Question

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

Answer 1

Let $x$ and $y$ crystals be in $5kg$ of sugar and $1.2kg$ of sugar, respectively.

Clearly, it is a case of direct variation.

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

$\Rightarrow x=\frac{5\times 9\times {10}^5}{2}$

$\Rightarrow x=\frac{9\times 10\times {10}^6}{4}-\frac{9}{4}\times {10}^7=2.25\times {10}^7$

Hence, 5kg of sugar contains $2.25\times {10}^7$ crystals.

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

$\Rightarrow y=\frac{1.2\times 9\times {10}^6}{2}$

$\Rightarrow y=0.6\times 9\times {10}^6=5.4\times {10}^6=5.4\times {10}^6$

Hence, 1.2 kg of sugar contains $5.4\times {10}^6$ crystals.