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**Hint:**(i)Here first we will assume the number of crystals in 5kg of sugar to be y and then use the direct proportion to find the value of y and get the desired value of a number of crystals.

(ii) Here first we will assume the number of crystals in 1.2kg of sugar to be $t$ and then use the method of direct proportion with respect to the given quantity to find the value of $t$ and get the desired value of the number of crystals.

A direct proportion is defined as a proportion of two variable quantities when the ratio of the two quantities is constant.

**Complete step by step solution:**

(i) It is given that :-

\[{\text{2kg of sugar}} = 9 \times {10^6}{\text{ crystals}}\]

Let 5kg of sugar contains y crystals.

Thus,

Quantity of sugar (in kg) | 2 | 5 |

Number of crystals | \[9 \times {10^6}\] | y |

Now as we know that as the quantity of sugar increases then the number of crystals also increases.

Therefore,

Quantity of sugar and number of crystals are in direct proportion

Hence,

\[\dfrac{2}{{9 \times {{10}^6}}} = \dfrac{5}{y}\]

Now we will solve for the value of y to get the number of crystals in 5kg of sugar.

Now we will solve for the value of y to get the number of crystals in 5kg of sugar.

\[ 2y = 5 \times 9 \times {10^6} \]

On simplification,

$\Rightarrow 2y = 45 \times {10^6}$

Now dividing the equation by 2 to get the value of y:

\[ \Rightarrow y = \dfrac{{45 \times {{10}^6}}}{2}\]

On simplfication of the above values,

$\Rightarrow y = 22.5 \times {10^6} $

Now on shifting the decimal point to the left we get:-

$ y = 2.25 \times 10 \times {10^6} $

\[\Rightarrow y = 2.25 \times {10^7} \]

**$\therefore$ There are \[2.25 \times {10^7}\] crystals in 5kg of sugar.**

**Note:**

Another approach for solving this question using the unitary method.

Here first we will find the number of crystals in 1 kg of sugar and then simplify multiply it by 5 and 1.2 to get the number of crystals in 5kg and 1.2kg of sugar respectively.

(i) Now since it is given that:-

\[{\text{2kg of sugar}} = 9 \times {10^6}{\text{ crystals}}\]

Then, 1 kg of sugar will contain:-

\[{\text{1kg of sugar}} = \dfrac{{9 \times {{10}^6}}}{2}{\text{ crystals}}\]

Now since we have to find the number of crystals in 5kg

Therefore we will multiply 5 in number of crystals of 1kg sugar

Hence on multiplying we get:-

\[

{\text{5kg of sugar}} = \dfrac{{9 \times {{10}^6}}}{2} \times {\text{5}} \\

{\text{ 5kg of sugar}} = \dfrac{{45 \times {{10}^6}}}{2} \\

{\text{5kg of sugar}} = 22.5 \times {10^6} \\

{\text{5kg of sugar}} = 2.25 \times {10^7}{\text{crystals}} \\

\]

(ii) Similarly,

it is give that :-

\[{\text{2kg of sugar}} = 9 \times {10^6}{\text{ crystals}}\]

Then, 1 kg of sugar will contain:-

\[{\text{1kg of sugar}} = \dfrac{{9 \times {{10}^6}}}{2}{\text{ crystals}}\]

Now since we have to find the number of crystals in 1.2kg

Therefore we will multiply 1.2 in number of crystals of 1kg sugar

Hence on multiplying we get:-

\[

{\text{1}}{\text{.2kg of sugar}} = \dfrac{{9 \times {{10}^6}}}{2} \times 1.2 \\

{\text{1}}{\text{.2kg of sugar}} = \dfrac{{10.8 \times {{10}^6}}}{2} \\

{\text{1}}{\text{.2kg of sugar}} = 5.4 \times {10^6}{\text{crystals}} \\

\]

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