### Theory:

**Detailed explanation about inverse proportion:**

Let us consider a group of social workers gathered at a place, and they planned to dig 100 pits to plant trees. If one person can dig one pit then,

If there are \(20\) workers, they will take, $\frac{100}{20}$ \(=\) 5

**days**.If there are \(10\) workers, they will take \(=\) $\frac{100}{10}$ \(=\) 10

**days**.If there are \(5\) workers, they will take \(=\) $\frac{100}{5}$ \(=\) 20

**days**Now, in this condition, are the number of pits, and the number of workers is in direct proportion?

If your answer is NO, that is correct. Because when the number of workers is increasing, the days are decreasing accordingly at the same rate.

Now we say that these quantities are in

**.****inverse proportion**Let us denote the number of social workers as \(X\) and the number of days as \(Y\). Now observe the following table,

Number of Social Workers \(X\) | \(20\) | \(10\) | \(5\) |

Number of days \(Y\) | 5 | 10 | 20 |

From the table, we can observe that when the values of \(X\) decrease the corresponding values of \(Y\) increases in such way that the ratio of $\frac{X}{Y}$ in each case has the same value which is a constant (say \(k\)).

**Derivation:**Consider each of the value of \(X\) and the corresponding value of \(Y\). Their products are all equal say $\mathit{XY}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}100\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}k\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}(k\phantom{\rule{0.147em}{0ex}}\mathit{is}\phantom{\rule{0.147em}{0ex}}a\phantom{\rule{0.147em}{0ex}}\mathit{constant})$, and it can be expressed as $\mathit{XY}=k(k\phantom{\rule{0.147em}{0ex}}\mathit{is}\phantom{\rule{0.147em}{0ex}}a\phantom{\rule{0.147em}{0ex}}\mathit{constant})$.

If ${X}_{1}{X}_{2}$ are the values of \(X\) is corresponding to the values of ${Y}_{1}{Y}_{2}$ of \(Y\), respectively.

Therefore,

$\begin{array}{l}{X}_{1}{Y}_{1}={X}_{2}{Y}_{2}=k\phantom{\rule{0.147em}{0ex}}\mathit{contant}.\\ \mathit{That}\phantom{\rule{0.147em}{0ex}}\mathit{is}\phantom{\rule{0.147em}{0ex}}\frac{{X}_{1}}{{X}_{2}}=\frac{{Y}_{2}}{{Y}_{1}}\\ \mathit{Thus}\phantom{\rule{0.147em}{0ex}}X\phantom{\rule{0.147em}{0ex}}\mathit{and}\phantom{\rule{0.147em}{0ex}}Y\phantom{\rule{0.147em}{0ex}}\mathit{are}\phantom{\rule{0.147em}{0ex}}\mathit{in}\phantom{\rule{0.147em}{0ex}}\mathit{inverse}\phantom{\rule{0.147em}{0ex}}\mathit{proportion}.\end{array}$

From the above table, we should take ${X}_{1}$ and ${X}_{2}$ from the values of \(X\). Similarly, take ${Y}_{1}$ and ${Y}_{2}$ from the values of \(Y\).

That is,

Number of Social Workers \(X\) | ${X}_{1}$ | ${X}_{2}$ | ${X}_{3}$ | ${X}_{4}$ |

Number of days \(Y\) | ${Y}_{1}$ | ${Y}_{2}$ | ${Y}_{3}$ | ${Y}_{4}$ |

From the above table, we can learn that we need at least \(3\) variables to find out the other value.

**Do you know how to find out the values of**${Y}_{2}$

**,**${Y}_{3}$

**and**${Y}_{4}$?

**Step 1**

Let's consider that ${X}_{1}$ and ${Y}_{1}$ are in \(1\) series, ${X}_{2}$ and ${X}_{3}$ are in \(2\) series and so on.

If ${X}_{1}$, ${X}_{2}$ and ${Y}_{1}$ values are given using these values, we can find out ${Y}_{2}$.

**Step 2**

Similarly to find ${Y}_{3}$ value first, you have to make sure that you know the values of ${X}_{2}$, ${X}_{3}$ and ${Y}_{2}$.

If you don't know that you have to find out that unknown value using previous series values.

After that using the \(3\) variables ${X}_{2}$, ${X}_{3}$ and ${Y}_{2}$, we can find out the value of ${Y}_{3}$.

**Step 3**

Now we know the value of ${Y}_{3}$ then using ${X}_{3}$, ${X}_{4}$ and ${Y}_{3}$ values we can calculate the value of ${Y}_{4}$.