Question

Suppose 2 kg of sugar contains 9 106 crystals. How many sugar crystals are there in (i) 5 kg of sugar? (ii) 1.2 kg of sugar?

Answer 1

Question :-

Suppose 2 kg of sugar contains 9 × ${10}^{6}$ crystals. How many sugar crystals are there in

(i) 5 kg of sugar?

(ii) 1.2 kg of sugar?

Solution :-

(1)

let x be the number of sugar crystal needed.

Directly proportional equation √

∴ $\dfrac{9×{10}^{6}}{x}$ = $\dfrac{2}{5}$

${➝\:2x\:=\:5×9×\:{10}^{2}}$

∴ $x$ = $\dfrac{5×9×{10}^{2}}{2}$

= ${ \dfrac{45}{2}}$ × ${10}^{6}$

= $2.25$ × ${10}^{7}$

hands required number of sugar crystal

= $2.25$ × ${10}^{7}$

(2)

since the two quantities are directly proportional

∴ $\dfrac{9×{10}^{6}}{y}$ = $\dfrac{2}{1.</p><p>2}$

${\hookrightarrow\:2y\:=\:1.2×9×\:{10}^{6}}$

∴ $y$ = $\dfrac{1.2×9×{10}^{6}}{2}$

= $y$ = ${5.4×{10}^{6}}$

hence the required number of crystals

= $y$ = ${5.4×{10}^{6}}$

Answer 2

(i) No. of crystals in 5 kg of sugar =

$( \frac{9106}{2}) \times5=22765 \: \: crystals$

(ii) No. of crystals in 1.2 kg of sugar =

$( \frac{9106}{2} ) \times 1.2=5463.6 \: \: crystals$