7.Suppose 2 kg of sugar contains \(9 \times 10^6\) crystals. How many sugar crystals are there in
(i) 5 kg of sugar?

(ii) 1.2 kg of sugar?

Given: we have 2 kgs of sugar containing \(9 \times 10^6\) crystals.So let x and y crystals are in 5 kg of sugar and 1.2 kg of sugar.
(i)We can see that this is the case of direct proportion,So we have

\(\frac{9\times 10^6}{2}=\frac{x}{5}\)

\(\Rightarrow x= \frac{5\times 9 \times 10^6}{2}\)

\(\Rightarrow x= \frac{ 9 \times 10^7}{4}\)

\(\Rightarrow x=2.25\times 10^7\)

Hence, for 5kg of sugar we have \(2.25\times 10^7\) crystals.

(ii) \(\frac{9\times 10^6}{2}=\frac{y}{1.2}\)

\(\Rightarrow x= \frac{1.2\times 9 \times 10^6}{2}\)

\(\Rightarrow x=0.6\times9\times 10^6=5.4\times 10^6\)

Hence, for 1.2kg of sugar, we have \(5.4\times 10^6\) crystals.