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)

Q30) How many 7 digit numbers abcdefg are there such that a > b ≥ c > d ≥ e > f ≥ g ? [OA : 13C7]

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

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

p = prob (overnight born = boy| Next day chosen = boy) = prob (overnight born = boy AND Next day chosen = boy) / prob (Next day chosen = boy) = .5 * 4/T / (.5*4/T + .5 * 3/T) = 2/3.5 = 4/7 ~ .57

14 to 28 : 15 students 30, 31, 32 and 1 to 12 : 15 students so total 32

x = 3 - 2 root 2 1/x = 3 + 2 root 2 x^6 + (1/x)^6 => 2 [ 3^6 + 6C2 * 3^4 * ( 2 root 2 )^2 + 6C4 * 3^2 * ( 2 root 2 )^4 + ( 2 root 2 )^6 ] 2 [ 729 + 15 * 81 * 8 + 15 * 9 * 64 + 512 ] = 39202

5x - 3y = 140 x=25 , x=22 , x=19 , x=16 , x=13 , x=10 , x=7 , x=4 , x=1 9 solutions

110 = 2 * 5 * 11 Number of ordered positive triplets = (1 + 3 - 1)C2 * (1 + 3 - 1)C2 * (1 + 3 - 1)C2 = 3 * 3 * 3 = 27 Ordered integer triplets = 4 * 27 = 108 (considering the possibilities (-a, -b, c), (-a, b, -c), (a, -b, -c), (a, b, c)) This includes cases like aab and aaa, which needs to be removed to get the unordered pairs. a = b = c : no cases possible a = b or b = c or c = a As all the prime factors only once, we cannot split it into cases where 2 entries are equal. So the only cases are (110, 1, 1) & (110, -1, -1) -> which can be arranged in 3!/2! = 3 ways each All the other triplets can be arranged in 3! ways So final unordered triplets = (108 - 6)/3! + 6/3 = 102/6 + 2 = 17 + 2 = 19 ways.

x^2 + y^2 - xy = x + y (x - y)^2 + (x -1)^2 + (y -1)^2 = 2 (0,0) (0,1) (1,0) (2,2) (1,2) (2,1) 6 integral solutions 3 positive integer solutions [credits : @Raman_Sharma]