@hemant_malhotra 8..............................................................

@gaurav_sharma answer option:a

Q30) Find the number of non negative integral solutions to the equation a + b + c = 13, such that a, b and c are distinct

@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)

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.

Each angle is 180(p-2)/p. 180-{360}/{p} = k So 360/p has to be an integer. 360 = 2^3 * 3^2 * 5^1 So there are 4 * 3 * 2 = 24 possibilities, but we exclude 1 and 2, because p > = 3 So , 24 -2 = 22 Hence, choice (c) is the right answer

2xy - 3x - y = 4 2x(y - 3/2) - y = 4 add 3/2on both sides 2x(y - 3/2) - y + 3/2 = 4 + 3/2 2x(y - 3/2) - 1(y - 3/2) = 11/2 (2x - 1)(y - 3/2) = 11/2 (2x - 1)(2y - 3) = 11 11 = 1 * 11 Case-1) 2x - 1 = 1 => x=1 2y - 3 = 11, y = 7 Case-2) 2x - 1 = 11 x = 6 2y - 3 = 1, y = 2 so 7/9

a^2 + ab + b^2 = n^2(a+b)^2 = n^2 + ab(a+b+n)(a+b-n) = ab as 'a' and 'b' are prime numbersab has 4 factors: 1, a, b, abobviously, a + b + n > a + b - n Case 1:a + b + n = aba + b - n = 1 Case 2:a + b + n = aa + b - n = b ... considering a>b But, case 2 is not possible.b = -na = n,a = -b ... which is impossible Case 1:adding the two equations2a + 2b = ab + 12a - ab + 2b = 1a(2-b) - 2(2-b) = -3(a-2)(b-2) = 3 only possibility for 'a' and 'b' being prime and satisfying the above condition is: 5 and 3 giving us he only possibility 49

Concept : numbers of consecutive natural numbers sum to n = (number of odd factors of n ) -1 1000 = 5^3 * 2^3here number of odd factors = 4 and required answer = 4 - 1 = 3

The clock loses 30 minutes per hour. i.e 30 minutes of this faulty clock = 60 minutes of the correct clock From 2 p.m to 6 p.m , total number of hours = 4 hours 4 hours of this faulty clock => 4 x 60/30= 8 hours of original clock So, The correct time when the clock show 6 p.m = 6 p.m + 4 = 10 p.m But The clock stopped 3 hours ago , So present time is 10 p.m + 3 hours = 1 a.m

1000 = 2^3 * 5^ When we say HCF of factors (x,y,z) = 1 ; then we mostly talk of 3 cases .. Case -1 ; ( 1, 2^3, 5^3) ..now here ordered arrangements = 3^2*3! = 54 Case - 2 ( 1, 1, 2^3) ; (1, 1, 5^3) ; (1, 1, 2^3 * 5^3)now total arrangement of these cases is 3!/2! ( 15) = 45 case -3 when (1 , 1, 1) ; as we are taking of factors of 1000 with HCF 1 , so this has to be a case ... And it can be arranged in only n only 1 way .. so 1 So finally 54 + 45 + 1 = 100

@sibanand_pattnaik number of scores possible = 251 - 15 + 1 ...= 237 Could anyone explain this last step ?

Here nf(x)= f(x^n) F(343sqrt3)= 3f(7)+1/2f(3) F(7) and f(3) can be found using given values.