Ex.13.1 Q7 Direct and Inverse Proportions Solution - NCERT Maths Class 8

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Question

 Suppose \(2\,\rm{kg}\) of sugar contains \(9 \times {10^6}\) crystals. How many sugar crystals are there in

(1)  \(5\,\rm{kg}\)  of sugar?

(2) \(1.2\,\rm{kg}\) of sugar?

Text Solution

 

(i) How many crystals are there in \(5\, \rm{kg}\) of crystals?

What is Known?

\(2\,\rm{kg}\) of sugar contains \(9 × 10^6 \) crystals.

What is Unknown?

(i) \(5\,\rm{kg}\) of sugar contains how many crystals?

Reasoning:

Two numbers \(x\) and \(y\) are said in direct proportion if,

\[\begin{align}\frac{x}{y} = k,\quad\;\; x = yk\,\end{align}\]

Where \(k\) is a constant.

Steps (i):

\({{\rm{Amount \,of \,sugar}}}\) \({{\rm{No}}{\rm{. of\, crystals}}}\)
\(2\) \({9 \times 10^6}\)
\(5\) ?

More the amount of sugar more will be the number of crystals. Hence this is a direct proportion.

\[\begin{align}\frac{{{x_1}}}{{{y_1}}} &= \frac{{{x_2}}}{{{y_2}}}\\\frac{2}{{9 \times {{10}^6}}}&= \frac{5}{{{y_2}}}\\2 \times {{\rm{y}}_2} &= 9 \times {10^6} \times 5\\{{\rm{y}}_2} &= \frac{{9 \times {{10}^6} \times 5}}{2}\\{y_2} &= 22.5 \times {10^6}\\{y_2} &= 2.25 \times {10^7}\end{align}\]

Hence there are \(2.25 \times {10^7}\) crystals.

(ii) How many crystals are there in \(1.2\,\rm{kg}\) of crystals?

What is Known?

\(2\,\rm{kg}\) of sugar contains \(9 × 10^6 \) crystals.

What is UnKnown?

How many crystals are there in \(1.2\,\rm{kg}\) of crystals?

Steps (ii):

\[\begin{align}\frac{{{x_1}}}{{{y_1}}} &= \frac{{{x_2}}}{{{y_2}}}\\\frac{2}{{9 \times {{10}^6}}} &= \frac{{1.2}}{{{y_2}}}\\2 \times {{\rm{y}}_2} &= 9 \times {10^6} \times 1.2\\{{\rm{y}}_{2  }}&=\frac{{9 \times {{10}^6} \times 5}}{2}\\{y_2} &= 5.4 \times {10^6}\end{align}\]

Hence there are \(5.4 \times {10^6}\) crystals.