hjafferi

IMO B ac/bd = a/d * c/b > c/b therefore a/d > 1. since a,b,c,d are postive integers therefore the expression must be greater than zero. in statment 2 a > d therefore a/d must be greater than 1 Hence statement II is sufficient. Hence B

Imo E

IMO E too Set No 14 says that the OA is C. but i like E best....

IMO D I got the answer. there was a misprint in the question. read the equation as Root {(2*root 63) + 2/(8 + (3*root 7))}.. the mistake is the positive sign. the question has mistakently been written as negative. the correct question would be with the positive sign here is how i solved it ....eventually realize that 3* root 7 = root 63 now let root 63 be A then the equation would read root {2*{A+1/(8+A)} ....taking 2 common root {2*{[A*(8+A)+1]/(8+A)} root {2*{[8A+A^2 +1]/(8+A) now put A = 3*root 7 then 8A+A^2-1 = 8*3*root 7 + 9*7 +1 = 8*3*root 7 + 63 +1 = 8*3*root7 +64 take 8 common 8(8+ 3* root 7) put 3*root 7 = A then = 8(8+A) put the expression back in the main equation we get root {2*{8*[8+A]/[8+A]}} = root 8*2 = root 16 = 4 hence D is the answer

Hi all just saw this thread. would appreciate it if we could also add Sets 1 to 20 in the same thread also. I have recently started doing sets and have reached set 7. Also i only do Quants section in the sets because in the verbal section there are too many mistakes and the questions are also wrong most of the time. would appreciate it if those who are going through verbal part of the sets could elaborate if they are worth doing. thanks.

Stuck on this one. would appreciate it if someone can assist Qs. Root {(2*root 63)-2/(8 + (3*root 7))} A. 8 + (3* root 7) B. 4 + (3* root 7) C. 8 D. 4 E. root 7 Thanks

IMO B Modelled after is the correct idiom

IMO E explanation given above.

GMAT set I with explanations.....

IMO C Statemetn I root x > y. now two possibilities could exist that x is less than y but both are fractions therefore its root is more than y for example let x = 1/4 and y = 1/3. now root x = 1/2 which is more than y. another possibility is that x is larger than y and is so large that even its root is greater than y for example let x = 16 and y = 2 therefore root 16 = 4 therefore it is greater than y. Statement II clarifies that point. It says that x is greater than y. This rules out the first possibility becasuse the premise of the possbility is that x is less than y and both are fractions. Hence that leaves us with possibility that x > y therefore x^3 must be greater than y. BTW it has already been pointed out that both x and y are positive integers.

This is a Thread Started to discuss the Sets. I am sure that this would help us all so here is the first installment. :tup: BTW answers are given at the end of the set but would appreciate it if someone can solve it along with explanations. Will try it myself and will try to post the solutions.

IMO C take 2^5 common from 2^5+2^5 we get 2^5(1+1) = 2^5*2 = 2^6 Do the same for 3^5 we get 3^6 hence answer is 2^6 + 3^6

IMO D. (A) indicate the possibility that everyone alive today might be descended from a single female ancestor who (wrong idiom. decedent of is the correct idiom) (B) indicate that everyone alive today might possibly be a descendant of a single female ancestor who had (might possibily is redundant both convey same meaning) © may indicate that everyone alive today has descended from a single female ancestor who had (D) indicates that everyone alive today may be a descendant of a single female ancestor who (E) indicates that everyone alive today might be a descendant from a single female ancestor who Hence D

IMO E North indians developed.....calanders based on..... is the correct usage of tense and idiom

Imo E

IMO E Pay off ....with is the correct idiom and "to profit" is parallel.

IMO A too

I had gone with E. But agree with Aqeel's reasoning for E be wrong is correct. hence D should be the correct answer.

Sura thanks for the correction ......Agree with none of the options are correct.

Question 1. IMO 66.67% or 2/3 total number of outcomes = 10C4 = 210 number of outcomes with one bad apple = 8C3*2C1 = 112 number of outcomes with two bad apples = 8C2*2C2 = 28 total probability = (112+28)/ 210 = 2/3 or 66.67% Question 2 Bionomial Theorem total probibilty with 3 heads = 5C3*(1/2)^3*(1/2)^2 Total probability with 4 heads = 5C4*(1/2)^4*(1/2) Total probability with 5 heads = 5C5*(1/2)^5 Sum it up to get total probability = 1/2

IMO C the difference between the prime numbers = 2 so there is only one number between them..... lets say x is one prime number and y is another prime number and A is a number between x and y (A = x+1 and y-1) then x-y = 2 The average would have to be A ....Why??? becacuse of the equidistance rule. i.e "if the numbers are evenly spaced then the average of the number is middle number. for example 77, 78, 79, 80, 81 then the average is 79 because it is in the middle." now apply that same rule here x,A,y are equidistant numbers we are given A (i.e the average or the middle number). For A. 10,11,12.....both 10 and 12 are not prime numbers B. 20,21,22.........both 20 and 22 are not prime numbers C. 51,52,53.........both 51 and 52 are prime numbers so it must be the correct answer D. 25,26,27.......both 25 and 27 are not prime numbers E.103,104,105......105 is not a prime number Hence the answer is C HTH

Imo B

Imo D

Imo E

Imo B