$$\frac{8}{{15}},\frac{{14}}{{33}},\frac{7}{{13}},\frac{{11}}{{13}}$$

(264) , (210)   ↑   ↑   $$\frac{8}{{15}}$$   ⤩   $$\frac{14}{{33}}$$  = $$\frac{8}{{15}}$$ > $$\frac{14}{{33}}$$ (182) , (231)   ↑   ↑   $$\frac{14}{{33}}$$   ⤩   $$\frac{7}{{13}}$$  = $$\frac{14}{{33}}$$ < $$\frac{7}{{13}}$$ < $$\frac{11}{{13}}$$
$${\text{So}}\frac{{14}}{{33}}\,{\text{is}}\,{\text{smallest}}$$

$$\eqalign{ & {\text{Let number of boys}} = x \cr & {\text{Number of 25 paise coins}} = {x^2} \cr & {\text{According to question,}} \cr & \frac{{{x^2}}}{4} = 400 \cr & {x^2} = 1600 \cr & x = 40\,\, \cr} $$

L.C.M of 12, 18, 21, 28 = 252
9999 is divided by 252
Then remainder is 171
= 9999 - 171
The number will be = 9828

$$\eqalign{ & \frac{4}{5}{\text{ of an estate = 16800}} \cr & {\text{1 of an estate = 16800}} \times \frac{5}{4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 21000 \cr & \frac{3}{7}{\text{ of an estate = 21000}} \times \frac{3}{7} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 9000 \cr} $$

$$\eqalign{ & \Rightarrow {\text{999}}\frac{{998}}{{999}} \times 999 \cr & \Rightarrow \left( {999 + \frac{{998}}{{999}}} \right) \times 999 \cr & \Rightarrow \left[ {\left( {1000 - 1} \right) + \frac{{998}}{{999}}} \right] \times 999 \cr & \Rightarrow 999000 - 999 + 998 \cr & \Rightarrow 998999 \cr} $$

For divisibility of 11
= 756896
= (9 + 6 + 7) - (6 + 8 + 5)
= 3
So, 3 should be added to number to make it divisible by 11

$$\eqalign{ & \left( {{3^{25}} + {3^{26}} + {3^{27}} + {3^{28}}} \right)\, \cr & = {3^{25}}\left( {{3^0} + {3^1} + {3^2} + {3^3}} \right)\, \cr & = {3^{25}} \times 40 \cr & = {3^{24}} \times 120 \cr & {\text{now check with option }} \cr & {\text{Only 30 can divide this}}{\text{.}} \cr} $$

    771   ×   663   ×   365     ↓   ↓   ↓ Unit Place   73   63   31     ↓   ↓   ↓ Unit Digit ⇒ 3   ×   6   ×   3 = 54   ⇒ 4 is answer

A number will divisible by 18 if it is divisible by 2 and 9
Clearly we can see 65043 is not divisible by 2
Because unit digit of 65043 is 3 so this will not be divisible by 18

Let the two number are P and Q
P + Q = a
PQ = b
⇒ $$\frac{{1}}{{P}}$$ + $$\frac{{1}}{{Q}}$$
⇒ $$\frac{{Q + P}}{{PQ}}$$
⇒ $$\frac{{a}}{{b}}$$

$$\root 3 \of 9 ,\,\,\root 4 \of {20} ,\,\,\root 6 \of {25} $$
Taking L.C.M of 3, 4 and 6
$$\eqalign{ & = \root {12} \of {{9^4}} ,\,\,\root {12} \of {{{20}^3}} ,\,\,\root {12} \of {{{25}^2}} \cr & = \root {12} \of {6561} ,\,\,\root {12} \of {8000} ,\,\,\root {12} \of {625} \cr} $$
So, ascending order
$$ = \root 6 \of {25} , < \root 3 \of 9 , < \root 4 \of {20} $$

Let number are a, b = 3x, 5x
So, 3x + 5x = 8x = 80
x = 10
So, number is :
3 × 10 = 30
5 × 10 = 50

$$\eqalign{ & {\text{Let the number is }}x{\text{}} \cr & {\text{According to question,}} \cr & {\text{3}}\left( {2x + 9} \right) = 75 \cr & 2x + 9 = 25 \cr & x = \frac{{16}}{2} \cr & x = 8 \cr} $$

The sum of all natural numbers upto 97 - sum of all natural numbers upto 74
sum of n natural numbers
$$\eqalign{ & = \frac{{n\left( {n + 1} \right)}}{2}\,\,\left( {{\text{Formula}}} \right) \cr & = \frac{{97 \times 98}}{2} - \frac{{74 \times 75}}{2} \cr & = 4753 - 2775 \cr & = 1978 \cr} $$

$$\eqalign{ & {\text{a + b = 3 }}.......{\text{ (i)}} \cr & {\text{(a and b are two numbers)}} \cr & {{\text{a}}^2}{\text{ + }}{{\text{b}}^2} = 12 \cr & {\text{On squaring}}\,\,{\text{equation (i)}} \cr & {\left( {{\text{a + b}}} \right)^2} = \,{3^2} \cr & {{\text{a}}^2}{\text{ + }}{{\text{b}}^2} + 2{\text{ab}} = 9 \cr & 12 + 2{\text{ab = 9}} \cr & {\text{2ab = - 3}} \cr & {\text{ab = }}-\frac{{3}}{2} \cr} $$

$$0.\overline {001} \,\, \Rightarrow \frac{1}{{999}}$$

$$\eqalign{ & \frac{{\text{n}}}{5} \Rightarrow {\text{remainder}}\,\,2 \cr & {\text{if we put n = 7}} \cr} $$
Then it satisfies above situation
$$\eqalign{ & {\text{so n = 7}} \cr & \frac{{{{\text{n}}^2}}}{5} = \frac{{{7^2}}}{5} = \frac{{49}}{5} \cr & \Rightarrow {\text{remainder = 4}} \cr} $$

$$\eqalign{ & = \frac{2}{3} \times \frac{3}{4} \times n\,\,\,\left( {{\text{n}}\,{\text{is}}\,{\text{a}}\,{\text{number}}} \right) \cr & = \frac{1}{2}n \cr} $$

$$\eqalign{ & {\text{According to the question,}} \cr & {{\text{n}}^4} + 6{{\text{n}}^3} + 11{{\text{n}}^2} + 6{\text{n + 24}} \cr & {\text{put n = 1}} \cr & = 1 + 6 + 11 + 6 + 24 \cr & = 48 \cr & {\text{put n = 2}} \cr & {\text{ = 16 + 48 + 44 + 12 + 24}} \cr & {\text{ = 144}} \cr & {\text{Clearly it is divisible by 48}} \cr} $$

Let the number = x
⇒ According to the question,
$$\eqalign{ & \Rightarrow x - \frac{2}{5}x = 75 \cr & \Rightarrow \frac{{5x - 2x}}{5} = 75 \cr & \Rightarrow 3x = 75 \times 5 \cr & \Rightarrow x = 125 \cr} $$