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:
IF X^2 + 5Y = 49 is y an integer? 1. 1 < x < 4 2. X^2 is an integer
E.
from i, x = 1.00001 to 3.999999999999. so not suff.
from ii, x = any integer.
from i and ii, x= 2,3,4,5,6,7,8,9,10,11,12,13,14, and 15. y could be integer or fraction.. still not enough...
From prior posts, it is obvious that Stmt 1 & 2 is insufficient. Combining, we could take X^2 is 7, hence x would be sqrt(7), which is between 1 & 4. However this would make 'y' a fraction (42/5).
From prior posts, it is obvious that Stmt 1 & 2 is insufficient. Combining, we could take X^2 is 7, hence x would be sqrt(7), which is between 1 & 4. However this would make 'y' a fraction (42/5).
Please explain this to me.
If 1 < x < 4 then for what values of X will we have x^2 as an integer? I think I am not understanding this point here.
From prior posts, it is obvious that Stmt 1 & 2 is insufficient. Combining, we could take X^2 is 7, hence x would be sqrt(7), which is between 1 & 4. However this would make 'y' a fraction (42/5).
Please explain this to me.
If 1 < x < 4 then for what values of X will we have x^2 as an integer? I think I am not understanding this point here.
There are 2 constraints here.
1) 1 < x < 4 and
2) x^2 is an integer
Lets take x = 1, then x^2 is also 1. If x = 4, then x^2 is 16. However the question did not mention that x is an integer. It mentions only x^2 is an integer, which means x^2 = 2,3,4 ... 15. In this case, x = sqrt2, sqrt3, sqrt4 = 2, and so on. We cannot take negative values here because of (1).
I got this one off a kaplan test- I disagree with the answer. here goes:
IF X^2 + 5Y = 49 is y an integer?
1. 1 < x < 4 2. X^2 is an integer
I think some of you guys are making this too complicated. Let's look at this in two parts by picking numbers:
1. If x=2, then 5Y = 45, and Y=9 (integer)
If x=3, then 5Y = 43, and y=43/9 (not an integer)
So statement 1 is insufficient
2. If x^2 is an integer, then you can once again use the numbers I picked for statement 1, and see that this is insufficient.
The numbers I picked for statement 1 can be used when looking at both statements at once, showing that both statements together are still insufficient, thus the answer is E.
Every time when you are told that x^2 is an integer, you need to aware that there may be a trap and need to check if x itself is an integer. _________________
Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.
actually bumbleeman, if X = 3 than x^2 = 9 and 5y = 49-9 = 40 and y would be 8
you have to think about radicals b/n 1 and 4 to find the correct answer
D'oh! This is the sort of stupid mistake that costs me points on the practice tests. In this case I will blame the way that x^2 appears on the screen here, as opposed to how it does on the exam.
Of course, despite my best efforts, I still got the correct answer of E. As HongHu pointed out, the fact that x^2 is an integer does not mean that x is an integer.