vaishar3 wrote:
There is a similar question 5 questions back on previous page,same situation.
How can it be A?ratios can be anything
Posted from my mobile device
I agree with that, however, in this case, there are several ingredients that allows finding a pattern which is always true:
Let's suppose that the relation between the total income of A and B can be written as:
\(3x + 2x = I\), with \(I = A_{I} + B_{I}\)
In this case, the total income of \(A\) and \(B\) are given by:
\(A_{I} = \frac{3*I}{5}\)
\(B_{I} = \frac{2*I}{5}\)
And the total expenditures: (\(E\))
\(4y + 3y = E\), with \(E = A_{E} + B_{E}\)
that could be written as:
\(A_{E} = \frac{4*E}{7}\)
\(B_{E} = \frac{3*E}{7}\)
Now, for each person, we will calculate their savings equation:
\(A_{S} = \frac{3*I}{5}-\frac{4*E}{7}\)
\(B_{S} = \frac{2*I}{5}-\frac{3*E}{7}\)
In order to work with friendly numbers, we are going to multiply each equation by 35:
\(35*A_{S} = 21*I-20*E\)
\(35*B_{S} = 14*I-15*E\)
Now, we can substract both equations:
\(35*A_{S} - 35*B_{S} = 21*I-14*I - 20*E-15*E\)
Solving:
\(35*A_{S} - 35*B_{S} = 7*I - 5*E\)
(the following step is not necessary)
\(A_{S} - B_{S} = \frac{7*I - 5*E}{35}\)
In this case, we now that I>E (they told us), therefore, this expression \(7*I - 5*E\) is always positive. Finally, we can say that no matter what kind of numbers do you pick, the savings of A will be always greater than B.