MATHS M.C.Q. [1 Marks Each]

$2\frac{17}{100}=2.17$

Requried number $= 301.01 - 0.101 - 198.01$
$= 301.01 - 198.111 = 102.899$

$417.25 - 41.72 - 4.53$
$= 371.$

Since $1m = 100\ cm$
$\therefore150\text{cm}=\frac{150}{100}=1.5\text{cm}$

$\because 5.01 - 3.6$
$= 5.01 - 3.60 = 1.41$

$\underline{\ \ \ \ 0.350\\-0.035}\\\underline{=0.315}$

$\frac{6}{1000}$
$= 6$ thousandths $= 0.006$

$2kg 56g = 2.056kg$
($\because$ $1000g = 1kg)$

$5.5, 6.6, 7.7, 8.8$ are like decimals, since the number of decimals is the same.

$2\frac{1}{25}$
$=2+\frac{1}{25}\times\frac{1}{4}$
$=2+\frac{4}{100}$
$=2.04$

$1.5625 + 0.75 - 0.3125$
$= 2.3125 - 0.3125 = 2$

$31.08=31+0.08$
$=31+\frac{8}{100}$
$=31+\frac{\frac{8}{4}}{\frac{100}{4}}$
$=31+\frac{2}{25}$
$=31\frac{2}{25}$

$\frac{3}{25}=\frac{3\times4}{25\times4}=\frac{12}{100}=0.12$

$\frac{6}{25}$
$\frac{6\times4}{25\times4}$
$\frac{24}{100}$
$=0.24$

 $24.8=24\frac{8}{10}$
$24\frac{4}{5}$

$1$ paisa $= 0.01 Rs$
$\Rightarrow 50$ paise $= 0.50$
$Rs. 47$ and $50$ paise can hence be written as $Rs. 47.50$

$1$ meter $= 100\ cm$
So, $88\text{cm}=\frac{88}{100}\text{ metar}$
$= 0.88 \text{ meter}$

$4+\frac{3}{100}=\frac{400+3}{100}=4.03$

The smallest possible decimal fraction upto three decimal places $=\frac{1}{1000}=.001$

Given $\frac{3}{13}$ If we divide $3$ by $13$ we get $0.230769$ which is repeating and non-terminating.

Weight of rice $= 4\ kg \ 500g = 4.500\ kg$
Weight of sugar $= 5\ kg \ 650g = 5.650\ kg$
Weight of atta $= 12\ kg \ 600g = 12.600\ kg$
$\therefore$ Total weight of all items $= (4.500 + 5.650 + 12.600)\ kg$
$= 22.750\ kg.$

$\because 2 - 0.7$
$= 2.0 - 0.7 = 1.3$

The given numbers are $0.4,15.46$ and $0.002$
$\Rightarrow 15.46 + 0.4 + 0.002 = 15.46 + (0.4 + 0.002)$
$= 15.46 + 0.402$
$= 15.862$
$\therefore$ Sum of $0.4,15.46$ and $0.002$ is $15.862$

$2+\frac{3}{10}+\frac{4}{100}$
$=2+\frac{30}{100}+\frac{4}{100}$
$=2.34$

$\frac{7}{10}=0.7$