9A is prime number then by option, 9A
Option A: A = 9, Number = 99 → Composite
Option B: A = 7, Number = 97 → Prime
Option C: A = 3, Number = 93 → Composite
Option D: A = 1, Number = 91 → Composite

Every real number is a rational number.

$$\eqalign{ & 1 + 2 + 3 + ...... + n = \frac{{n\left( {n + 1} \right)}}{2} \cr & \therefore 91 + 92 + ...... + 110 \cr & = \left( {1 + 2 + ...... + 110} \right) - \left( {1 + 2 + ...... + 90} \right) \cr & = \frac{{110 \times 111}}{2} - \frac{{90 \times 91}}{2} \cr & = 6105 - 4095 \cr & = 2010 \cr} $$

$$\eqalign{ & 2 + 4 + 6 + ...... \cr & {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \cr & = \frac{{25}}{2}\left[ {2 \times 2 + 24 \times 2} \right] \cr & = \frac{{25}}{2} \times 52 \cr & = 25 \times 26 \cr & {\text{Clearly it is divisible by }}5,\,13 \cr} $$

8 + 4 + 3 + 9 + x + 5 + 3 = 9
or multiple of x + 5 = 9
x = 4

Number which is divisible by 3
$$\eqalign{ & \frac{{700}}{3} - \frac{{950}}{3} \cr & = 233 - 316 \cr & = 83......\left( {\text{i}} \right) \cr} $$
⇒ Number which is divisible by 7
$$\eqalign{ & \frac{{700}}{7} - \frac{{950}}{7} \cr & = 100 - 135 \cr & = 35 + 1 \cr & = 36......\left( {{\text{ii}}} \right) \cr} $$
(∴ If 1st number is totally divisible by 7 then we add +1 in final result)
⇒ Number which is divisible by 21 (LCM of 3 or 7)
$$\eqalign{ & \frac{{700}}{{21}} - \frac{{950}}{{21}} \cr & = 33 - 45 \cr & = 12......\left( {{\text{iii}}} \right) \cr} $$
∴ Number which is divisible by 3 or 7
(i) + (ii) - (iii)
(83 + 36) - 12 = 119 - 12 = 107
Total number from 700 to 950
n = (950 - 700) + 1 = 251
∴ Number which is neither divisible by 3 nor 7
⇒ 251 - 107 = 144

$$\eqalign{ & \because {1^3} + {2^3} + {3^3}\,......\, + {10^3} = 3025 \cr & {\text{To find}} = {2^3} + {4^3} + {6^3}\,......\, + {20^3} = ? \cr & \Rightarrow {2^3}\left( {{1^3} + {2^3} + {3^3}\,......\, + {{10}^3}} \right) \cr & \left[ {{{\left( {\frac{{n\left( {n + 1} \right)}}{2}} \right)}^2} = {{\left( {\frac{{10\left( {11} \right)}}{2}} \right)}^2} = {{\left( {55} \right)}^2} = 3025} \right] \cr & \Rightarrow 8 \times 3025 \cr & \Rightarrow 24200 \cr} $$

In such type of question always go through option to save your valuable time.
By option D: 4, 2
4 + 2 × 2 = 8
4 - 2 = 2 (Satisfied)

A + B + AB = 65
(A, B ≤ 15)
Assume value of A & B
A = 10
B = 5
10 + 5 + 10 × 5 = 65
65 = 65 value satisfied
Difference between (A - B) = 10 - 5 = 5

121 to 999 - 3 digit
Total number of terms = 999 - 121 + 1 = 878 + 1 = 879
Total keys from 121 to 999 = 879 × 3 = 2637
1000 to 1346 - 4 digit
Total number of terms = 1346 - 1000 + 1 = 347
Total keys from 1000 to 1346 = 347 × 4 = 1388
Total keys = 2637 + 1388 = 4025

$$\eqalign{ & x + y = \frac{1}{7}\left( {10x + y} \right) \cr & 7x + 7y = 10x + y \cr & 3x = 6y \cr & x:y = 2:1 \cr & 1{\text{ unit}} \to {\text{4}} \cr & {\text{Number will be}} = 84 \cr & \frac{{48}}{7}R \to 6 \cr} $$

Let the number of oranges with Nathu be x
Number of oranges with Buchku = y
Case I,
x + 10 = 2(y - 10)
⇒ x + 10 = 2y - 20
⇒ 2y - x = 20 + 10 = 30 . . . . . . (i)
Case II,
y + 10 = x - 10
⇒ x - y = 10 + 10 = 20 . . . . . . (ii)
On adding equations (i) and (ii)
2y - x + x - y = 30 + 20
⇒ y = 50
From equation (ii),
x - 50 = 20
⇒ x = 50 + 20 = 70

I. N = 72 = 23 × 32
Total factors = (3 + 1)(2 + 1) = 4 × 3 = 12
True
II. Sum of n odd number = n2
Sum of 20 odd number = 202 = 400
True
III. Largest 2 digit prime number is 97
True
All are true

$$\eqalign{ & \frac{{A - 3}}{{B + 3}} = \frac{1}{3} \cr & \Rightarrow 3A - 9 = B + 3 \cr & \Rightarrow 3A - B = 12\,......\left( 1 \right) \cr & \frac{{A + 2}}{{B - 2}} = \frac{1}{2} \cr & \Rightarrow 2A + 4 = B - 2 \cr & \Rightarrow 2A - B = - 6\,......\left( 2 \right) \cr & {\text{Solving equation}}\left( 1 \right){\text{and}}\left( 2 \right) \cr & A = 18 \cr & B = 42 \cr & {\text{Hence, Team B have 42 members}}{\text{.}} \cr} $$

Number from 1 to 200 which is divisible by 3
3, 6, . . . . . . . . ., 198
$$\eqalign{ & = \frac{{198 - 3}}{3} + 1 \cr & = 66 \cr} $$
Total number which divisible by 3 & 7
$$\eqalign{ & = \frac{{189 - 21}}{{21}} + 1 \cr & = \frac{{168}}{{21}} + 1 \cr & = 9 \cr} $$
Total number which is divisible by 3 but not 7
= 66 - 9 = 57

Alternate solution
Total number which is divisible by 3 & 7 both from 1 to 200
LCM (3 & 7) $$ \to \frac{1}{{21}} - \frac{{200}}{{21}} = 0 - 9 = 9$$
Total number which is divisible by only 3, from 1 to 200
$$\frac{1}{3} - \frac{{200}}{3} = 0 - 66 = 66$$
∴ Total number which is divisible by 3 but not 7
= 66 - 9 = 57