• # question_answer 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?

Suppose the amount of sugar is $x\,\,kg$ and the number of crystals is $y$. As the amount of sugar increases, the number of crystals also increases in the same ratio. It is a case of direct proportion. We make use of the relation of the type $\frac{{{x}_{1}}}{{{y}_{1}}}=\frac{{{x}_{2}}}{{{y}_{2}}}$. (i) Here, ${{x}_{1}}=2$ ${{y}_{1}}=9\times {{10}^{6}}$ ${{x}_{2}}=5$ Therefore, $\frac{{{x}_{1}}}{{{y}_{1}}}=\frac{{{x}_{2}}}{{{y}_{2}}}$ gives $\frac{2}{9\times {{10}^{6}}}\,=\frac{5}{{{y}_{2}}}$ $\Rightarrow$               $2{{y}_{2}}=5\times 9\times {{10}^{6}}$ $\Rightarrow$               ${{y}_{2}}=\frac{5\times 9\times {{10}^{6}}}{2}$ $\Rightarrow$               ${{y}_{2}}=22.5\,\times {{10}^{6}}$ $\Rightarrow$               ${{y}_{2}}=2.25\,\times {{10}^{7}}$ Hence, there are $225\times {{10}^{5}}$ crystals. (ii) Here, ${{x}_{1}}=2$ ${{y}_{1}}=9\times {{10}^{6}}$ ${{x}_{3}}=1.2$                 Therefore, $\frac{{{x}_{1}}}{{{y}_{1}}}=\frac{{{x}_{3}}}{{{y}_{3}}}$ gives $\frac{2}{9\times {{10}^{6}}}=\frac{1.2}{{{y}_{3}}}$ $\Rightarrow$               $2{{y}_{3}}=1.2\times 9\times {{10}^{6}}$ $\Rightarrow$               $2{{y}_{3}}=10.8\times {{10}^{6}}$ $\Rightarrow$               ${{y}_{3}}=\frac{10.8\times {{10}^{6}}}{2}$ $\Rightarrow$               ${{y}_{3}}=5.4\times {{10}^{6}}$ Hence, there are $5.4\times {{10}^{6}}$ crystals.