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

→( 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.

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

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

(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.

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

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

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
$$\eqalign{ & \therefore {S_n} = \frac{n}{2}\left( {a + I} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{15}}{2} \times \left( {2 + 30} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 15 \times 16 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 240 \cr} $$

$$\eqalign{ & = {17^{200}} \div 18 \cr & = {\left( {18 - 1} \right)^{200}} \div 18 \cr} $$
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
$$\eqalign{ & = \frac{{{{\left( {18} \right)}^0}{{\left( { - 1} \right)}^{200}}}}{{18}} \cr & = 1 \times {1^{200}} \cr & = 1 \cr & {\text{So, remainder}} = 1 \cr} $$

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

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

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

$$\frac{7}{6},\,\frac{7}{9},\,\frac{4}{5},\,\frac{5}{7}$$
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

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

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

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

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

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