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:

Statment 2: x^2-y^2=8
<=> (x+y)(x-y) = 8 = 4*2 (=8*1 not working)
Since x & y are integers, we can say than:
o x+y = 4 (1)
and
o x-y = 2 (2)
Thus (1)+(2) <=> 2*x= 6 <=> x=3
=> y=1

Q1: Is "m" a multiple of 6?
1) More than 2 of the first 5 positive integer multiples of m are multiples of 3

2) Fewer than 2 of the first 5 positive integer multiples of m are multiples of 12

Plugging in numbers should give it to you.
1) If more than 2 of the first 5 positive integer multiples of m are multiples of 3, then the number itself must be a multiple of 3. However, this does not mean it is a multiple of 6. Insufficient.
e.g. m = 9, then first 5 integer multiples are 9, 18, 27, 36, 45.
or m = 6, then we have 6, 12, 18, 24, 30. Both satisfy the criteria, but 9 is not a multiple of 6.

2) If you take m=6, then 2 of the first 5 integer multiples are multiples of 12. Not so if you take any other number. Sufficient.

Hence B.

Q2: What is the sum of positive integers x and y?
1) x^2 + 2xy + y^2 = 16
2) x^2 - y^2 = 8

Exactly as Fig has explained.

Q3: If x is an integer, is x even?
1) x^2 - y^2 = 0
2) x^2 + y^2 = 18

---------------------------------------------------------------------------------------
Q1: Is "m" a multiple of 6?
1) More than 2 of the first 5 positive integer multiples of m are multiples of 3

2) Fewer than 2 of the first 5 positive integer multiples of m are multiples of 12

M to be a multiple of 6 it takes the form = 3*2*k where k is positive intiger

from one

first five multiples are M ,m*2 , M*3 , m*4 , m*5

worst case scenario m*5,m*3,m are multiples of 3 , and m is thus a multiple of three but we need a factor of 2 at least to make sure m is multiple of 6........insuff

from 2

clearly insuff

both together still insuff

my answer is E

Q2: What is the sum of positive integers x and y?
1) x^2 + 2xy + y^2 = 16
2) x^2 - y^2 = 8
from one

(x+y)^2 = +or -ve sqrt 16 ie (x+y) might be +ve or -ve 4 but stem say they are positive thus =4.....suff

from two

(x-y)(x+y) = 8( 8 = 8*1 or 4*2 )

it is obvious that there is no two intigers that when added together =8 and when subtracted = 1

Q3: If x is an integer, is x even?
1) x^2 - y^2 = 0
2) x^2 + y^2 = 18

from one

x^2 = y^2 we have no clue about values of y....... and zero is even

even - even = even or odd - odd = even .insuff

from two

x^2+y^2 = 18

18 is even ie even+even or odd +odd.......insuff

one and two together

2x^2 = 18

ie x^2 = 9 thus x = odd.....suff

most probably i should go to sleep
_________________
Nothing is impossible and nothing is as bad as it seems
------------------------------------------------------------------------------------

Guys im still not clear....will shoot my questions on the in the evening....

[/quote]S2: < 2 multiples of 3 ,in first 5 multiples of m.
m, 2m,3m, 4m, 5m

Again as only 3 is a multiple and less than 2 multiples then m is not a multiple of 3.

If m is not a multiple of 3 then enough to say that m is not divisible by 6 (to be divisble m to be divisible by both 3 and 2).

Sufficient.

Answer: B.
---------------------------
The second statement was
2) Fewer than 2 of the first 5 positive integer multiples of m are multiples of 12 (not of 3)

Hass can you explain the second point with respect to 12...
Are there any tricks to get to these questions.......

Thanks guys I got the other two questions.....as per your explanations. Answer for Q2 is D and for Q3 it is C.......