Decimal Fraction

$$0.125125....$$
$$ = 0.\overline {125} $$
$$ = \frac{{125}}{{999}}$$

$$\frac{2}{3} = 0.666$$
$$\frac{3}{5} = 0.6$$
$$\frac{2}{5} = 0.4$$
$$\frac{1}{3} = 0.333$$
$$\frac{1}{{15}} = 0.066$$
$$\frac{{31}}{{50}} = 0.62$$
Clearly, 0.62 lies between 0.6 and 0.666
So, $$\frac{31}{50}$$ lies between $$\frac{2}{3}$$ and $$\frac{3}{5}$$

Let the third number be 2960
∵ Second number = $$\frac{1}{4}$$ of the third number = $$\frac{1}{4}$$ × 2960 = 740
$$\frac{5}{9}$$ of first number = 25% of second number
$$\frac{5}{9}$$ first number = $$\frac{25 × 740}{100}$$  = 185
⇒ First number = $$\frac{185 × 9}{5}$$  = 333
∴ 30% of 333 = $$\frac{30}{100}$$ × 333 = 99.9

Given expression
= $$\frac{35 × 1.4}{7}$$
= 5 × 1.4
= 7

$$\therefore \frac{{1999}}{{2111}} = 0.946$$

$${\text{Given,}}$$
$$\frac{{52416}}{{312}} = 168$$
$$ ⇒ \frac{{52416}}{{168}} = 312$$
$${\text{Now,}}$$
$$\frac{{52.416}}{{0.0168}}$$
$$ = \frac{{524160}}{{168}}$$
$$ = \frac{{52416}}{{168}} \times 10$$
$$ = 312 \times 10$$
$$ = 3120$$

Given expression :
(1.25)³ - (0.75)³ - 3 × (1.25)² × 0.75 + 3 × 1.25 × (0.75)²
= (1.25 - 0.75)³
= (0.5)³
= $${\left( {\frac{1}{2}} \right)^3}$$
= $$\frac{1}{8}$$
[∵ (a - b)³ = a³ - b³ - 3a² b + 3ab²]

Let the missing number be x
$$\frac{{21.5}}{5} + \frac{{21}}{6} - \frac{{13.5}}{{15}} = \frac{{{{\left(×\right)}^{\frac{1}{3}}}}}{4} + \frac{{17}}{{30}}$$
$$\frac{{21.5}}{5} + \frac{{21}}{6} - \frac{{13.5}}{{15}} - \frac{{17}}{{30}} = \frac{{{{\left(×\right)}^{\frac{1}{3}}}}}{4}$$
L.C.M of 5, 6, 15 and 30 is 30
$$\frac{{129 + 105 - 27 - 17}}{{30}} = \frac{{{{\left(×\right)}^{\frac{1}{3}}}}}{4}$$
$$\root 3 \of x = \frac{{190 \times 4}}{{30}}$$
$$\root 3 \of x = 25.33 \approx 25 \cr &×= {25^3} \cr &×= 15625$$
Hence, the numbers 15625

Converting each of the given fractions into decimal form, we get:
$$\frac{9}{31}$$ = 0.29
$$\frac{3}{17}$$ = 0.176
$$\frac{6}{23}$$ = 0.26
$$\frac{4}{11}$$ = 0.363
$$\frac{7}{25}$$ = 0.28
Clearly, 0.363 > 0.29 > 0.28 > 0.26 > 0.176
So, $$\frac{4}{11}$$ > $$\frac{9}{31}$$ > $$\frac{7}{25}$$ > $$\frac{6}{23}$$ > $$\frac{3}{17}$$

$$0.\overline {142857} \div 0.\overline {285714} $$
$$ = \frac{{142857}}{{999999}} \div \frac{{285714}}{{999999}}$$
$$ = \left( {\frac{{142857}}{{999999}} \times \frac{{999999}}{{285714}}} \right)$$
$$ = \frac{1}{2}$$