$$\eqalign{ & {\text{ }}\frac{2}{5},\frac{5}{6},\frac{{11}}{{15}}{\text{ & }}\frac{7}{8} \cr & \frac{7}{8}\,{\text{is largest among }}\frac{2}{5},\frac{5}{6},\frac{7}{8} \cr & {\text{now compare}} \cr } $$ (105)   (88) $$\frac{7}{8}$$   ⤧   $$\frac{{11}}{{15}}$$
So $$\frac{{7}}{{8}}$$ is largest

$$\eqalign{ & {{\text{7}}^{105}} \Rightarrow \cr & {{\text{7}}^1} \to 7 \to 7 \cr & {{\text{7}}^2} \to 49 \to 9 \cr & {{\text{7}}^3} \to 343 \to 3 \cr & {{\text{7}}^4} \to 2401 \to 1 \cr & {\text{Divided power of 7 by 4}} \cr & \frac{{105}}{4} \to {\text{ remainder = 1}} \cr & \Rightarrow {{\text{7}}^1}\,{\text{is left unit digit = 1}} \times 7 = 7 \cr} $$

Let the number of cows be x.
Then, number of hens would be (180 - x)
According to the question
(180 - x) × 2 + 4x = 420
⇒ 360 - 2x + 4x = 420
⇒ 2x = 420 - 360 = 60
⇒ 2x = 60
∴ x = $$\frac{{60}}{2}$$ = 30

L.C.M. of powers (3, 2, 1 and 6) = 6
$$\eqalign{ & {\left( 3 \right)^{\frac{1}{3}}} \Rightarrow {\left( {{3^2}} \right)^{\frac{1}{6}}} = {{\bf{9}}^{\frac{{\bf{1}}}{{\bf{6}}}}} \cr & {\left( 2 \right)^{\frac{1}{2}}} \Rightarrow {\left( {{2^3}} \right)^{\frac{1}{6}}} = {8^{\frac{1}{6}}} \cr & (1) \Rightarrow {\left( {{1^6}} \right)^{\frac{1}{6}}} = {1^{\frac{1}{6}}} \cr & {\left( 6 \right)^{\frac{1}{6}}} \Rightarrow {\left( {{6^1}} \right)^{\frac{1}{6}}} = {6^{\frac{1}{6}}} \cr} $$

$$\frac{1}{{20}} + \frac{1}{{30}} + \frac{1}{{42}} + \frac{1}{{56}} + \frac{1}{{72}} + \frac{1}{{90}}$$      $$ + \frac{1}{{110}}$$   $$ + \frac{1}{{132}}$$
$$ = \left( {\frac{1}{{4 \times 5}}} \right) + \left( {\frac{1}{{5 \times 6}}} \right) + \left( {\frac{1}{{6 \times 7}}} \right)$$       $$ + ..... + $$   $$\left( {\frac{1}{{11 \times 12}}} \right)$$
$$ = \left( {\frac{1}{4} - \frac{1}{5}} \right) + \left( {\frac{1}{5} - \frac{1}{6}} \right) + \left( {\frac{1}{6} - \frac{1}{7}} \right)$$       $$ + ..... + $$   $$\left( {\frac{1}{{11}} - \frac{1}{{12}}} \right)$$
$$ = \frac{1}{4} - \frac{1}{5} + \frac{1}{5} - \frac{1}{6} + \frac{1}{6} - \frac{1}{7} \, + .....$$      $$ + \frac{1}{{11}}$$  $$ - \frac{1}{{12}}$$
$$\eqalign{ & = \frac{1}{4} - \frac{1}{{12}} \cr & = \frac{{3 - 1}}{{12}} \cr & = \frac{2}{{12}} \cr & = \frac{1}{6} \cr} $$

According to the question,
As we know that the square of 252 is which that is near to the value of 63520.
Therefore, 63520 - x = 63504
∴ x = 16

$$\eqalign{ & \frac{{{\text{remainder of number}}}}{{16}} = \frac{{54}}{{16}} \cr & \Rightarrow {\text{Remainder = 6}} \cr} $$

n2(n2 - 1)
put n = 2
= 22(22 - 1)
= 4(4 - 1)
= 4 × 3
= 12
Check the option it is divisible by 12.
$$\eqalign{ & {\text{Take n = 3}} \cr & \Rightarrow {{\text{3}}^3}\left( {{3^3} - 1} \right) \cr & \Rightarrow 9\left( {9 - 1} \right) \cr & \Rightarrow 9 \times 8 = 72 \cr & {\text{It is divisible by 12}} \cr & {\text{So 12 is answer}} \cr} $$

$$\eqalign{ & x:2x:3x \cr & {x^3} + 8{x^3} + 27{x^3} = 4500 \cr & 36{x^3} = 4500 \cr & {x^3} = \frac{{4500}}{{36}} = 125 \cr & x = 5 \cr & {\text{Smallest number is 5}} \cr} $$

$$\eqalign{ & {\text{Let the unit digit }} \cr & {\text{Ten digit }} = x - 2 \cr & \therefore {\text{ Number :}} \cr & = {\text{ }}10(x - 2){\text{ }} + {\text{ }}x \cr & = {\text{ }}10x{\text{ }} - {\text{ }}20{\text{ }} + {\text{ }}x \cr & = {\text{ }}11x{\text{ }} - {\text{ }}20 \cr} $$
New number obtained after reversing the digits
$$\eqalign{ & = 10x + x - 2{\text{ }} \cr & = {\text{ }}11x - 2 \cr & {\text{According to the question,}} \cr & {\text{3}}\left( {11x - 20} \right) + \frac{6}{7}\left( {11x - 2} \right) = 108 \cr & 7\left( {11x - 20} \right) + 2\left( {11x - 2} \right) = 36 \times 7 \cr & 77x - 140 + 22x - 4 = 252 \cr & 99x = 252 + 144 \cr & x = \frac{{396}}{{99}} = 4 \cr & {\text{Number :}} \cr & = {\text{ }}11x - 20{\text{ }} \cr & = {\text{ }}11 \times 4 - 20 \cr & = {\text{ }}24 \cr & {\text{Sum of digit }} = 2 + 4 = 6 \cr} $$

$$\eqalign{ & \frac{6}{7},\frac{5}{6},\frac{7}{8},\frac{4}{5} \cr & \frac{6}{7} = 0.85, \cr & \frac{5}{6} = 0.83, \cr & \frac{7}{8} = 0.87, \cr & \frac{4}{5} = 0.80 \cr & \frac{7}{8}\,\,{\text{is}}\,{\text{largest}} \cr} $$

Explanation :
When the fractions are given in this order when the difference between the numerator and denominator is same so largest fraction will be largest numerator fraction and the smallest will be smallest numerator fraction.
Note: - Fraction should be in the order where denominator will be larger than numerator.
In this question : largest is $$\frac{7}{8}$$and smallest is $$\frac{4}{5}$$

\[\begin{gathered} {\text{ unit place}} \hfill \\ \left[ \begin{gathered} {7^1} \to 7 \to 7 \hfill \\ {7^2} \to 49 \to 9 \hfill \\ {7^3} \to 343 \to 3 \hfill \\ {7^4} \to 2401 \to 1 \hfill \\ \end{gathered} \right]{\text{cycle }} \hfill \\ \end{gathered} \]
$$\eqalign{ & {\left( {4387} \right)^{245}}\,\,\,\,\, \times \,\,\,\,\,{\left( {621} \right)^{72}} \cr & \,\,\,\,\mathop {{{\left( 7 \right)}^{245}}\,\,\,\,\,\,\,\,}\limits^ \downarrow \times \,\,\,\,\,\,\,\,\,{\left( 1 \right)^{72}} \cr & \,\,\,\,\mathop {{{\left( 7 \right)}^{4 \times 61 + 1}}\,\,\,}\limits^ \downarrow \times \,\,\,\,\,\,\,\,\,\,\,1 \cr & \,\mathop {{{\left( 7 \right)}^{4 \times 61}} \times 7\,\,}\limits^ \downarrow \times \,\,\,\,\,\,\,\,\,\,\,1 \cr & \,\,\,\,\,\,\mathop {{{\left( 1 \right)}^{61 }\times 7}}\limits^ \downarrow \,\,\,\,\, \times \,\,\,\,\,\,\,\,\,\,\,1 \cr} $$
unit digit = 7

4x2 - 12x + k
for perfect square must be
b2 = 4ac
12 × 12 = 4 × 4 × k
9 = k
k = 9

$$\eqalign{ & {\text{Number at (1) = 643 - 175}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ = 468}} \cr & {\text{Number at (2) = 1752 - 114}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ = 1638 }} \cr & {\text{Number at (3) = 1149 - 213 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ = 936}} \cr & {\text{HCF of 468, 1638, 936 = 234}} \cr & {\text{The divisor is 234}}{\text{.}} \cr} $$

Let the remainder in each case be x
Then, (2272 - x) and (875 - x) are exactly divisible by three digit number
Difference :
= (2272 - x) - (875 - x)
= 1397
Factor of 1397 = 11 × 127
Since, both 11 and 127 are prime number
Three digit number is 127
Sum of digits = 1 + 2 + 7 = 10
Note: - In this type of questions the number which divide the given number and leaves no remainder is either difference of number or a factor of difference.

  (1570)2  +  (1571)2  +  (1572)2  +  (1573)2 unit digit → ↓   ↓   ↓   ↓   02 + 12 + 22 + 32   ↓   ↓   ↓   ↓   0 + 1 + 4 + 9 = 14
unit digit = 4

Let the numerator and denominator br x and 11 - x
According to question,
$$\eqalign{ & {\text{Fraction }} = \frac{x}{{11 - x}} \cr & \Rightarrow \frac{{x + 2}}{{11 - x + 2}} = \frac{x}{{11 - x}} + \frac{1}{{24}} \cr & \Rightarrow \frac{{x + 2}}{{13 - x}} - \frac{x}{{11 - x}} = \frac{1}{{24}} \cr} $$
$$ \Rightarrow \frac{{11x + 22 - {x^2} - 2x - 13x + {x^2}}}{{\left( {13 - x} \right)\left( {11 - x} \right)}}$$       $$ = \frac{1}{{24}}$$
$$\eqalign{ & \Rightarrow \frac{{22 - 4x}}{{143 - 24x + {x^2}}} = \frac{1}{{24}} \cr & \Rightarrow 528 - 96x = 143 - 24x + {x^2} \cr & \Rightarrow {x^2} + 72x - 385 = 0 \cr & \Rightarrow \left( {x + 77} \right)\left( {x - 5} \right) = 0 \cr & \Rightarrow x = 5 \cr & {\text{Numerator }}(x) = 5 \cr & {\text{Denominator }} = 11 - 5 = 6 \cr & {\text{Difference }} = 6 - 5 = 1 \cr} $$

$$\eqalign{ & {\text{0}}{\text{.7 + }}\sqrt {0.16} = 0.7 + 0.4 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1.1 \cr & 1.02 - \frac{{0.6}}{{24}} = 1.02 - 0.025 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.995 \cr & 1.2 \times 0.83 = 0.996 \cr & \sqrt {1.44} = \sqrt {\frac{{144}}{{100}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{12}}{{10}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1.2 \cr & {\text{ So }}\sqrt {1.44} {\text{ is the greatest}}{\text{.}} \cr} $$

Number of pants can be made = $$\frac{{252}}{{\frac{5}{2}}}$$ = 100

Number of shirts can be made = $$\frac{{141}}{{\frac{7}{4}}}$$ = 80
Number of pants and shirts = 100, 80

$${{\text{2}}^{31}} \div 5$$
\[\begin{gathered} {\text{power }}\,\,\,\,\,\,\,\,\,{\text{ remainder}} \hfill \\ \left[ \begin{gathered} {2^1}\,\,\, \to \,\,\,\frac{2}{5} \to \,\,\,2 \hfill \\ {2^2}\,\,\, \to \,\,\,\frac{4}{5} \to \,\,\,4 \hfill \\ {2^3}\,\,\, \to \,\,\,\frac{8}{5} \to \,\,\,3 \hfill \\ {2^4}\,\,\, \to \,\,\,\frac{{16}}{5} \to \,\,1 \hfill \\ \end{gathered} \right]{\text{cycle 1}} \hfill \\ \left[ {{2^5}\,\,\, \to \,\,\,\frac{{243}}{5} \to \,\,2} \right]{\text{cycle 2}} \hfill \\ \end{gathered} \]

$$\eqalign{ & {\text{Divided = }}\frac{{{2^{31}}}}{5} = \frac{{{2^{4 \times 7}} \times {2^3}}}{5} \cr & \Rightarrow {\text{remainder = 3}} \cr & {\text{So }}{{\text{2}}^3}{\text{ has 3 remainder }} \cr & {{\text{2}}^{31}}{\text{ has 3 remainder}} \cr} $$