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

$1.02\div6=\frac{1.02}{6}=0.17$

To round $13.73$ to nearest tenth means to round the numbers so you only have one digit in the fractional part.
So, If the last digit in the fractional part of $13.73$ is less than $5,$ then we simply remove the last the digit of fractional part.
So the correct answer will be $13.7.$

The decimal which should be subtracted from $0.1$ to get $0.06$ can be obtained by subtracting $0.06$ from $0.1$
Converting given decimals into like decimals, we have $0.10$ and $0.06$
Now,
$= 0.10 - 0.06$
$= 0.04$
$\therefore$ Required decimal $= 0.1 - 0.06$
$= 0.04$

Given decimal number is $15.526$
In this, $2$ lies on one-hundredth place.
$\therefore$Place value of $ \text{2}=\frac{2}{100} =0.02$

$ \frac{1}{\text{k}} = \frac{1}{3} + \frac{1}{4}$ Multiply by $k,$
$\therefore 1=\frac{\text{k}}{3}+\frac{\text{k}}{4}\Rightarrow12=4\text{k}+3\text{k}$
$\Rightarrow7\text{k} =12\text{k}\Rightarrow\text{1}\frac{5}{7}$

 $=5\text{kg }6\text{g}=5$
$\frac{6}{1000}\text{kg}=5.006\text{kg}$

 Divide $13$ by $7$.
The quotient is $1$ and remainder is $6.$
$ ∴ \frac{13}{7}=\text{1}\frac{6}{7} $

$ 256.36,256.56,256.26,256.46$
As integral part of all the numbers is same $(256),$ we compare them by fractional part, greater the fractional part, greater is the number. In ascending order$,256.26,256.36,256.46,256.56$

$ 69=2.5^2+3.5 + 4.1={234}_5 ($​that is,$ 234$ in the base $5$ system$).$

A fraction$\frac{\text{a}}{\text{b}}$ is said to be irreducible or in its lowest terms if the $HCF$ of $a$ and $b$ is $1$
$46 = 2 \times 23 \times 1$
$63 = 3 \times 3 \times 21 \times 1$
Clearly, the $HCF$ of $46$ and $63$ is $1.$
Hence$\frac{46}{63}$ is an irreducible fraction.

We have,
$12 \times 15 = 180$
It can be seen that the sum of the decimals in the given decimals is $3 + 2 = 5$
So, the product must contain $5$ places of decimals.
$\therefore 0.012 \times 0.015$$= 0.00180$
$= 0.0018$

$0.004=\frac{0.004}{1}$​
Here, we have three numbers after decimal point.
So, we multiply by both numerator and denominator by $1000.$
$=\frac{0.004\times1000 }{1\times 1000}$
$=\frac{4}{1000}$
$\therefore$ Fraction for $0.004$ is $\frac{4}{1000}$

In one day, he reads $=1\frac{3}{4}=\frac{7}{4}\text{ ​​​​hours}$
and in $6$ days he will read $=\frac{7}{4}\times6=\frac{21}{2}\text{ hours}$
$=10\frac{1}{2}\text{ hours}.$

 Consider the fractions $\frac{3}{4},\frac{2}{3}$ and $\frac{12}{15}$
$LCM$ of $4, 3$ and $15 = 60$
Firstly, convert the fractions into equivalent fractions with denominator $60$
$\Rightarrow\frac{3}{4}=\frac{3\times15}{4\times15}=\frac{45}{60}$
$\Rightarrow\frac{2}{3}=\frac{2\times20}{3\times20}=\frac{40}{60}$
$\Rightarrow\frac{12}{15}=\frac{12\times4}{15\times4}=\frac{48}{60}$
Now,
$40<45<48$
$\therefore\ \frac{40}{60}<\frac{45}{60}<\frac{48}{60}$
$\frac{2}{3}<\frac{3}{4}<\frac{12}{15}$

 Place Value Definition: It is the value of the digit with reference to its position in the number. For example, $238$ has $2$ hundred, $3$ tens and $8$ ones. Therefore ans is $90,00,000$

$ \Rightarrow\frac{(0.96)^3-{(0.1)^3}}{(0.96)^2+(0.096) + 0.01}$
$\Rightarrow\frac{(0.96-0.1)[(0.96)^2+(0.96 \times 0.1 + (0.1)^2]}{(0.96)^2 + (0.096) + (0.01)]} $
$ \Rightarrow\frac{0.86[(0.96)^2 + (0.096 + (0.01)]}{[(0.96)^2 + (0.096) + (0.01)]}$
$ \Rightarrow{ 0.86}$

$89.1\div2.2=\frac{89.1}{2.2}$
$=\frac{891\times10}{22\times10}=\frac{81}{2}=40.5$

Consider the fractions $\frac{2}{3},\frac{4}{7},\frac{8}{11}$ and $\frac{5}{9}$
$LCM$ of $3, 7, 9$ and $11 = 693$
Firstly, convert the fractions into equivalent fractions with denominator $693$
$\Rightarrow\frac{2}{3}=\frac{2\times231}{3\times231}=\frac{462}{693}$
$\Rightarrow\frac{4}{7}=\frac{4\times99}{7\times99}=\frac{396}{693}$
$\Rightarrow\frac{8}{11}=\frac{8\times63}{11\times63}=\frac{504}{693}$
$\Rightarrow\frac{5}{9}=\frac{5\times77}{9\times77}=\frac{385}{693}$
Now,
$385<396<462<504$
$\therefore\ \frac{385}{693}<\frac{396}{693}<\frac{462}{693}<\frac{504}{693}$
$\Rightarrow\frac{5}{9}<\frac{4}{7}<\frac{2}{3}<\frac{8}{11}$
Thus, the smallest of the given fractions is $\frac{5}{9}$

$\frac{5}{8}$ is a vulgar fraction, because its denominator is other than $10, 100, 1000,$ etc.

Given, $7\times6\frac{3}{4}$
$\because\ 6\frac{3}{4}=\frac{(6\times4+3)}{4}=\frac{24+3}{4}=\frac{27}{4}$
$\therefore7\times6\frac{3}{4}=7\times\frac{27}{4}=\frac{189}{4}=47\frac{1}{4}$
Hence, the product of $7$ and $6\frac{3}{4}\ \text{is}\ 47\frac{1}{4}$

Let $x = 1.263263263 [$we multiply it by $1000]$
Here $3$ digits are repeated
$1000x = 1263.263263.....$
$x = 1.263263...$
$999x = 1262$
$\Rightarrow\text{x}=\frac{1262}{999}$
The recurring decimal $1.263$ in a fraction form is equal to $\frac{1262}{999}$

$2\frac{1}{25}=\frac{51}{25}=2.04$

$ 95.75 \times0.2554=24.45455$
Sum of decimal places $= 7$
Since the last digit to the extreme right will be zero $( $since $5\times 4=20),$
 so there will be $5$ significant digits to the right of the decimal point.

Consider the fractions $\frac{2}{3},\frac{5}{7},\frac{3}{4}$ and $\frac{5}{6}$
$LCM$ of $3, 4, 5, 6$ and $7 = 420$
Firstly, convert the fractions into equivalent fractions with denominator $420$
$\Rightarrow\frac{2}{3}=\frac{2\times140}{3\times140}=\frac{280}{420}$
$\Rightarrow\frac{5}{7}=\frac{5\times60}{7\times60}=\frac{300}{420}$
$\Rightarrow\frac{3}{4}=\frac{3\times105}{4\times105 }=\frac{315}{420}$
$\Rightarrow\frac{4}{5}=\frac{4\times84}{5\times84}=\frac{336}{420}$
$\Rightarrow\frac{5}{6}=\frac{5\times70}{6\times70}=\frac{350}{420}$
Now,
$280<300<315<336<350$
$\therefore\ \frac{280}{420}<\frac{300}{420}<\frac{315}{420}<\frac{336}{420}<\frac{350}{420}$
$\Rightarrow\frac{2}{3}<\frac{5}{7}<\frac{3}{4}<\frac{4}{5}<\frac{5}{6}$
Thus, none of the fractions $\frac{3}{4},\frac{4}{5},\frac{5}{6}$ lies between the fractions $\frac{2}{3}$ and $\frac{5}{7}$

Denominator in $(a)$ and $(b)$ is $10$ these are decimal fractions But denominator of $(c)$ is $3\frac{10}{3}$ is a vulgar fraction.

Converting fraction to decimal $=\frac{3}{10}=0.3$

$ 1.1 \times .1 \times .01 = 0.0011$

Let $x = 1.263263263 [$we multiply it by $1000]$
Here $3$ digits are repeated
$1000x = 1263.263263.....$
$x = 1.263263....$
$999x = 1262$
$\Rightarrow\text{x}=\frac{1262}{999}$
The recurring decimal $1.263$ in a fraction form is equal to $\frac{1262}{999}.$

$21.236$
There are two $2s$ in this number, one at Tens place and other at one-tenth place.
$\Rightarrow$ Place values of $ 2=2\times{10}$ and $ 2\times{0.1}$
Sum $ =20+0.2=20.2$

$ 0.43=\frac{43}{100}​ ($As it is expressed a fraction.$)$

$3.5 - 3.07 = 3.50 - 3.07 = 0.43$

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

Given, $3\times4\frac{2}{5}$
$\because4\frac{2}{5}=\frac{(4\times5)+2}{5}=\frac{22}{5}$
$\therefore3\times4\frac{2}{5}=3\times\frac{22}{5}=\frac{66}{5}=13\frac{1}{5}$
Hence, the product of $3$ and $4\frac{2}{5}\ \text{is}\ 13\frac{1}{5}$