AsterixCyprus

IMO A. A uses appositive appropriately.

The OA is E.

There is nothing like a minimum score. The Top schools look for your GMAT score and not your AWA score. But you would be better off if you can pull out a 5 atleast. Just master the basic fallacies in AWA and you should easily get a 5.5. HTH! PS: i got a 6 in my AWA on G-Day.. :)

IMO A. Agree with LegoLife. No issues in A.

IMO D. D is clear and concise. "Stimulus to a broader appreciation" doesn't sound right. Hence A, B and E are out.

IMO A. I agree with LegoLife's explanation. Here "without any original thought" is modifying the "Leader of the Opposition" and NOT the "Prime Minister".

Hey ravindra mate!..Awesome score man!!!...You really nailed it..:) I was waiting for your debrief..:)..Congrats man!

Gave CAT1 of MGMAT: Score 700(Q48,V38) Quant was tough and so was the verbal. But MGMAT is definitely a great source for practice. The assessment reports are excellent!

Congrats TBAY...My story is similar to yours..and I just hope that this time I am able to join the elite 700 Club!...You have been an awesome contributor to this forum..:)

Both the above mentioned idioms are correct. In this question the idiom used is :: require X to Y

monu: Whats the OA?

Are you sure the OA is E? IMO the answer should be C. Stmt1: xy > 0 From this, we get that both x and y are either +ve or -ve so x 0 --> --INSUFF-- Stmt2: x + y > 0 No info on y. --INSUFF-- But combining both stmts, we get that x >0 since x + y > 0 will hold true only if x > 0 (Because if x The inequality(x + y > 0) will not hold true) Hence IMO C.

It didn't take me more than 2 mins to solve it. Its a question on plain concepts only. But I would be curious to know to see if there is an easy soln to this problem.

Imo D. 7c2 . 3c1 --------- 10c3 = 21/40

aham/TBAY: can you explain why would you chose 3? Consider an example: Set s = {1, -3} Now for this set, AM = -1, SD = 4 but if you consider, S' = {|1|, |-3|} AM = 2, SD = 1 So the Std Dev of both the sets are different. Let me know if I am doing something wrong. Given the options, I believe only 1.) holds true.

whats the OA spoud?

IMO A. All nos that are not divisible by 3 are divisible by only one prime number. Stmt2 runs into problem for '21' since it is divisible by 2 different prime numbers(3,7) while the other numbers are divisible by only 1 prime no. Whats the OA?

IMO E. 1) squaring both sides of stmt1 (x^2+y^2)^2 > z^4 => x^4 + y^4 +2x^2*y^2>z^4 --INSUFF-- since we need 2x^2*y^2 for inequality to hold e.g : 1+3+4>5 so if we take out 1 number from the left side inequality may or may not hold. consider x=sqrt(2) y=sqrt(3) z =sqrt(4) So from stmt1, you get 2+ 3 > 4 When you square both sides, you get 4 + 9 +12 > 16 Now if we remove 12, the inequality doesn't hold. But if we remove 4, the inequality still holds. Hence, we need 2x^2*y^2 for inequality to hold. 2) --INSUFF--. Doesn't help in any way. HTH!

IMO D. here's my approach: w - trainees who passed the written test f - trainees who passed the flight test Given: n(w) = 70 n(f) = 80 n(w n f) = ?? stmt1: 70 + 80 - n(w n f) + 10 = 100 Hence suff Stmt2: Only n(f) = 20 = n(f) - n(w n f) => 20 = 80 - n(w n f) Hence suff.

Agree with givemeanid. Answer should be 2.

IMO D. Given: 4s + t ---> Odd 3s + 5t ---> even If we add the above two, we know that the result will be odd, since odd+ even is odd. By adding, We get 7s + 6t --> odd for this to be odd, 7s should be odd since 6t is even. Hence s is odd. since, 4s + t is odd and s is odd,therefore t is also odd. So 's' and 't' are both odd. For any of the given choices to be a multiple of 8, we are looking for an eqn which gives us three even numbers. -Choice b & c can be eliminated since they generate odd nos. -Choice 'a' generates only 2even nos. Considering the min possible value of an even no as 2, this eqn will not 'always' generate a number that will be a multiple of 8. -Choice 'd' can be re-written as:: (t-1)(t+1)(s+1) ==> All three are even numbers. So if we consider the min possible even no 2, we get that 8 will 'always' be a factor of this eqn. -Choice 'e' generates only 2 even nos and hence doesn't solve the purpose. Whats the OA Kevin?

IMO B. Its a question on conditional probability. p(p) = Probability of accurate paper records p(e) = Prob of accurate electronic records Given: P(e'|p') = 0.6 = p(e' n p')/p(p') = .03/p(p') => p(p') = 0.05 also, p(p) = 1 - p(p') = 0.95 p(p'|e') = 0.75 = p(p' n e')/p(e') = .03/p(e') => p(e') = 0.04 Also, p(e) = 1 - 0.04 = .96 p(p' n e') = 0.03 = p((p u e)') --> Using De Morgan's law => 1 - p(p u e) = 0.03 => p(p u e) = 0.97 p(p u e) = p(p) + p(e) - p(p n e) => 0.97 = .95 + .96 - p(p n e) => p(p n e) = .95 + .96 - .97 = .94 Hence B. HTH!

IMO A. Its a mixed series, which is a combination of 3 AP series, with a1 of each series being 64, 66 and 67. The difference for each of the series is d =8 nth term of the AP series is:: a(n) = a(1) + (n-1)d => an = a1 + (n-1)8 => an -a1 = (n-1)8 All we need to find is which of the given numbers will belong to anyone of the 3 series. In order to do this, subtract 'a1'(which can be 66,67 or 68) from the given number and see if its divisible by 8. Only 723 fits the bill (it belongs to the AP series with a1=67). HTH!

TBAY, even if we consider K as -ve, we get the total no of values greater than 10 as 7 (a1, a2,....,a7). The question stem is precisely asking the same thing: "How many of the terms in the sequence are greater than 10 ?" so basically for stmt2, it doesn't matter if k is -ve or +ve as long as we know whats the value of the middle term (a8). HTH!