Swiss_boy

No need for GP. It is clear that s = r + 1/81 after substituing the value of r in the expression for s.

777 = n*k + 77 Since reminder is 77, 77 n*k = 700 (with values of k between 1 to 9 & n > 77) k = 1, 2, 4, 5, 7 satisfy the above equation So 5 values possible?? Whats the OA? I am sure there is a simpler way to solve

200 / 3 = 66 (number divisible by 3) 200/4 = 50 (divisible by 4) 200/12 = 16 (divisible by both 3 & 4) 66 + 50 - 16 = 100

First determine the values of X where the expression involving x changes from +ve to -ve. (inother words where the expression becomes ZERO) X = -2 and X = 2 are those points. Now consider the three scenarios. 1. X 2. X > 2 3. -2 This is what kundan did.

do you know the OA

Yeah. for AP the median is equal to mean.

It is D. x-y = 6 4x-4y = 24 => y = 24 (as 4x = 5y) - suff y + z = 36 => z = 12 (as 5y = 10 z) - suff

But question asks if the value always less than 4 or not.

Stmt 1 is NOT sufficient. for x = -1/2 and x = -7 we get different results. Statment 2 is sufficient. Answer is B.

Good one Answer is A. statement 1 ---------------- 99+99 = 198 (maximum value) 50+50 = 100 (minium value) With in this range of maximum to minimum the digit at 100 is always 1. Statment 2 ------------ maximum 3 digit number is 999 - maximum 2-digit numbers can be 31*32 minumum 3 digit number is 100 - minimum 2-digit nr can be 10*10 With in this rage 100 digit can vary from 1 to 9.

arithematic mean equal to median for "Arithmetic Progression". Stmt 1 -indicates Arithmetic progression - sufficient Stmt 2 - Ranges does not give us if the numbers are in arithmetic progression or not. So answer is A

Refer Math Forum - Ask Dr. Math If you factor a number into its prime power factors, then the total number of factors is found by adding one to all the exponents and multiplying those results together. Example: 108 = 2^2*3^3, so the total number of factors is (2+1)*(3+1) = 3*4 = 12.

IMO answer is B.

Agree with hasanth

I think correct answer is A. statment 1 - number of factors is (4+1) * (3+1) = 20 - sufficient Statment 2 cannot be sufficient. n can 5^2 * 7^3 (factors are 12) or 5^3 * 7^4 (factors are 20)

Even though graph passes through 3rd quadrant, no values satisfy both the equations. Both the equations are satisified ONLY in 1st quadrant. So m+n > 0

IMO for the first one answer is A. (1) px/qy= p/q -> x = y (2) xy = p does not give me anything. Second one answer is clear A √n+k = 2√n when k = 3n So A is sufficient B is not sufficient. Had it said n+k > 4n then it would have been sufficient.

Answer is C. One way to answer these type of question is (if you have co-ordinate geometry background) consider m = y, z= x y > 3x y Try to draw these in a x-y plane and you can see that both equations are satisfied in the 1st quadrant, where x and y are positive. so, m + z should be +ve

It should be C I think. Arriving at 8:10 does not necessary mean that he has left by 7:08. He could have left on yesterday evening's train as well.. Interesting one..

unit and tenth digit combination repeats for every 100 numbers. Aqueels has correctly stated the answer

IMO answer is A. 1/k > 1/2 - gives k Bit rusty after few months of rest.

Surprised to see that my thread is still active... I am starting now again and hope to learn a lot of from TM members to get 750 this time

Excellent score.. All the best for your Apps. WHich schools are u targetting. I am giving a shot at GMAT again next month.

Chris - Not sure why did you watch the second half of that game. Match over after the first session. Noway Kiwis can chase 290 against SL. You are not playing in NZ and opposition is not Aussies.

Dear All, Nice to see my thread being alive so-long. Thanks for the wishes. Regarding Sets, please enroll in Yahoo groups - you will find them in Files section. there are 30 sets in total and verbal part of the sets closely matches with real GMAT. For quant I used to get max 3 or 4 wrong, still managed 47 in real GMAT.