Let, the number be 'x'
According to the question
$$\eqalign{ & \frac{1}{5}x - \frac{1}{7}x = 10 \cr & \frac{{7x - 5x}}{{35}} = 10 \cr & 2x = 350 \cr & x = \frac{{350}}{2} \cr & x = 175 \cr} $$

(0.1)2   →   .01   $$\sqrt {0.0121} $$   →   0.11   0.12   →   0.12   → Largest $$\sqrt {.0004} $$   →   0.02  

63592 is not a square of any natural number
∴ Any number that have 2, 3, 7, 8 on its units place , It could not be perfect square of any number.

$$\eqalign{ & \frac{1}{3}{\text{,}}\frac{4}{7},\frac{2}{5} \cr & \frac{1}{3} = 0.33,{\text{ }} \cr & \frac{4}{7} = 0.5, \cr & \frac{2}{5} = 0.4 \cr & \frac{1}{3} < \frac{2}{5} < \frac{4}{7} \cr} $$

$$\eqalign{ & = {17^{200}} \div 18 \cr & = {\left( {18 - 1} \right)^{200}} \div 18 \cr} $$
Apply Binomial theorem
$$ = {\left( {18} \right)^{200}}{\left( { - 1} \right)^0} + {\left( {18} \right)^{199}}{\left( { - 1} \right)^1} + $$       $$..... + \, {\left( {18} \right)^1}{\left( { - 1} \right)^{199}}$$   $$ \, + \, {\left( {18} \right)^0}{\left( { - 1} \right)^{200}}$$

$$ \Rightarrow $$ Remainder always comes from last term is (18)0 (- 1)200
$$\eqalign{ & = \frac{{{{\left( {18} \right)}^0}{{\left( { - 1} \right)}^{200}}}}{{18}} \cr & = 1 \times {1^{200}} \cr & = 1 \cr & {\text{So, remainder}} = 1 \cr} $$

Let Sn = (2 + 4 + 6 + ..... + 30). This is an A.P. in which a = 2, d = 2 and l = 30
Let the number of terms be n Then,
a + (n - 1)d = 30
⇒ 2 + (n - 1) x 2 = 30
⇒ n = 15
$$\eqalign{ & \therefore {S_n} = \frac{n}{2}\left( {a + I} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{15}}{2} \times \left( {2 + 30} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 15 \times 16 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 240 \cr} $$

$$\frac{7}{6},\,\frac{7}{9},\,\frac{4}{5},\,\frac{5}{7}$$
Step 1 : Compare two fractions
$$\frac{{7}}{{6}}$$ ⤩ $$\frac{{7}}{{9}}$$
Cross multiply
63 , 42
 ↑      ↑
$$\frac{7}{6}$$ ⤩ $$\frac{7}{9}$$
42 is smaller than 63
So, $$\frac{7}{9}$$ is smaller than $$\frac{7}{6}$$
Step 2 : Compare
(i) $$\frac{7}{9},\frac{4}{5}$$
(ii) $$\frac{7}{9}$$ ⤩ $$\frac{4}{5}$$
Cross multiply it
35 , 36
 ↑      ↑
$$\frac{7}{9}$$ ⤩ $$\frac{4}{5}$$
35 is smaller than 36
So, $$\frac{7}{9}$$ is smaller than $$\frac{4}{5}$$
Step 3 : Compare
(i) $$\frac{7}{9},\frac{5}{7}$$
(ii) $$\frac{7}{9}$$ ⤩ $$\frac{5}{7}$$
Cross multiply it
49 , 45
 ↑      ↑
$$\frac{7}{9}$$ ⤩ $$\frac{5}{7}$$
45 is smaller than 49
S0, $$\frac{5}{7}$$ is the smallest

Alternate :
$$\eqalign{ & \frac{7}{6} = 1.166, \cr & \frac{7}{6} = 0.777 \cr & \frac{4}{5} = 0.8, \cr & \frac{5}{{7\,}} = 0.714 \cr & {\text{Thus }}\frac{5}{7}{\text{ is the smallest}}{\text{.}} \cr} $$

Unit digit in 34 = 1 ⇒ Unit digit in (34)16 = 1
Therefore Unit digit in 365 = Unit digit in [ (34)16 × 3 ] = (1 × 3) = 3
Unit digit in 659 = 6
Unit digit in 74 ⇒ Unit digit in (74)17 is 1
Unit digit in 771 = Unit digit in [(74)17 × 73] = (1 × 3) = 3
Therefore Required digit = Unit digit in (3 × 6 × 3) = 4

$$\eqalign{ & {\bf{Series :}} \cr & 103 + 108\, + ........ + \,998 \cr & a{\text{ }} = {\text{ }}103 \cr & d{\text{ }} = {\text{ }}5 \cr & {\text{Last term = 998}} \cr & {\text{Number of term }} \cr & = \frac{{998 - 103}}{5} + 1 \cr & = \frac{{895}}{5} + 1 \cr & = 180 \cr & {\text{Sum of n terms}} \cr & = \frac{n}{2}\left[ {2a{\text{ }} + \left( {n - 1} \right)d} \right] \cr & = \frac{{180}}{2}\left[ {2 \times 103 + \left( {180 - 1} \right)5} \right] \cr & = 99090 \cr} $$

437 > 22
All prime numbers less than 22 are : 2, 3, 5, 7, 11, 13, 17, 19
161 is divisible by 7, and 221 is divisible by 13
373 is not divisible by any of the above prime numbers.
Therefore 373 is prime

→( 1, 3, 5, 7. . . . . 99 ) × 128
5   →   51 15   →   51 25   →   52 35   →   51 45   →   51 55   →   51 65   →   51 75   →   52 85   →   51 95   →   51 ×     128   →   27
512 + 27 will make zero but since 2 comes 7 times, so only 7 zero will come.

Let numbers are 3x, 6x, 2x
Average : $$\frac{11x}{3}$$ = 44
x = 12 = Difference

4 | x             y = (5 × 1 + 4) = 9
------------
5 | y - 1         x = (4 × y + 1) = (4 × 9 + 1) = 37
------------
   | 1 - 4

Now, 37 when divided successively by 5 and 4, we get

5 | 37
-------------
4 | 7 - 2
-------------
   | 1 - 3
Respective remainders are 2 and 3

For divisibility by 8, last 3 number should be divisible by 8.
Now, put the value 0, 1 ... 2 ... and check.
When use * = $$\frac{632}{8}$$ = 79
So, hence it is divisible by 8 answer is 3

$$\eqalign{ & {\left( {2153} \right)^{167}} \cr & {\text{unit digit = }}{{\text{3}}^{167}} \cr & {\text{unit digit }} \cr & {{\text{3}}^1} \to {\text{3}} \to {\text{3}} \cr & {{\text{3}}^2} \to 9 \to 9 \cr & {{\text{3}}^3} \to 27 \to 7 \cr & {{\text{3}}^4} \to 81 \to 1 \cr} $$
This cycle will continue
⇒ divide the power of 3 by 4
$$\eqalign{ & \frac{{167}}{4} \Rightarrow {\text{remainder is 3}} \cr & {{\text{3}}^3} \Rightarrow 7 \cr & {\text{unit digit = }}1 \times 7 = {\text{ }}7 \cr} $$

$$\eqalign{ & \sqrt {a + b\,} \,{\text{and }}\sqrt a + \sqrt b \cr & {\text{Squaring both sides}} \cr & {(\sqrt {a + b\,} )^2}{\text{and (}}\sqrt a + \sqrt b {)^2} \cr & \Rightarrow a + b\,\,{\text{and}}\,a + b + 2\sqrt a \sqrt b \cr & So\,\sqrt {a + b\,} \, < {\text{ }}\sqrt a + \sqrt b \cr} $$

Let the two number are a and b
According to question,
$$\eqalign{ & {a^2} + {b^2} = 289 \cr & ab{\text{ }} = {\text{ }}120 \cr & {\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab \cr & {\left( {a + b} \right)^2} = {\left( {289} \right)^2} + \left( {2 \times 120} \right) \cr & {\left( {a + b} \right)^2} = 529 \cr & \left( {a + b} \right) = 23 \cr} $$

$$\eqalign{ & {\text{Let the one number be }}x \cr & {\text{And the other number be }}\frac{1}{5}x \cr & {\text{According to question}} \cr & x \times \frac{1}{5}x = 0.008 \cr & {x^2} = 0.040 \cr & {\text{ }}x = 0.2 \cr & {\text{Other number}} \cr & = \frac{1}{5} \times 0.2\, \cr & = 0.04 \cr & {\text{Smaller number is 0}}{\text{.04}} \cr} $$

Always do these types of question by option to save time
Pick up the option and follow the the question instruction
Take option 3
64 ⇒ Divided 3 it gives remainder 1
Now add 1 to 64
$$\frac{{65}}{5}$$ ⇒ remainder '0' is satisfies
So K = 64 this is answer

$$\eqalign{ & 20 \times x = \left( {64 + 36} \right)\left( {64 - 36} \right) \cr & 20 \times x = 100 \times 28 \cr & \therefore {\kern 1pt} x = \frac{{100 \times 28}}{{20}} \cr & \,\,\,\,\,\,\,\,\,\,\, = 140 \cr} $$