Decimal Fraction

4×162 = 648. Sum of decimal places = 6.
So,
0.04×0.0162
= 0.000648
= 6.48×10^-4

$$\text{Let } a = 4.2 \text{ and } b = 1.9 $$
$$ {\text{Given Expression}}$$
$$ = \frac{{ {{a^2} - {b^2}} }}{{\left( {a + b} \right)\left( {a - b} \right)}}$$
$$ = \frac{{ {{a^2} - {b^2}} }}{{ {{a^2} - {b^2}} }}$$
$$ = 1$$

$$\frac{{144}}{{0.144}} = \frac{{14.4}}{x}$$
$$ ⇒ \frac{{144 \times 1000}}{{144}} = \frac{{14.4}}{x}$$
$$ ⇒ x = \frac{{14.4}}{{1000}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.0144$$

Suppose commodity X will cost 40 paise more than Y after z years
Then,(4.20 + 0.40z) − (6.30 + 0.15z) = 0.40
⇒ 0.25z = 0.40 + 2.10
$$ ⇒ z = \frac{{2.50}}{{0.25}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{250}}{{25}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\, = 10$$
∴ X will cost 40 paise more than Y 10 years
After 2001 i.e., 2011

Converting each of the given fractions in to decimal form, we get
$$\frac{1}{3} = 0.33$$
$$\frac{2}{5} = 0.4$$
$$\frac{3}{7} = 0.42$$
$$\frac{4}{5} = 0.8$$
$$\frac{5}{6} = 0.83$$
$$\frac{6}{7} = 0.85$$
Clearly, 0.85 > 0.83 > 0.8 > 0.42 > 0.4 > 0.33
So, $$\frac{6}{7}$$ > $$\frac{5}{6}$$ > $$\frac{4}{5}$$ > $$\frac{3}{7}$$ > $$\frac{2}{5}$$ > $$\frac{1}{3}$$

$$ = \frac{{3.157 \times 4126 \times 3.198}}{{63.972 \times 2835.121}}$$
$$ \approx \frac{{3.2 \times 4126 \times 3.2}}{{64 \times 2835}}$$
$$ = \frac{{32 \times 4126 \times 32}}{{64 \times 2835}} \times \frac{1}{{100}}$$
$$ = \frac{{66016}}{{2835}} \times \frac{1}{{100}}$$
$$ = \frac{{23.28}}{{100}}$$
$$ = 0.23$$
$$ \approx 0.2$$

477 × 124 × 86 = 5086728
Sum of decimal place = 3
∴ 47.7 × 12.4 × 8.6 = 5086.728

Given, 168 × 32 = 5376 or 5376 ÷ 168 = 32
Now,
$$ = \frac{{5.376}}{{16.8}}$$
$$ = \frac{{53.76}}{{168}}$$
$$ = \left( {\frac{{5376}}{{168}} \times \frac{1}{{100}}} \right)$$
$$ = \frac{{32}}{{100}}$$
$$ = 0.32$$