$ 10 $ crores $ = 10,00,00,000 $

$ 1 $ billion $ = 1,000,000,000 $

So, $ 1 $ billion  $ = 100 \times 10,00,00,000 = 100 $ crore

Since,  $1,00,002$ is six digit number and $10,00,002$ is seven digit number.
$\therefore 1,00,002 < 10,00,002$
Option $B$ is correct.

In a subtraction minuend is a greater number and subtrahend is smaller number.
$\therefore$ greater number in the subtraction is called minuend.
Option A is correct.

Place value and face value are always equal for $ 0 $

In international numbering system
The 1st period consists of - ones, tens and hundred.
The 2nd period consists of - thousand, $10$ thousand and $100 $ thousands.
The 3rd period consists of - million, $10$ million and $100$ million.
$\therefore $  Numeral for sixty million and sixty six $=60,000,066$
Option B is correct.

As $8,08,08,079+1=8,08,08,080$

$\Rightarrow$  $Addend+Addend=Sum$

46)57804(1256
     46
     118
      92
      260
      230
        304
        276
           28

$\Rightarrow$  $3535\times 78$

Given number is $1,78,762$ 
First we see $10's$ place place value which is $62$ and it is more than $50$.
Now we will round up means to round off $762$ it will become $800$
So the correct answer is $1,78,800$.

Since the digit at hundreds place of $ 38,400 $ is $ 4 < 5 $,
Rounding of $ 38,400 $ to nearest thousand we get $ 38,000 $

Minuend is the first number in the subtraction and subtrahend is the number that is to be subtracted.
$minuend - subtrahend = difference$
$\therefore $  Option B is correct.

The given no is $8974$ for rounding for nearest 1s will see the no at
ones place it is 4 . 

It is clear that  $888+88+8+8+8=1000$
 

(B) $10:crores=100:millions$.

(B) Ten lakhs & lakhs places are in the lakhs period.

$1 crores=100000000$

(D) $27012=20000+7000+0+10+2$

$\Rightarrow$  The given number is $78.059$

$8t^3$ $- 27 = (2t)$$^3$ $- (3)$$^3$
$= (2t - 3)$$[(2t)$$^2$ $+ 2t \times 3 + 3$$^2$]
$= (2t - 3)(4t$^2$ + 6t + 9)$
$\because$ $(2t - 3)(4t$$^2$ $+ 6t + 9)$
$= (2t - 3)(a + 6t + 9)$
$\Rightarrow$ $4t$$^2$ $+ 6t + 9 = (a + 6t + 9)$
$\therefore$ $a = 4t$$^2$

There are three special numbers in the Indian numbering system – lakh, crore, and arab. A lakh is 1,00,000, 1 , 00 , 000 , a crore is equal to a hundred lakhs and is expressed as 1,00,00,000. An arab is 100 crores and is expressed as 1,00,00,00,000.

Consider the sum of the reciprocals of ,

$\dfrac{x+3}{{{x}^{2}}+1}$ and $\dfrac{{{x}^{2}}-9}{{{x}^{2}}+3}$

  $ \Rightarrow \dfrac{{{x}^{2}}+1}{x+3}+\dfrac{{{x}^{2}}+3}{{{x}^{2}}-9}=\dfrac{{{x}^{2}}+1}{x+3}+\dfrac{{{x}^{2}}+3}{\left( x-3 \right)\left( x+3 \right)} $

 $ \Rightarrow \dfrac{\left( {{x}^{2}}+1 \right)\left( x-3 \right)+{{x}^{2}}+3}{\left( x-3 \right)\left( x+3 \right)} $

 $ \Rightarrow \dfrac{{{x}^{3}}-3{{x}^{2}}+x-3+{{x}^{2}}+3}{\left( x-3 \right)\left( x+3 \right)} $

 $ \Rightarrow \dfrac{{{x}^{3}}-2{{x}^{2}}+x}{\left( x-3 \right)\left( x+3 \right)} $

 $ \Rightarrow \dfrac{{{x}^{3}}-2{{x}^{2}}+x}{\left( x^2-9 \right)} $

Expanded form of 27012 is

Expanded form of 920,831

$\Rightarrow$  The number which is multiple of $2$ is called an even number.

$999 = 100 \times 9 + 9\times10+9$

$99 = 10 \times 9 + 9$

$72 = 10 \times 7 + 2$

$11 = 10 \times 1 + 1$

The general form of any three digit number is, $abc = a \times 100 + b \times 10 + c$
So, $753 = 7 \times 100 + 5 \times 10 + 3$

The usual form: $6 \times 10 +9 = 60+9 = 69$

The general form of any two digit number is, $ab = a \times 10 + b$
So, $9 \times 10 + 0$ is in general form.

The general form of any three digit number is, $abc = a \times 100 + b \times 10 + c$
Here, $2 \times 10 + 3 \times 10 + 1$ and $3 \times 100 + 2 \times 100 + 2$ are not in generalised form.

The general form of any three digit number is, $abc = a \times 100 + b \times 10 + c$
So, $1 \times 100 + 2 \times 10 + 3$ and $4  \times 100 + 3 \times 10 + 8$ are in general form.

The general form of any three digit number is, $abc = a \times 100 + b \times 10 + c$
Here, $ 3 \times  100 + 3 \times  10 + 3 = 333 $
Hence, the solution is $333.$

The general form of any three digit number is, $abc = a \times 100 + b \times 10 + c$
So, $6 \times 100 + 4 \times 10 + 0$ is in general form.

The general form of any three digits numbers is, $abc = a \times 100 + b \times 10 + c$
So, $875 = 8 \times 100 + 7 \times 10 + 5$

The general form of any three numbers will be, $abc = a \times 100 + b \times 10 + c$
Here, $4 \times 100 + 5 \times 10 + 0 = 450 $
So, option C is correct.

The general form of any three digit number is, $abc = a \times 100 + b \times 10 + c$
Here, $3 \times 10 + 2 + 2 \times 10$ and $2 \times 100 + 2 - 0 \times 10 $ are not in generalised form.

The general form of any three digit number is, $abc = a \times 100 + b \times 10 + c$
Here, $300 = 3 \times 100 + 0 \times 10 + 0$

Given that $P=EI$ and $E=IR$
which implies $P=(IR)I = {I}^{2}R$
Since $I=\dfrac {E}{R}$, we get $P={E}^{2}/R$
So, the correct option is $C$.

$302$ can genarally expressed as
$302$= $300 \times 100 + 2 \times 1$

Given $T = 2\pi \sqrt{\dfrac{L}{g}}$
Now square it on both sides

$1$ million = $10$ lakhs $= 10,00,000$

Original number $= 8727$
After interchanging tens and thousands of place digits, we get $2787$.
So, new member is smaller than original number.

$2 + ( 8 × 0.1 ) + ( 6 × 0.01 ) + ( 4 × 0.001 )$

$= 2 + ( 0.8 ) + ( 0.06 ) + ( 0.004 )$

$= 2.864$

$\Rightarrow$  Let two positive integers are $x=17$ and $y=71$.

$\Rightarrow$  The given number is $67$.

The expanded form of $78.059$ is

General form of a number is denoted by expansion of itself written as sum of multiplication of the digit by its place value.

Let the number in one's place be $ u$ and the number in ten's place be $10t$.

$\because $ The number is greater than the number obtained in reversing the digits and the ten's digit is greater than the unit's digit
Let ten's and unit's digit be $2x$ and $x$ respectively
Then, $(10\times 2x+x)-(10x+2x)=36$
$9x=36$
$x=4$
Required difference$=(2x+x)-(2x-x)=2x=8$

$202\times 315=63,630$

$\therefore$  In word $699$ can be written as "Six hundred and ninety nine."