Problem related to ratio amd proportion

Question

If expenditure cannot exceed the income for either of them find the condition under which we can ascertain who saves more?

Eg: Incomes of A and B are in the ratio 4:5 and expenditures are in the ratio 3:4. If incomes are 400 and 500 and expenditures are 30 and 40 respectively, B saves more. In the case incomes are 400 and 500 and expenditure are 330 and 440 respectively, A saves more.

However, in case of A and B are in the ratio 4:5 and expenditures in the ratio 5:6, whatever be the absolute value of their incomes and expenditures, B will always be saving more. What generalization we can draw from here.?? This question I found in an MBA. book.

Answer 1

Let me start with some notations:

• $I_A$ and $I_B$ are the respective incomes of $A$ and $B$
• $E_A$ and $E_B$ are the respective expenditures of $A$ and $B$
• $r_I$ is the ratio between the incomes of $A$ and $B$, i.e. $I_A = r_I I_B$
• $r_E$ is the ratio between the expenditures of $A$ and $B$, i.e. $E_A = r_E E_B$

Using the notations above $A$ saves $S_A = I_A - E_A$ and $B$ saves $S_B = I_B - E_B$.

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Now that we have some proper notations we can easily answer several questions. For example we can wonder what are the conditions such that $A$ saves more than $B$ ? It translates as: \begin{align} && S_A >{}& S_B \\ \tag{Definitions of $S_A$ and $S_B$.} \\ &\Leftrightarrow& I_A - E_A >{}& I_B - E_B \\ \tag{Definitions of $I_A$ and $E_A$.} \\ &\Leftrightarrow&r_I I_B - r_E E_B >{}& I_B - E_B \\ &\Leftrightarrow&(1-r_E) E_B >{}& (1-r_I)I_B \text{.} \end{align} We have three cases:

• If $1-r_E > 0$ then $A$ saves more than $B$ if $E_B > \frac{(1-r_I)}{(1-r_E)}I_B$.
• If $1-r_E = 0$ then $A$ saves more than $B$ if $0 > (1-r_I)I_B$.
• If $1-r_E < 0$ then $A$ saves more than $B$ if $E_B < \frac{(1-r_I)}{(1-r_E)}I_B$.

Using the ratios given in your first example

We have $r_I = \frac{4}{5}$ and $r_E = \frac{3}{4}$. Hence we are in the case where $1-r_E > 0$ so we compute $\frac{(1-r_I)}{(1-r_E)} = \frac{(1-\frac{4}{5})}{(1-\frac{3}{4})} = \frac{4}{5}$. We can then deduce that $A$ will save more than $B$ if and only if $E_B > \frac{4}{5}I_B$. Also, note that if $E_B = \frac{4}{5}I_B$ then $A$ will save as much as $B$ and if $E_B < \frac{4}{5}I_B$ than $A$ will save less than $B$.

Using the ratios given in your second example

We have $r_I = \frac{4}{5}$ and $r_E = \frac{5}{6}$. Hence we are in the case where $1-r_E > 0$ so we compute $\frac{(1-r_I)}{(1-r_E)} = \frac{(1-\frac{4}{5})}{(1-\frac{5}{6})} = \frac{6}{5}$. We can then deduce that $A$ will save more than $B$ if and only if $E_B > \frac{6}{5}I_B$ which is impossible since, by assumption, $E_B < I_B$. Hence, in this case, $B$ will always save more than $A$.