udaymathapati wrote:
This year Henry will save a certain amount of his income, and he will spend the rest. Next year Henry will have no income, but for each dollar that he saves this year, he will have 1 + r dollars available to spend. In terms of r, what fraction of his income should Henry save this year so that next year the amount he was available to spend will be equal to half the amount that he spends this year?
A. \(\frac{1}{(r+2)}\)
B. \(\frac{1}{2r+2}\)
C. \(\frac{1}{3r+2}\)
D. \(\frac{1}{r+3}\)
E. \(\frac{1}{2r+3}\)
1. Read and understand. We are looking for the fraction \((\frac{x}{I})\)saved such that the amount available to spend \((y)\) this year would be half of the disposed income\((I)\) of yesteryear.
2. Note that the amount saved can be deducted from the amount disposed of and we can then find the amount required to be saved (x), i.e \(I-y\)
Now the best approach can be either algebraic or arithmetic, I will go for a combination of both. I will start with arithmetic, i.e plugging in numbers as this can allow for a quick backsolve if running out of time.
Now to choose numbers just a cursory look at the problem would suggest that multiples of 2 and 3 would make for straightforward (I). Why we are looking for three parts in two periods( I know this may be confusing, but it's not necessary to solve the problem), however, it does help to backsolve. Then I will choose 0 or 1 for \(r\) (easy to manipulate).
Now onto the solution:
Say\(I = 6\) and\(r=0\)
Then \(x= \frac{y}{2}= \frac{(6-x)}{2}\)
\(2x= 6-x; 3x= 6; x=2\)
Our fraction is \(\frac{2}{6}= \frac{1}{3}\)
Only option \(E\) fulfills this when r= 0. Hence, \(E\)