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

$ 0.25=\frac{25}{100}=\frac{1}{4}$

$1.007 - 0.7 = 1.007 - 0.700 = 0.307$

Consider the fractions $\frac{7}{12}$ and $\frac{4}{21}$
Prime factorisation of $12 = 2 \times 2 \times 3$
Prime factorisation of $21 = 3 \times 7$
$\therefore LCM$ of $12$ and $21 = 2 \times 2 \times 3 \times 7 = 84$
Firstly, convert the fractions to equivalent fractions with denominator 84
$\Rightarrow\frac{7}{12}=\frac{7\times7}{12\times7}=\frac{49}{84}$
$\Rightarrow\frac{4}{21}=\frac{4\times4}{21\times4}=\frac{16}{84}$
Now,
$49>16$
$\therefore\ \frac{49}{84}>\frac{16}{84}$
$\frac{7}{12}>\frac{4}{21}$

$ \frac{232}{990}$ is equal to $0.234$

In order to find the product, we first multiply $2$ by $5$
We have, $2 \times 5 = 10$
Now, $0.02$ has $2$ decimal places is $ 2+ 2 = 4$
So, the product must contain $4$ places of decimals.
$\therefore 0.02 \times 0.05= 0.0010$
$= 0.001$

To find the decimal value of a fraction we just divide the numerator of the fraction by the denominator.
Here, numerator $= 48$ And denominator $= \frac{100048}{1000}=48;0.048$

 $9275$ meters in $km,$ as decimal fraction $\frac{9275}{1000}\text{km}=\text{9.275 km}$ 

 Apples used on monady $=\frac{4}{5}\ \text{of}\ 5=\frac{4}{5}\times5=4\text{kg}$
Remaining apples $= 5 - 4 = 1\ kg$
Apples used next day $=\frac{1}{3}$ of remaining apples
$=\frac{1}{3}\times1\text{kg}=\frac{1}{3}\text{kg}$
So, weight of aplles left now
= Total apples-Apples use monday-Apples used next day
$=\big(-4-\frac{1}{3}\big)$
$=\frac{15-12-1}{3}$ $\big[\text{taking LCM}\big]$
$=\frac{2}{3}\text{kg}$

 Since, $ 0.004=\frac{0.004}{1}$
Now, Multiply both numerator and denominator by $1000.$
$ \frac{0.004\times1000}{1\times1000}$$= \frac{4}{1000}$

$ 25.369$
In this, $5$ lies on ones place.
$ \therefore$ Place value of $ 5=5\times{1}=5$

To find the percent of a given number, we will multiply it with $100$.
Let the number in decimal be $x$ When converted to percent, the number becomes $ \text{x}\times 100=62.9\Rightarrow\text{x}=\frac{62.9}{100}$
$ \Rightarrow\text{x}=0.629$

$=3 \times 0.3 \times 0.03 \times 0.003 \times 30$
$=3\times\frac{3}{10}\times\frac{3}{100}\times\frac{3}{1000}\times3\times10$
$=\frac{3\times3\times3\times3\times3}{100\times1000}$
$=\frac{243}{100000}$
$=0.00243 ($Decimal point is shifted to left by $5$ places$)$

Since, $ 0.012=\frac{0.012}{1}$
Now, Multiply both numerator and denominator by $1000$
$ \frac{0.012\times1000}{1\times1000}$ $= \frac{12}{1000}$

$2\frac{2}{25}=\frac{52}{25}=\frac{52\times4}{25\times4}$
$=\frac{208}{100}=2.08$

We have
$0.012\div0.15=\frac{0.012}{0.15}=\frac{0.012\times100}{0.15\times100}$
$=\frac{1.2}{15}=0.08$

 We know that,
$1\text{g}=\frac{1}{1000}\text{kg}$
Now,
$5\text{kg }5\text{g}=5\text{kg}+5\text{g}$
$=5\text{kg}+\frac{5}{1000}\text{kg}$
$=5\text{kg}+0.005\text{kg}$
$=5.005\text{kg}$
$\therefore\ 5\text{kg }5\text{g}$
$=5.005\text{kg}$

$0.4 \times 0.4 \times 0.4 = 0.064$

We know that a fraction is in its lowest terms if its numerator and denominator have no common factor other than $1.$
Thus, if the fraction $\frac{\text{a}}{\text{b}}$ is in its lowest terms, then the $HCF$ of a and b is $1.$

$\frac{105}{112}$ is reducible fraction because $HCF$ $112$ of $105$ and $112$ is $7$

$\Big(5\frac{1}{4}-3\frac{1}{3}\Big)$
$=\frac{21}{4}-\frac{10}{3}$
$=\frac{21\times3}{4\times3}-\frac{10\times4}{3\times4} (LCM$ of $3$ and $4$ is $12)$
$=\frac{63}{12}-\frac{40}{12}$
$=\frac{63-40}{12}$
$=\frac{23}{12}$
$=1\frac{11}{12}$

Decimal means divide the numerator by the denominator.
Here, Numerator $= 4999$ and
Denominator $= 1000\Rightarrow \frac{4999}{1000}=4.999$

The first decimal digit from the decimal point is the tenth.
$4.4$ has $4$ on the ones, after decimal point on the tenths is $4$ tenths.
$4.4$ is the sum of $4$ and $\frac{4}{10}$ or $\frac{44}{10}.$

$0.5\times0.05=\frac{5}{10}\times\frac{05}{100}=\frac{25}{1000}$
$=0.025$

$ \text{4}\frac{7}{11} = \frac{4\times11 + 7}{11} = \frac{51}{11}$

$2\text{kg} 5\text{g} = (2 \times 1000)\text{g} + 5\text{g} = (2005)\text{g}$
$=\Big(\frac{2005}{1000}\Big)\text{kg}=2.005\text{kg}$

A decimal is a fractional number and is indicated by digits after a period which is called a decimal point. Tenths have one digit after the decimal point. The decimal $0.8$ is pronounced as eight tenths. Hundredths have two digits after the decimal point. The decimal $0.06$ is pronounced as six hundredths. Thousandths follow a similar pattern. They have three digits after the decimal point. The decimal $0.005$ is pronounced as five thousandths. Hence, five thousandths is $0.005.$

$2.73\div1.3=\frac{2.73}{1.3}$
$=\frac{273\times10}{13\times100}=\frac{273}{130}=\frac{21}{10}=2.1$

$=24.125$
$=24+0.125$
$=24+0.1+0.02+0.005$
$=24+\frac{1}{10}+\frac{2}{100}+\frac{5}{1000}$
Comparing this the given expression, we get
$\text{A}=1,\text{B}=2$ and $\text{C}=5$
$\therefore\ \text{A}+\text{B}+\text{C}=1+2+5$
$\text{A}+\text{B}+\text{C}=8$

To find the percent of a given number, we will multiply it with $100$ Let the number in decimal be $x$
When converted to percent,
the number becomes $ \text{x} \times100=0.002,$
which gives $ \text{x}=0.00002.$

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

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 $4\frac{2}{5}\ \text{is}\ 13\frac{1}{5}$

$6\text{cm}=\frac{6}{100}\text{m}$
$=\frac{6}{100\times1000}\text{km}$
$=\frac{6}{100000}=0.00006\text{km}$

$0.04=\frac{104}{100}=\frac{26}{25}=1\frac{1}{25}$

We know that,
$1\text{ml}=\frac{1}{1000}\text{l}$
$\therefore\ 8\text{ml}=\frac{8}{1000}\text{l}$
$=0.008\text{l}$

Let the length of the platform be $x$ metres
Length of the platform $= 300\ m$
Speed of the train$ =\frac{300}{18}$
$ =\frac{50}{3}\text{m/s}$
$ \frac{50}{3}=\frac{X+ 300}{39}$
$ \text{50}\times{39}=3\text{x} + 900$
$ \text{1950}=3\text{x} + 900$
$ 3\text{x} = 1950 - 900$
$ 3\text{x} = 1050$
$ \text{x} = \frac{1050}{3}$
$ \text{x} = 350$
So, the length of of the platform, $x = 350\ m$

We have, $\frac{1}{3}\ \text{of}\ \frac{1}{6}=\frac{1}{3}\times\frac{1}{6}$
Note of represents multiplication$(×).$

 To write it as a decimal we divide the numerator from the denominator.
$\frac{9}{1000}= 0.009$
So, $0.009$ is the decimal representation for $\frac{9}{1000}.$

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

In order to find the product, we first multiply $22$ by $2$
We have, $22 \times 2 = 44$
Now, $2.2$ has $1$ decimal place, $0.2$ has $1$ decimal place and $0.001$ has $3$ decimal places.
The sum of the decimal places is $1 + 1 + 3 = 5$
So, the product must contain 5 places of decimals.
$\therefore 2.2 \times 0.2 \times 0.001 = 0.00044$
Thus, the value of $2.2 \times 0.2 \times 0.001 is 0.00044$

Since we are given a mixed fraction so the resulting fraction will be: $ =\frac{(3\times10) +6 }{10}=\frac{36}{10}$

To write it as a decimal we divide the numerator with the denominator.
$\frac{78}{100}= 0.780.78$ is the decimal representation for $\frac{78}{100}.$

$2.73\div1.3=\frac{2.73}{1.3}$
$=\frac{273\times10}{13\times100}=\frac{273}{130}=\frac{21}{10}=2.1$

It is given that,
$14 × 4 = 56$
Now, $0.014$ has $3$ decimal places.
So, the required product must contain $3$ places of decimals.
$\therefore 0.014 × 4 = 0.056$

Given, $3\frac{3}{4}\div\frac{3}{4}$
$\because\ 3\frac{3}{4}=\frac{(3\times4)+3}{4}=\frac{12+3}{4}=\frac{15}{4}$
$\therefore3\frac{3}{4}\div\frac{3}{4}=\frac{15}{4}\times\frac{4}{3}=5$
$\bigg[\because\text{reciprocal of} \frac{3}{4}=\frac{4}{3}\bigg]$

$ 0.585=0.5 +0.08+0.005=\frac{5}{10}+\frac{8}{100}+\frac{5}{1000} $
$ 0.585=\frac{585}{1000}$

$2.08\div(0.16)=\frac{2.08}{0.16}=\frac{2.08\times100}{0.16\times100}$
$=\frac{208}{16}=13$

First, we will find the product $11 \times 1 \times 1$
i.e., $11 \times 1 \times 1 = 11 \times 1 = 11$
Sum of decimal places in the given decimals $= (1 + 1 + 2) = 4$
$1.1 \times 0.1 \times 0.01 = 0.0011 [4$ places of decimal$]$

 Here, $4$ is in tens place. Hence, its place value is $40.$

$1.008=\frac{1.008\times1000}{1\times1000}=\frac{1008}{1000}$
$=\frac{126}{125}=1\frac{1}{125}$

$0.23 \times 0.3 = 0.069$

 $=0.002\times0.5$
$=\frac{2}{1000}\times\frac{5}{10}$
$=\frac{2\times5}{1000\times10}$
$=\frac{10}{10000}$
$=\frac{1}{1000}$
$=0.01$

 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.

To convert a decimal to a fraction, write it over the appropriate power of $10$ and simplify:
$ 0.008=\frac{8}{1000}=\frac{1}{125}$

Multiplying the numerator and denominator by $10$ we get, $ 0.8=\frac{8}{10}$

$0.02 \times 30 = 0.60 = 0.6$

 Given,$ \frac{2}{3}\text{p - }\text{2}\frac{1}{2} = \text{3}\frac{1}{2}$
$\Rightarrow\frac{2}{3}\text{p = }$
$ \text{3}\frac{1}{2} + \text{2}\frac{1}{2}$
$\Rightarrow\frac{2}{3}\text{p = }\frac{7}{2} + \frac{5}{2}$
$\Rightarrow\frac{2}{3}$ $ \text{p} = \frac{12}{2}$
$\Rightarrow  \text{p} = \text{6}\times\frac{3}{2}$
$\Rightarrow\text{p} = {3} \times3 $
$\Rightarrow\text{p} = 9$

$ 0.25 × 0.8 = 0.200 = 0.2$

 Number of pieces
$=\frac{\text{Total length of ribbon}}{\text{Length of one piece}}=\frac{\big(5\frac{1}{4}\big)}{\big(\frac{3}{4}\big)}$
$=\Bigg(\frac{\frac{(5\times4)+1}{4}}{\frac{3}{4}}\Bigg)=\bigg(\frac{\frac{21}{4}}{\frac{3}{4}}\bigg)$
$=\frac{21}{4}\times\frac{4}{3}=7$
$\big[\because\text{reciprocal of}\ \frac{3}{4}=\frac{4}{3}\big]$

The correct statement will be
$\frac{2}{3},\frac{3}{5},\frac{14}{15}$
$=\frac{10,9,14}{15}$
$LCM$ of $3, 5, 15, = 15$
or $\frac{3}{5}<\frac{2}{3}<\frac{14}{15}$

$3\times\frac{2}{3}$ means $3$ times the two-third part of anything.
$\therefore$ Option $(b)$ is correct.

We have
$0.1 - x = 0.04$
$\Rightarrow x = 0.1 - 0.04$
Converting the given decimals into like decimals, we get
$0.10$ and $0.04$
Thus, required number $= (0.10 - 0.04) = 0.06$
Hence, 0.06 should be subtracted from $0.1$ to get $0.04$

 Consider the fractions $\frac{4}{9},\frac{2}{5}$ and $\frac{1}{4}$
$LCM$ of $4, 5, 7$ and $9 = 1260$
Firstly, convert the fractions into equivalent fractions with denominator $1260$
$\Rightarrow\frac{4}{9}=\frac{4\times140}{9\times140}=\frac{560}{1260}$
$\Rightarrow\frac{2}{5}=\frac{2\times252}{5\times252}=\frac{504}{1260}$
$\Rightarrow\frac{3}{7}=\frac{3\times180}{7\times180}=\frac{540}{1260}$
$\Rightarrow\frac{1}{4}=\frac{1\times315}{4\times315}=\frac{315}{1260}$
Now,
$315<504<540<560$
$\therefore\ \frac{315}{1260}<\frac{504}{1260}<\frac{540}{1260}<\frac{560}{1260}$
$\Rightarrow\frac{1}{4}<\frac{2}{5}<\frac{3}{7}<\frac{4}{9}$
Thus, the smallest fraction is $\frac{1}{4}$

 Consider the fractions $\frac{3}{4},\frac{5}{6},\frac{2}{3},\frac{1}{2},\frac{4}{5}$ and $\frac{9}{10}$
$LCM$ of $2, 3, 4, 5, 6$ and $10 = 60$
Firstly, convert the fractions into equivalent fractions with denominator $60$
$\Rightarrow\frac{3}{4}=\frac{3\times15}{4\times15}=\frac{45}{60}$
$\Rightarrow\frac{5}{6}=\frac{5\times10}{6\times10}=\frac{50}{60}$
$\Rightarrow\frac{2}{3}=\frac{2\times20}{3\times20}=\frac{40}{60}$
$\Rightarrow\frac{1}{2}=\frac{1\times30}{2\times30}=\frac{30}{60}$
$\Rightarrow\frac{4}{5}=\frac{4\times12}{5\times12}=\frac{48}{60}$
$\Rightarrow\frac{9}{10}=\frac{9\times6}{10\times6}=\frac{54}{60}$
Now,
$30<40<45<48<50<54$
$\therefore\ \frac{30}{60}<\frac{40}{60}<\frac{45}{60}<\frac{48}{60}<\frac{50}{60}<\frac{54}{60}$
$\Rightarrow\frac{1}{2}<\frac{2}{3}<\frac{3}{4}<\frac{4}{5}<\frac{5}{6}<\frac{9}{10}$
Thus, the fraction $\frac{4}{5}$ is greater than $\frac{3}{4}$ and less than $\frac{5}{6}$

The number $0.257$ can be represented as $0.2 + 0.05 + 0.007 .$
Therefore we can see that digit $7$ represents
$ 0.007=7\times\frac{1}{1000}$

Total quantity of both ingredients in one packet of biscuits
$=$ Quantity of flour $+$ Quantity of sugar
$=2\frac{1}{2}\ \text{cups}+1\frac{2}{3}\text{cups}$
$=\frac{(2\times2)+1}{2}+\frac{(1\times3)+2}{3}$
$-\frac{4+1}{2}+\frac{3+2}{3}$
$=\frac{5}{2}+\frac{5}{3}$
$=\frac{5\times3+2\times5}{6}$ $\big[\because\ \text{LCM of 2 and 3 = 6}\big]$
$=\frac{15+10}{6}$
$=\frac{25}{6}$
$\therefore$ Total quantity of both ingredients used in $10$ packets
$= 10\ ×$ Total quantity of ingredients in one packet
$=10\times\frac{25}{6}=\frac{250}{6}$
Since, $\frac{250}{6}$ lies between $40$ and $50.$

 We have,
$3 \times 3 \times 3 = 27$
The sum of the decimal places in the given decimals is $1 + 1 + 1 = 3$
So, the product must contain 3 places of decimals.
$\therefore 0.3 \times 0.3 \times 0.3 = 0.027$

$626.235$
There are two $6s$ in this number, one at Hundreds place and other at Ones place.
$ \Rightarrow$ Place values of $ 6=6\times100$ and $ 6\times1$
Difference $= 600 - 6 = 594$

 The given fractions are $\frac{2}{3},\frac{3}{5}$ and $=\frac{11}{15}$
$[LCM$ of $5, 3$ and $15 = 15]$
Now, we have:
$\frac{3}{5}\times\frac{3}{3}=\frac{9}{15},\frac{2\times5}{3\times5}$
$=\frac{10}{15}$ and $\frac{11}{15}$
Clearly,
$\frac{9}{15}<\frac{10}{15}<\frac{11}{15}$
$\therefore\frac{3}{5}<\frac{2}{3}<\frac{11}{15}$

Given, $\frac{2}{3},\frac{6}{7},\frac{13}{21}$
$LCM$ of $(3, 7, 21) = 21$
$\therefore\frac{2}{3}=\frac{2}{3}\times\frac{7}{7}=\frac{14}{21}$
$\frac{6}{7}=\frac{6}{7}\times\frac{3}{3}=\frac{18}{21}$
and $\frac{13}{21}=\frac{13}{21}$
Now, compare $\frac{14}{21},\frac{18}{21}\text{and}\frac{13}{21}$
$\text{So},\frac{13}{21}<\frac{14}{21}<\frac{18}{21}$
Hence, $\frac{13}{21}<\frac{2}{2}<\frac{96}{7}$ (ascending order)
Note with same denominators, fraction with larger numerator is greater.

$ 0.34=\frac{34}{100}$

$0.06=\frac{06}{100}=\frac{3}{50}$

The given expression can be simplified as follows:
$ \frac{(0.0036)(2.8)}{(0.04)(0.1)(0.003)}$
$ =\frac{0.0036\times2.8}{0.04\times0.1 \times 0.003}=$
$ \frac{0.01008}{0.000012}=840$
Hence, the value of $ \frac{(0.0036)(2.8)}{(0.04)(0.1)(0.003)}$ is $ 840$

Factors of $84: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84$
Factors of $98: 1, 2, 7, 14, 49, 98$
Common factors of $84$ and $98: 1, 2, 14$
$\therefore HCF$ of $84$ and $98 = 14$
Now,
$\frac{84}{98}=\frac{84\div14}{98\div14}=\frac{6}{7} ($Dividing numerator and senominator by the $HCF$ of $84$ and $98$ i.e., $14)$

$=75.57\div0.01$
$=\frac{75.57}{0.01}$
$=\frac{75.57\times100}{0.01\times100} ($Multiply numerator and denominator by $100$ to convert the divisor$)$
$=\frac{7557}{1}$
$=7557$

Given, $0.03 × 0.9$
Here, $3 × 9 = 27$
Sum of the decimal places to the right of the decimal point is $0.03$ and $0.09$ is $3$
So, $0.03 × 0.9 = 0.027$

$=0.012\div1.5$
$=\frac{0.012}{1.5}$
$=\frac{0.012\times10}{1.5\times10}($ Multiply the numberator and denominator by $10$ to convert the divison$)$
$=\frac{0.12}{15}$

$\therefore\ 0.012\div1.5=0.008$

To round to the nearest tenth, write down the number with a decimal point, and find the tenths place directly to the right of the decimal. Then, to the right of the tenths place, look at the number in the hundredths place. In $6050.287$ the number in the hundredths is $2.$
So, $6050.287$ rounded to the nearest tenths will be $6050.3$

We know that a fraction is irreducible (or is in its lowest terms) if the $HCF$ of its numerator and denominator is $1.$
Consider the fraction $\frac{91}{104}$
$HCF$ of $91$ and $104=13\neq1$
So, the fraction $\frac{91}{104}$ is reducible.
Consider the fraction $\frac{105}{112}$
$HCF$ of $105$ and $112=7\neq1$
So, the fraction $\frac{105}{112}$ is reducible.
Consider the fraction $\frac{51}{85}$
$HCF$ of $51$ and $85=17\neq1$
So, the fraction $\frac{51}{85}$ is reducible.
Now,
Consider the fraction $\frac{43}{83}$
$HCF$ of $43$ and $83 = 1$
So, the fraction $\frac{43}{83}$ is irreducible (or is in its lowest terms).

$ \text{10}\frac{2}{10} ⇒\text{10}\frac{2}{10}=\frac{102}{10}=10.2$

$2\text{km }5\text{m}=2\frac{5}{1000}\text{km}=2.005\text{km}$

$0.1 - 0.03 = 0.10 - 0.03 = 0.07$

Let the required number be $x.$
Now,
$1\frac{3}{4}\div\text{x}=2\frac{1}{2}$
$\Rightarrow\frac{7}{4}\times\frac{1}{\text{x}}=\frac{5}{2}$
$\Rightarrow\text{x}=\frac{7}{4}\times\frac{2}{5}$
$\Rightarrow\text{x}=\frac{7\times1}{2\times5}$
$\Rightarrow\text{x}=\frac{7}{10}$
Thus, the required number is $\frac{7}{10}$