option (b)

x=(-b±√D)/2a

D=n²-4mp<n² (as m>0 and p>0), the roots of this equation will be –ve. If D<0, then the roots will have negative real parts

Shortcut: Assumption

Option (c)

Product of roots=1 is the only constraint.

Assume values that satisfy the constraint and substitute, take a= -1 and b= -1, then p= 2

Assume c= - 1/2 and d= - 2; q= 5/2

Question is = (a-c)(b-c)(a+d)(b+d)=9/4

Only option (c) satisfies this => q² –p² = 25/4 – 4 =9/4

WXYZ = WXY * XZ => WXY (10) + Z = WXY (10x) + WXYZ.

Comparing we get => X = 1 and Z = 0

WXYZ = WXY * XZ => WXY (10) + Z = WXY (10x) + WXYZ.

Comparing we get => X = 1 and Z = 0

Option (c)

Go from answer options

1) Take p=2

a² –a +2=0 and a² – a + 6=0

Roots are =  $\frac{1+\sqrt{(1+24)}}{2}$  and   $\frac{\sqrt{(1-24)}}{2}$

Hence, this option is ruled out

Take p = -2

a² –a -2=0 and a² – a -6=0

Roots of 1^st equation = 2,-1 and roots of second equation = 3, -2

OPTION (d)

for n≥10, n! is divisible by 100, hence the 10s digit will be zero for numbers greater than 10!

Hence we will have to find the 10's digit of numbers form 1! to 9!

we see that 1! ends with 1

2! ends with 2.... hence sum of 1! to 9! the last two digits will be 13 hence the option is a

Option(b)

To find the average of all numbers, we need to find the sum of all the possible numbers and divide it by the total such numbers possible.The sum of all possible numbers can be found using the formula

(n-1)!(sum of digits)(1111…n times)

here n=5

(5-1)!(15)(11111)= 4!(15)(11111)

Total number of such possible cases= 5!

Average= 3999960/5! = 33333

Short cut: - If we arrange given numbers, These numbers will be in A.P.

Ao, A.M. of these numbers = (First term + last term)/2

(12345+54321 )/2 = 33333

Similar to different grouping( permutation and combination)

a,b,c,d will all be some power of 5 from 0 to 12.
This question is same as a + b + c + d = 12
Solutions = ¹⁵C₃ 

12 identical things to 4 distinct groups = ¹⁵ C₃ = 455

Option(c) 

This is S -> D type of grouping with a lower limit of 1

N+L+EL+ES= 20

N,L,EL,ES ≥ 1

Removing 4 from each side

N+L+EL+ES=16

Now, N, L, EL, ES ≥ 0

This is the arrangement of 16 zeroes and 3 ones. Answer = ¹⁹C₃

option (d)

50 runs can be scored from 12 balls as follows 

Case 1:- 11 fours and 1 six = ¹²C₁₁^{. 1}C1

Case 2:- 1 zero + 8 fours + 3 sixes = ¹²C₈ . ⁴C₃

Case 3:- 2 zeroes+ 5 fours+ 5 sixes = ¹²C₅.⁷C₅

Case 4: - 3 zeroes + 7 sixes + 2 fours = ¹²C₃.⁹C_{2 .}⁷C₇

Answer = sum of all 3. option- d