@kamal_lohia I THINK THE ANSWER SHOULD BE 1 WHRE X=2 AND Y=5 CUZ OTHER VALUES WILL BE SURELY NEGATIVE

Q30) Given f(x) = x - 1/x, solve the equation f(f(x)) = x.

Q30) There are 10 consecutive positive integers written on a blackboard. One number is erased. The sum of remaining nine integers is 2011. Which number was erased?

Let students be x (X/2)-1 /x >=0.47 ( for x even)(X/2 -0.5)/x >=0.47 (x being odd case) It satisfies at x=17( min)

The roots are a and b: a + b = p and ab = 12 (a + b)^2 = p^2 (a - b)^2 = (a + b)^2 - 4ab => (a - b)^2 = p^2 - 12 * 4 = p^2 - 48 If |a - b| ≥ 12 { Difference between the roots is at least 12} then, (a - b)^2 ≥ 144 p^2 - 48 ≥ 144 p^2 ≥ 192 P ≥ 8√3 or P ≤ -8√3

@rajesh_balasubramanian Hello. I was just looking for an online CAT coaching and stumbled upon this. The answer seems to be 7. And i don't think 10 can be a value for B. Eg: Take the Track length as 300m. A will complete it in 50secs. And B will take 75 secs to overlap A for the first time(300/4). I can email you my solution if you'd like to give it a look. Thank you.

1/x + 1/y = 1/n=> xy = nx + ny=> (x - n)(y - n) = n² Now, n² must have 5 factors to get 5 ordered pairs=> n must be of form p², where p is prime number Hence all primes less than 20

@sumit-agarwal In such scenarios like finding distinct values of [x^2/n] where x can be from 1, 2, 3 ... n[1^2/n], [2^2/n] ... [(n/2)^2/n] will yield all numbers from 0 to [n/4] (means [n/4] + 1 distinct integers)Then the next set (from [(n/2 + 1)^2/n] till [n^2/n] will be all different integers (means [n/2] distinct integers)So the number of distinct integers would be [n/2] + [n/4] + 1 if n = 100,number of distinct integers would be [100/2] + [100/4] + 1 = 76 if n = 2014,number of distinct integers would be [2014/2] + [2014/4] + 1 = 1511 if n = 13number of distinct integers would be [13/2] + [13/4] + 1 = 10 Just trying to generalize a solution shared by Kamal sir (Quant Boosters - Set 1 - Q2). You can try out with various numbers (may be smaller numbers) so that this can be verified.

@anubhav_sehgal Hey! Can you please tell me that why is power of 3 inside the bracket 0? (In the question 58!-38!)

General term - nCr * (x^2)^(n-r) (3y)^r It is given that 2n - 2r + r = 52n - r = 5n = 4 => r = 3 So the term we have here is 4C3 * (x^2)^1 (3y)^3Coefficient = 4C3 * 3^3 = 4 * 27 = 108

@rowdy-rathore 22

Q100) A and B play a game of dice between them. The dice consist of colors on their faces (instead of numbers). When the dice are thrown, A wins if both show the same color; otherwise B wins. One dice has 4 red face and 2 blue faces. How many red and blue faces should the other dice have for both the to players have same chances of winning? a) 3 red and 3 blue faces b) 2 red and remaining blue c) 6 red and 0 blue d) 4 red and remaining blue

where are the answers and answer key ?

@Rowdy-Rathore 90+30x=60+60x X=1 Total steps=120

@Rowdy-Rathore what is the solution?

@rowdy-rathore 78

@Rowdy-Rathore maximum value 152 and minimum 38