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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.)))
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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!!
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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

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.

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:(
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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.

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..

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.

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|

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|

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.
_________________

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|

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.
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