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A. 36n - n^2 >= 324 n(36-n)>=324.. Now 18*18 =324....If we keep n = 18...36-n = 18...satisfying !!!!!! If we increase to 19...then 19*17 = 323...going on 20*16 = 320.....Not satisfying. Also n can't be negative too....Because eqn will always be -ve*+ve so never satisfying. Only conclusion n =18. Sufficient
B. 325 > n^2 > 323 Now n can be -18 or +18 not sure...
Can anyone look at my solution and point out my mistakes. Thank you!
(1) Sufficient. 36n ≥ n^2 + 324 36n -n^2 - 324 ≥ 0 /*(-1) n^2 -36n + 324 <=0 (n-18)*(n-18)<=0 n<=18, however checking values we find that only n=18 satisfies the inequality, because anything that is < or > than 18 makes the inequality invalid. Even the value of 17.99 does not satisfy the inequality.
(2) Insufficient. 325 > n^2 > 323, by number testing we find that n can be between 18.01 and 17.99
Looks to me like the A's have it. Note that we end up with a perfect square that is <=0. A perfect square can't be less than 0, so it has to *be* 0, making x=18.
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Dmitry Farber | Manhattan Prep GMAT Instructor | San Diego
My answer is E. Solving equation 1, we get (n-18)(n-18) is <=0 therefore, n is less than or equal to 18. no solid value for n.
2) same reason, n can be any number (non integers too) which satisfy both sides of the inequality.
Even putting them together gives us nothing, therefore E
I guess you are right....E is the correct answer....Messed up the first eqn
hi stayarth and user123.. statement one is sufficient because simplifying we get (x-18)^2=<0.. only 0 satisfies the value and any other value will result in some +ive number/fraction <0 which will be wrong... statement two is wrong not only because it can be 18 or -18 but it can also be a fraction, say 18.001 etc so ans A
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Statement 1: 36n>=n^2+324 n^2-36n+324<=0 (n-18)^2<=0 A square number cannot be < 0 So, (n-18)^2=0 and n-18 = 0 n=18 Sufficient
Statement 2: n^2 can be any number (not just integer) between 323 and 325. So there are multiple values for n^2 and hence for n. Even if n^2=324, n = +18 or -18. Insufficient
quadratic equation, solve for n, when n=18, equation = 0... when n < 18, equation > 0 (i.e not within the boundary), When n >18 also equation > 0. Therefore n=18 (sufficient)
(2) n^2 is between 323 and 325. (This is a short cut and should be used carefully) since we already know that n=18, we can assume that this statement means n^2 = 324. However, we know roots of a perfect square can be either positive or negative, i.e n = +/- 18. (Insufficient)
A. You should immediately recognize the familiar symptoms of a quadratic equation in statement 1, even if it's a "quadratic inequality or equation." If you arrange that statement to look like a quadratic so that you can factor, you'll find:
0≥n^2−36n+324 Which factors to:
0≥(n−18)^2 And here's where you need to think strategically and logically - the only way for 0 to be "greater than or equal to" anything squared is for that squared term to be 0, as nothing squared can be negative. So this statement, while it may not immediately look like it, guarantees that (n−18)=0, so n=18. Statement 1 is sufficient.
Statement 2 is not sufficient, as it allows for multiple noninteger values of n, so the answer is A.
One big lesson from this problem - Data Sufficiency questions are written very precisely; had "greater than or equal to" been "less than or equal to" statement 1 would not have been close to sufficient. Precision-in-language (and symbology) matters!!
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51% (02:10) correct 
