Given expression = (11.98)2 + (0.02)2 + 11.98 x $x$.
For the given expression to be a perfect square, we must have
11.98 x $x$ = 2 x 11.98 x 0.02 or $x$ = 0.04
Given expression =$ \dfrac{(0.3333)}{(0.2222)} $ x $ \dfrac{(0.1667)(0.8333)}{(0.6667)(0.1250)} $
=$\dfrac{\dfrac{1}{6}\times\dfrac{5}{6}}{\dfrac{2}{3}\times\dfrac{125}{1000}}$
=$ \left(\dfrac{3}{2} \times\dfrac{1}{6} \times\dfrac{5}{6} \times\dfrac{3}{2} \times 8\right) $
=$ \dfrac{5}{2} $
= 2.50
Let 3889 + 12.952 - $ x $ = 3854.002.
Then $ x $ = [3889 + 12.952] - 3854.002
= 3901.952 - 3854.002
= 47.95.
Given Expression =$\dfrac{(a^2 - b^2)}{(a+b)(a-b)}$=$\dfrac{(a^2 - b^2)}{(a^2 - b^2)}$=1
Suppose commodity X will cost 40 paise more than Y after $ z $ years.
Then, [4.20 + 0.40$z$]- [6.30 + 0.15$ z $] = 0.40
$\Rightarrow$ 0.25$ z $ = 0.40 + 2.10
$\Rightarrow z $ =$ \dfrac{2.50}{0.25} $=$ \dfrac{250}{25} $= 10.
$\therefore$ X will cost 40 paise more than Y 10 years after 2001 i.e., 2011.
$ \dfrac{3}{4} $= 0.75, $ \dfrac{5}{6} $= 0.833, $ \dfrac{1}{2} $= 0.5, $ \dfrac{2}{3} $= 0.66, $ \dfrac{4}{5} $= 0.8, $ \dfrac{9}{10} $= 0.9.
Clearly, 0.8 lies between 0.75 and 0.833.
$\therefore \dfrac{4}{5} $lies between$ \dfrac{3}{4} $and$ \dfrac{5}{6} $.
$ \dfrac{489.1375 \times 0.0483 \times 1.956}{0.0873 \times 92.581 \times 99.749} \approx \dfrac{489 \times 0.05 \times 2}{0.09 \times 93 \times 100} $
=$ \dfrac{489}{9 \times 93 \times 10} $
=$ \dfrac{163}{279} $x$ \dfrac{1}{10} $
=$ \dfrac{0.58}{10} $
= 0.058 $\approx$ 0.06.
1÷0.2 =$\dfrac{1}{0.2} $=$ \dfrac{10}{2} $= 5;
0.2 = 0.222...;
(0.2)2 = 0.04.
0.04 < 0.2 < 0.22....<5.
Since 0.04 is the least, so (0.2)2is the least.
Given Expression =$ \dfrac{8 - 2.8}{1.3} $=$ \dfrac{5.2}{1.3} $=$ \dfrac{52}{13} $= 4.
Sum of decimal places = 7.
Since the last digit to the extreme right will be zero [since 5 x 4 = 20], so there will be 6 significant digits to the right of the decimal point.