$$\frac{8}{{25}},\frac{7}{{23}},\frac{{11}}{{23}},\frac{{14}}{{53}}$$

(184) , (175)   ↑   ↑   $$\frac{{8}}{{25}}$$  ⤩  $$\frac{{7}}{{23}}$$  = $$\frac{{8}}{{25}}$$ > $$\frac{{7}}{{23}}$$         (371) , (322)   ↑   ↑   $$\frac{{7}}{{23}}$$  ⤩  $$\frac{{14}}{{53}}$$  = $$\frac{{7}}{{23}}$$ > $$\frac{{14}}{{53}}$$
$${\text{So}}\frac{{14}}{{53}}\,{\text{is smallest}}$$

Let 'a' be the smallest even number
According to question
$$\eqalign{ & (a) + (a + 2) + (a + 4) + (a + 6) = 748 \cr & 4a + 12 = 748 \cr & 4a = 736 \cr & a = 184 \cr & \cr & {\bf{Alternate :}} \cr & {\text{Middle term }} = \frac{{748}}{4} = 187 \cr & \underline {184} \,\,\,\,\underline {186} \,\,\,\underline {188} \,\,\,\underline {190} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathop {187}\limits^ \downarrow \cr & {\text{Smallest number = 184}} \cr} $$

$$\eqalign{ & {\text{2, 4, 6, 8, 10}}......{\text{198, 200 }} \cr & {2^{100}}(1 \times 2 \times 3 \times ..... \times 99 \times 100) \cr} $$
$$\eqalign{ & \underline {5\,\,\,|\,\,100} \cr & \underline {5\,\,\,|\,\,\,20} \,\,\,\,\, = 24{\text{ zero's (20 + 4)}} \cr & \,\,\,\,\,|\,\,\,4 \cr} $$
When we multiply the series of 1 × 2 × 3 ×..... × 100 then we find 24 zero at the end of the product

$$\eqalign{ & {\text{Let number are }}x,{\text{ }}x + 5,{\text{ }}x + 10 \cr & {\text{So, }}x{\text{ + }}x + 5 + x + 10 = 225 \cr & 3x + 15 = 225 \cr & 3x = 210 \cr & x = 70 \cr & {\text{largest number is }}70 + 10 = 80 \cr} $$

$$\eqalign{ & {\text{Let the number be x}} \cr & {\text{According to question}} \cr & \frac{x}{{7/8}} - \frac{7}{8}x = 15 \cr & \frac{8}{7}x - \frac{7}{8}x = 15 \cr & \frac{{64x - 49x}}{{56}} = 15 \cr & 15x = 15 \times 56 \cr & x = 56 \cr & {\text{Sum of digits }} \cr & = 5{\text{ }} + {\text{ }}6 \cr & = 11 \cr} $$

Given :
$$\eqalign{ & \Leftrightarrow \frac{{51.84}}{{4.32}} = 12 \cr & \Leftrightarrow \frac{{5184}}{{432}} = 12 \cr & \therefore \frac{{0.005184}}{{0.432}} \cr & = \frac{{5.184}}{{432}} \cr & = \frac{{5184}}{{432}} \times \frac{1}{{1000}} \cr & = 12 \times \frac{1}{{1000}} \cr & = 0.012 \cr} $$

$$\eqalign{ & \frac{3}{4} = 75\% ,\frac{5}{6} = 83.33\% , \cr & (1)\frac{2}{3} = 66.66\% , \cr & (2)\frac{1}{2} = 50\% , \cr & (3)\frac{4}{5} = 80\% , \cr & (4)\frac{9}{{10}} = 90\% \cr & {\text{Hence , }}\frac{4}{5}{\text{ is between the fraction}}{\text{.}} \cr} $$

$$\eqalign{ & = \left( {999\frac{{999}}{{1000}} \times 7} \right) \cr & = \left( {999\frac{{999}}{{1000}}} \right) \times 7 \cr & = 6993 + \frac{{6993}}{{1000}} \cr & = 6993 + 6\frac{{993}}{{1000}} \cr & = 6993 + 6 + \frac{{993}}{{1000}} \cr & = 6999\frac{{993}}{{1000}} \cr} $$

Take LCM of 10 and 13 = 130
Any number divisible by 130 will be divisible by 13 and 10
i.e., 130, 260, 390, 520, 650, 780 & 910
⇒ total numbers are 7

$$\frac{3}{5},\frac{7}{9},\frac{{11}}{{13}}$$
$$\eqalign{ & \frac{3}{5} = 0.60, \cr & \frac{7}{9} = 0.7, \cr & \frac{{11}}{{13}} = 0.8 \cr & \frac{{11}}{{13}}{\text{ > }}\frac{7}{9} > \frac{3}{5} \cr} $$

0.34676767 ..... + 0.13333333 .....
= 0.480101 .....
= $$0.48\overline {01} $$

$$\eqalign{ & {\text{No}}{\text{. of digits}} \cr & {\text{0 - 9 }} \to {\text{ 1}} \times {\text{9 = 9}} \cr & {\text{10 - 19 }} \to {\text{ 2}} \times {\text{10 = 20}} \cr & {\text{20 - 29 }} \to {\text{ 2}} \times {\text{10 = 20}} \cr & {\text{30 - 39 }} \to {\text{ 2}} \times {\text{10 = 20}} \cr & 40{\text{ - 49 }} \to {\text{ 2}} \times {\text{10 = 20}} \cr & 50\, \to \,2 \times {\text{1 = 2 = 91}} \cr} $$

$$\frac{{41\frac{2}{3}}}{{\frac{1}{6}}} = \frac{{125}}{3} \times \frac{6}{1} = 250{\text{ times}}$$

Let numbers are = 3x, 4x
(3x)3 + (4x)3 = 5824
27x3 + 64x3 = 5824
91x3 = 5824
x3 = $$\frac{5824}{91}$$
x3 = 64
x = 4
So, numbers are
3x = 3 × 4 = 12
4x = 4 × 4 = 16
Sum = 12 + 16 = 28

$$\eqalign{ & 1.\overline {27} \, = \frac{{127 - 1}}{{99}}{\text{ = }}\frac{{126}}{{99}}{\text{ = }}\frac{{14}}{{11}} \cr & {\text{ }}\therefore {\text{ }}\frac{{\text{p}}}{{\text{q}}} = \frac{{14}}{{11}} \cr} $$

$$\eqalign{ & \left( {{n^3} - n} \right){\text{ and n is any integer}} \cr & {\text{put n = 2 so, }}{{\text{2}}^3} - 2 = 6 \cr & {\text{It will be always divisible by 6}} \cr & {\text{(put n = 2, 3, 4}}....{\text{)}} \cr & {\text{(n = 2, 3, 4}}.....{\text{)}} \cr} $$

$$\eqalign{ & \Leftrightarrow \left( {{2^{71}} + {2^{72}} + {2^{73}} + {2^{74}}} \right)\, \cr & = {2^{71}}\left( {{2^0} + {2^1} + {2^2} + {2^3}} \right)\, \cr & = {2^{71}}(1 + 2 + 4 + 8) \cr & = {2^{71}} \times 15 \cr & = {2^{70}} \times 30 \cr & {\text{It is divisible by 10}} \cr} $$

Let first number is x and second number is 2x.
Then, according to the question
x × 2x = 2048
x2 = 1024
x = 32
Smaller number = 32

$$\eqalign{ & = 999\frac{{98}}{{99}} \times 99 \cr & = (1000 - 1)\frac{{98}}{{99}} \times 99 \cr & = \left[ {\left( {1000 - 1} \right) + \frac{{98}}{{99}}} \right] \times 99 \cr & = 99000 - 99 + 98 \cr & = 99000 - 1 \cr & = 98999 \cr} $$

12345679 × 72
Shortcut method :
Since 72 is factor of 9, So the answer should be divisible by 9.
Now check with option by adding the digits of option and check which one is divisible by 9.
Here option (c) 888888888 is divisible by 9, so this is the multiple of 12345679 × 72

Alternate
12345679 x 72
= 12345679 x (70 +2)
= 12345679 x 70 + 12345679 x 2
= 864197530 + 24691358
= 888888888