I took an equation Say $p \times q = k $
When $p$ and $q$ changes by some percentages $a%, b%$ we can calculate the percentage change in $k$ as $\left( {a + b + \dfrac{{ab}}{{100}}} \right)\% $
But I am interested in finding $k$ when $p$, $q$ changes by some absolute value.
For example, when $p$ changed $+5$, then $q$ got changed by $ - \dfrac{1}{3}$ and when $p$ changed $-5$, then $q$ got changed by $\dfrac{1}{2}$ where $p \times q = k $
This is easy to solve by using equations.
$$p \times q = k$$ $$\left( {p + 5} \right) \times \left( {q - \dfrac{1}{3}} \right) = k$$ $$\left( {p - 5} \right) \times \left( {q + \dfrac{1}{2}} \right) = k$$
I got $p = 25$, $q = 2$
I observed that
$$\frac{{p + 5}}{{p - 5}} = \frac{{\left( {\frac{{\partial {p_1}}}{{\partial {q_1}}}} \right)}}{{\left( {\frac{{\partial {p_2}}}{{\partial {q_2}}}} \right)}}$$
Here $\partial {p_1},\partial {p_2},\partial {q_1},\partial {q_2}$ are absolute changes in $p, q$.
$$\frac{{p + 5}}{{p - 5}} = \frac{{\left( {\frac{5}{{1/3}}} \right)}}{{\left( {\frac{5}{{1/2}}} \right)}} = \frac{3}{2}$$
I would like to know what is the theory behind these ratios. I used this concept for wide range of problems but don't know why the above expression holds good. Can any one throw some light on this concept?
You have set up a system of simultaneous equations that can be solved by high-school algebra.
Let's write the equations with letters $a, b, c, d$ instead of the numbers $5$, $\frac13$, and so forth. That way we can hope to come up with formulas that will explain how these systems of equations work in general. The equations can then be written \begin{gather} pq = k \\ (p+a)(q+b) = k \\ (p+c)(q+d) = k \\ \end{gather} Here the letters $a, b, c, d$ are assumed to be constants that will be replaced by known numeric values when you solve a particular problem. We'll assume that $a\neq 0$ and $c \neq 0$.
The equations can be rewritten as follows by expanding the products of the terms in parentheses: \begin{gather} pq = k \\ pq + bp + aq + ab = k \\ pq + dp + cq + cd = k \\ \end{gather} Since $pq = k$, then $k - pq = 0$, so if we subtract $pq$ from each side of the last two equations we get \begin{gather} bp + aq + ab = k - pq = 0\\ dp + cq + cd = k - pq = 0\\ \end{gather} So we're down to just two equations in two unknowns. We can multiply the first equation by $\frac1a$ and the second by $\frac1/c$ to get \begin{gather} \frac ba p + q + b = 0\\ \frac dc p + q + d = 0\\ \end{gather} Subtract the second equation from the first, and we have $$ \frac ba p - \frac dc p + b - d = 0, $$ therefore $$ \left(\frac ba - \frac dc\right) p = d - b, $$ so $$ p = \frac{d - b }{\frac ba - \frac dc} = \frac{d - b}{\left(\frac {bc - ad}{ac}\right)} = \frac{acd - abc}{bc - ad}. $$ In your example, $a = 5$, $b = -\frac13$, $c = -5$, and $d = \frac12$, so \begin{align} p &= \frac{5(-5)\frac12 - 5\left(-\frac13\right)(-5)} {\left(-\frac13\right)(-5) - 5\left(\frac12\right)} \\ &= \frac{-5(5)\frac12 - 5\left(\frac13\right)(5)} {\left(\frac13\right)(5) - 5\left(\frac12\right)} && \text{resolving the negative signs}\\ &= \frac{-5\left(\frac12\right) - 5\left(\frac13\right)}{\frac13 - \frac12} && \text{canceling factors of $5$}\\ &= \frac{\left(-\frac{25}6\right)}{\left(-\frac16\right)}\\ &= 25. \end{align}
Your formula with $\frac{p+5}{p-5}$ on the left hand side is more generally \begin{align} \frac{p+a}{p+c} &= \frac{\frac{acd - abc}{bc - ad} + a}{\frac{acd - abc}{bc - ad} + c} \\ &= \frac{\left(\frac{acd - abc + a(bc - ad)}{bc - ad}\right)} {\left(\frac{acd - abc + c(bc - ad)}{bc - ad}\right)} \\ &= \frac{acd - a^2d}{acd - abc + bc^2 - acd} \\ &= \frac{acd - a^2d}{bc^2 - abc}\\ &= \frac{ad(c - a)}{bc(c-a)}\\ &= \frac{ad}{bc} \\ &= \frac{\left(\frac ab\right)}{\left(\frac cd\right)}. \end{align}
Your formula is almost the same: $$ \frac{p+a}{p+c} = \frac{\left(\frac{|a|}{|b|}\right)}{\left(\frac{|c|}{|d|}\right)} $$ where $|b|$ is the absolute value of $b$, so we take $\frac13$ instead of $-\frac13$. This works out because if $p$, $q$, and $k$ are positive then $a$ and $b$ have opposite signs, and $c$ and $d$ have opposite signs, so the signs end up canceling out.