Number of minutes = LCM of 18 and 12
Prime Factorisation:
12 = 2 $\times$ 2 $\times$ 3

18 = 2 $\times$ 3 $\times$ 3

L.C.M of the two numbers = $2^2\times 3^2=36$

We have,
Largest 4-digit number = 9999
88) 9999 (113
      9944
   -----------
      55
$\Rightarrow$ When 9999 is divided by 88, remainder is 55.
$\therefore$ Required number = (9999 - 55) = 9944

On dividing 1056 by 23, we get,
$\begin{array}{cc}& 45\\ 23& 1056\\ & 1035\downarrow \\ & 21\end{array}$
1056 = 23 $\times$ 45 + 21
Here, the remainder is 21 and divisor is 21.
The diffrenece between Divisor and remainder = 23 - 21 = 2.
$\Rightarrow$ If we add 2 to the dividend 1056, we will get a number completely divisible by 23.
$\therefore$ The least number to be added = 2

Let the smaller number be $x.$
$\Rightarrow$ Larger number = $x+1365$
By Euclid's division lemma, the larger number can also be written as $6x+15.$
$\Rightarrow x+1365=6x+15$
$\Rightarrow 5x=1350$
$\Rightarrow x=270$
$\therefore$ The smaller number is 270 and the larger number is 270 + 1365 = 1635.

The largest number that can divide both 324 and 144 is the HCF of both the numbers.
Prime factorising the two numbers we get,
$324=2\times 2\times 3\times 3\times 3\times 3$
$144=2\times 2\times 2\times 2\times 3\times 3$
Now, taking the common factors between them will give us the HCF.
$\therefore$ HCF $=2\times 2\times 3\times 3=36$

$\sqrt{2}=\mathrm{1.414213562373095...}$, is non-recurring and non-terminating. So it is an irrational number.
The value of $\pi$ is 3.14159265358..., the digits neither terminate nor recur and hence it is an irrational number.
$\sqrt{3}$=1.732142857…, is non-recurring and non-terminating. So it is an irrational number..
4.8 = $\frac{48}{10}$ satisfies all the properties of a rational number. Therefore, it is a rational number.

Two numbers 1848 and 3058, where 3058 > 1848
3058 = 1848 × 1 + 1210
1848 = 1210 × 1 + 638 [Using Euclid's division algorithm to the given number 1848 and 3058]
1210 = 638 × 1 + 572
638 = 572 × 1 + 66
572 = 66 × 8 + 44
66 = 44 × 1 + 22
44 = 22 × 2 + 0
Therefore HCF of 1848 and 3058 is 22.
HCF (1848 and 3058) = 22
Let us find the HCF of the numbers 1331 and 22.
1331 = 22 × 60 + 11
22 = 11 × 2 + 0
HCF of 1331 and 22 is 11
HCF (22, 1331) = 11
Hence the HCF of the three given numbers 1848, 3058 and 1331 is 11.
HCF (1848, 3058, 1331) = 11

If $\frac{p}{q}$ is a rational number with terminating decimal expansion where $p$ and $q$ are coprimes, then the prime factorisation of $q$ will be in the form of $2^n{5}^m$, where n and m are non - negative integers.
Consider the fraction $\frac{17}{210}$.
Here 210 = 2 $\times$ 3 $\times$ 5 $\times$7
Hence 210 is not in the form of $2^n{5}^m$.
Consider the fraction $\frac{23}{9}$.
Here 9 = 3 $\times$ 3 = $3^2$
Hence 9 is not in the form of $2^n{5}^m$.
Consider the fraction $\frac{17}{80}$.
Here 80 = 2 $\times$ 2 $\times$ 2 $\times$ 2 $\times$ 5 = $2^4\times 5$ 
Hence 80 is in the form of $2^n{5}^m$ with n = 4 and m = 1.
Consider the fraction $\frac{35}{50}=\frac{7}{10}$.
Here 10 = 2 $\times$ 5 = $2^1\times 5^1$ 
Hence 10 is in the form of $2^n{5}^m$ with n = 1 and m = 1.
Therefore, $\frac{17}{210}$ and $\frac{23}{9}$ will have non - terminating decimal expansion.
 

Given rational number $\frac{141}{120}$

Here, $120=2^3\times 3\times 5$

$141=3\times 47$

$\Rightarrow$ $\frac{141}{120}$ = $\frac{3\times 47}{2^3\times 3\times 5}$

=  $\frac{47}{2^3\times 5}$

Multiply and divide by $5^2$.

=$\frac{47\times 5^2}{2^3\times 5\times 5^2}$
= $\frac{47\times 25}{(2\times 5{)}^3}$

= $\frac{1175}{1000}$

= 1.175

Therefore, $\frac{141}{120}$ will terminate after three decimal places.