bhagyad

Friends, I took the TOEFL on 24th March 2007. I didn't get any score after completing it. I believe it takes sometime to get the score frm ETS. Can any one tell me when do I get my official score ?? And what is this unofficial score? How do I get this ?? Someone, please come for my help. It is very urgent. One of the B-Schools is asking me for this unofficial score. And I don't know, how can I provide that. -Bhagya.

OA z D. GMAT Gurus, plz come for help.

OA z C Thanks for the discussion.

Yeah, it shud be 5.5 * 5.5 = 30.25, i.e. C But, dunno how some body arrives at D

Well ans is 47, i.e E h(100) = 2*4*6*8*..............92*94*96*98*100 100 = 2^2*5^2 98 = 2 * 7^2 96 = 2^5 * 3 94 = 2*47 so, 47 is the largest prime in h(n). SO, h(n) is divisible by all the primes below 47 including itself. But, when u add 1 to it, i.e. h(n) + 1, it becomes in-divisible by all these primes. But, it may be divisble by a prime greater than 47. Hence, p now is greater than 47, E HTH.

C it is.

looks like C. OA Plz.

Got to be A.

This is a good question. The ans shud be E: 6 & 6. standard deviation,s.d = ROOT{[(mean - item1)^2 + (mean - item2)^2 + ......mean - itemn)^2]/n} now n increases by 2. but no addition in the numerator. So, this decreases the s.d. HTH.

I wud have gone for B,D,E,C,A , had I not seen Bob's answers.... miles to go.... for sure.

Ankur, let's look at it this way. We know that 2 = ROOT2 * ROOT 2 Area of left,a = s * small perpendicular, let's say p = s*p Area of right,A = S * large perpendicular,let's say P = S*P A = 2a Hence, S * P = 2 * s * p This can also be written as, S * P = ROOT2(s) * ROOT(p) So, S = ROOT2(s). Similarly had the que. asked the relation betn the perpendiculars, the same can be said about them too. HTH.

Itz got to be E Only 10% shows that the the amount is very less. And this is possible only when the the toal no. of smokers in the industry is taken into calculation. C can't be proved.

Ankur, Intially I was little shocked with the ans that I had got. Then I re-read the que. It says the medallion consists of the circular glass + circular metal frame surrounding the glass. Radius of the medallion is r. rad. of the metal fram is s. Hence, Area of teh circular glass = pi * (r-s)^2 Ans = [pi * r^2] - [pi * (r-s)^2] Solve this, u'll get the ans.

bscout, u clarified by putting the original que. Otherwise, I was wondering, what kind of a que is this.... :-) Thanx.

[(1001 + 999) * (1001 - 999)]/[(101 + 99)(101 - 99)] solve it further, u'll get 10 as ans.

can be written as 2^34 - 2^28 = 2^28(2^6 -1)= 2^28 * 3^2 * 7 Greatest prime factor is 7.

2nd m/c's rate is double that of the 1st m/c so, when they wrk together, if 1st m/c seals x cartons, 2nd m/c seals 2x cartons. so, the wrk done by the 2nd m/c = [2x/(2x + x)]*100 = 66.67%. HTH.

IMO E yeah, G Unit z correct. the slope just tells us that the line is descending and descending sharply on the left side, as the slope is -ve. if k = 1.1, it will definitely cut the circle. but if k = 1000, it will cut the y-axis at a very high point and will go a lot above the circle. Hence, it can't be said. E

The best way to solve this kind of que. is by elimination(POE). D appears to be the only choice left when u go on checking and ticking out each of the other options. IMO D

B & E respectively.

d = 0.XYZ 10th digit of d = X, 100th is Y and 1000th is Z que is is Y > 5?? stmnt 1: 10th digit of 10d is 7 => 10d = X.YZ and Y = 7 Suff. stmnt 2 : 100th digit of d/10 is 7 => d/10 = 0.0XYZ and Y = 7. Suff. Hence, D. HTH.

Imo e

IMO C. No. of choco bars = C No. of toffee bars = T stmnt 1.: C = T - 1 or T = 1+ C Not suff. stmnt 2 : 2/3C = 3/5T Not suff. Combine them both, 2/3C = 3/5(C+1) u can solve to get a value of C. Hence, C

Though, C is not the strongest choice, but by POE, it comes out to be the best choice. OA plz? IMO C

D E C