Answer: Option D. -> (−∞,0)∪(3,4)
:
D
Check for values which are included in the range in 2 options and not there in two options i.e. substitute x=0, equation is not satisfied at x=0; thus option (a) and option (c) can be eliminated next, check for values which are there in option (b) and not there in option (d) or vice versa. Put x=1The equation is not satisfied at x=1, LHS= 6 and RHS=1. Thus the correct range is option (d)

Answer: Option D. -> 4
:
D
Let $u=\frac{a}{b}.$ Then the problem is equivalent to finding all possible rational numbers 'u' such that
$u+\frac{14}{9u}=k$, for some integer k.
This equation is equivalent to $9u^2-9uk+14=0$
$9u^2-9uk+14=0$ whose solutions are,
$u=\frac{9k\pm \sqrt{81k^2-504}}{18}$
$=\frac{k}{2}\pm \frac{1}{6}\sqrt{9k^2-56}$
Hence u is rational if and only if:
$\sqrt{9k^2-56}$ is rational, which is true if and only if $9k^2-56$ is a perfect square.
Suppose that $9k^2-56=S^2$ for some positive integer S.
(3k+S)(3k-S)=56
The only factors of 56 are 1,2,4,7,8,14,28 and 56.
So, (3k-S) and (3k+S) is one of the ordered pairs (1,56), (2,28), (4,14) and (7,8). The cases (1,56) and (7,8) yield no integral solutions.
The cases (2, 28) and (4, 14) yield k = 5 and k = 3 respectively.
If k=5, then $u=\frac{1}{3}$ or $u=\frac{14}{3}.$
If k=3, then $u=\frac{2}{3}$ or $u=\frac{7}{3}.$
Therefore there are four pairs (a,b) that satisfy the given conditions, namely (1,3), (2,3), (7,3) and (14,3).

Answer: Option A. -> 3
:
A
abc leaves remainder of 1 when divided by $10⟹c=1$.
bca leaves remainder of 1 on dividing by $8⟹4b+2c+a=8m+1$ and similarly a+b+c = 3n+1.
On solving we get 3b = 8m -1 - 3n $⟹$ m = 2, 5, 8,... or 3p-1 format.
Now, 4b+2+a = 8m+1 $⟹$ 4b+a = 8m-1
For p=3, m=8, 4b+a = 63 means b > 8. So, p cannot be greater than 2.
For p=2, m=5, 4b+a = 39, if a = 3, 7 means b$\ge$8 which is not possible.
So, p=1, m=2, 4b+a = 15, possible value of a = 3.
Hence, choice (a) is the right option.

Answer: Option B. -> 3
:
B
Let the number of correct answers be 'x', number of wrong answers be 'y' and number of questions not attempted be 'z'.
Thus,x+y+z=50 .............(i)
And $x-\frac{y}{3}-\frac{z}{6}=32$
The second equation can be written as,
6x-2y-z=192 ...................(ii)
Adding the two equations we get,
$7x-y=242\Rightarrow x=\frac{242}{7}+\frac{y}{7}=\frac{242+y}{7}$
Since, x and y are both integers, y cannot be 1 or 2. The minimum value that y can have is 3.
Hence option (b)

:
A be the set of natural numbers less than 100.
$\therefore$ A = {1, 2, 3, 4, ... , 99}
Now, B is the subset of A such that B contains two digit numbers which are divisible by exactly two numbers from the set A .
$\because$ Only prime numbers are the numbers which are divisible by 1 and itself only.
$\therefore$ B = {11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}
Now, C is the subset of B such that C contains numbers whose sum of digits is divisible by 2.
$\therefore$ C = {11, 13, 17, 19, 31, 37, 53, 59, 71, 73, 79, 97}
Now, D is the subset of C such that D contains numbers whose sum of the digits is a perfect square or a perfect cube.
$\therefore$ D = {13, 17, 31, 53, 71, 79, 97}
$\therefore$ Median of the set D = 53

Answer: Option A. -> 3
:
A
For remainder when a number is divided by 10, we need to check only the last digit.
We can write 33=4k+1. The first digit in the cycle of 3 = 3 (the cycle of 3 is 3,9,7,1)
${123}^{33}={123}^{4k+1}=...3$. Thus, the last digit of this number will be 3.

Answer: Option A. -> 11
:
A
It is given that the number has 11 digits which add up to 2
When the first digit is 1,
The number can be formed from 9 zeroes and 2 ones which can be arranged in $\frac{10!}{9!}\times$1! = 10 ways
When the first digit is 2, we can have 1 more number. Total cases = 11. Answer is option (a)

Answer: Option C. -> 30
:
C
Mark a particular face X. (Later on, we will have to divide whatever answer we get by 6 because all 6 faces are identical at this point, and we have chosen one out of 6 randomly).
Put a colour on face X. There are 6 ways to do this.
Put another colour on the face opposite X. There are 5 ways to do this.
Arrange the remaining 4 colours in the remaining four places - there are 3! ways to do this (circular permutation of 4 colours = (4-1)!
Total =$\frac{\left[\right(6\times 5\times 3!\left)\right]}{6}$= 30

Answer: Option C. -> 17!2!15!
:
C
Non negative integral solutions means $x_1,x_2,x_3\ge 0$
This is also a "similar to different” type of question.
Applying 0's and 1's method:
Total no. of ways $=\frac{17!}{2!15!}.$ Hence option (c) is the correct answer.

Answer: Option B. -> 6,12
:
B
A cubic equation can be expressed in terms of its roots as
$x^3$-(sum of the roots) $x^2$+(sum of the product of the roots taken 2 at a time) x-(product of the roots)=0
thus,
p= sum of roots taken one at a time
q= sum of roots taken two at a time
Product of roots =8 (given in question)
'if the product of 2 or more numbers is constant, the sum will be minimum when they are equal'
i.e.If abc=constant, the minimum value of a+b+c will be obtained at a=b=c
Thus, For the sum of roots (p) to be minimum, 8 should be the product of 3 numbers that are equal to each other (difference between them should be equal). This number is 2. The roots are thus 2,2,2
P= 2+2+2= 6
And q= 2(2)+2(2)+2(2)= 12
Answer= option b