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

## 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.