Question

Suppose 2 kg of sugar contains 9x 10° crystals. How many sugar crystals are there in () 5 kg of sugar? (ii) 1.2 kg of sugar?​

Answer 1

Answer:

Crystals contain in 2 kg of sugar = 9×10⁶ .

(i) Given, 2Kg of sugar contain 9 × 10⁶ crystals.

Let 5Kg of sugar contains y crystals.

As Quantity of sugar increases, No.of crystals also increases.

So, Quantity of sugar and No.of crystals are in Direct pro ortion.

$\sf \: \begin{gathered} \frac{2}{9 \times {10}^{6} } = \: \frac{5}{y} \\ y \times 2 \: = \: 5 \times 9 \times {10}^{6} \: \\ y \times 2 \: = \: 45 \times {10}^{6} \end{gathered}$

$\sf \: \begin{gathered}y = \frac{45}{2} \times {10}^{6} \\ y = 2.25 \times {10}^{4} \times {10}^{6} \\ y = 2.25 \times {10}^{1 - 6} \\ y = 2.25 \times {10}^{7} \\ \end{gathered}$

Hence, 5Kg of sugar contains 2.25 × 10⁷ Crystals.

(ii) Let 1.2 of sugar contains x crystals.

We know, Quantity of sugar and number of crystals are in direct proportion.

$\sf \: \begin{gathered} \frac{2}{9 \times {10}^{6} } = \frac{1.2}{x} \\ x \times 2 = 1.2 \times 9 \times {10}^{6} \\ x \times 2 = \frac{12}{10} \times 9 \times {10}^{6} \\ x = \frac{1}{2} \times \frac{12}{10} \times 9 \times {10}^{6} \\ x = \frac{6}{10} \times 9 \times {10}^{6} \\ x = \frac{54}{10} \times {10}^{6} \\ x = 5.4 \times {10}^{6} \end{gathered}$

Hence, 1.2Kg of sugar contains 5.4 × 10⁶ crystals.

Answer 2

Answer:

Suppose 2 kg of sugar contains 9x 10° crystals. How many sugar crystals are there in () 5 kg of sugar? (ii) 1.2 kg of sugar?