\) is a non-zero integer --> \(a\) can be 1 and \(b\) any integer, positive or negative. Not sufficient.

(1)+(2) If a=1 and b=2, then |a|b > 0, but if a=1 and b=-2, then |a|b <0. Not sufficient.

Answer: E.

This is the easiest way to solve this problem. In my previous solution I just tried to show what each statement means algebraically. Please tell which part needs more clarification. _________________

Interested in applying for an MBA? In the fourth and final part of our live QA series with guest expert Chioma Isiadinso, co-founder of consultancy Expartus and former admissions...