Experts:

In this type of problems (long word problems with variables in the choices) I face 2 challenges:
1) Speed: It usually takes me more than 60 secs just to read and understand what is going in it. Is the reading speed lower than the normal for a guy who aims Q50/51 or maybe for anyone? Do i need more practice so that I can improve the speed?
2) choosing between algebra and vic(using numbers) method: how much time should one spend to decide which way one should go- algebra or vic? What indicators should we look for each approach i.e. algerbra or vic? I usually know that I have taken an inefficient approach only when I am well past half way around 3mins.

Any advice much appreciated. Thanks

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 has available to spend will be equal to half the amount that he spends this year?

1/(r + 2)
1/(2r + 2)
1/(3r + 2)
1/(r + 3)
1/(2r + 3)
Explanation:

Let the total income of Henry be H and let the amount he saves be s.
⇒ Amount he spends = H − s

Given that for each dollar that he saves this year, he will have 1 + r dollars available to spend.
⇒ Amount Henry will have next year = s(1 + r)

The amount Henry has available to spend next year should be half the amount that he spends this year.
⇒ s(1 + r) = ½ × (H − s)
⇒ 2s(1 + r) = H − s
⇒ H = 2s(1 + r) + s
⇒ H = s(3 + 2r)

We have to find the fraction of his income should Henry save this year, i.e. we have to find s/H. From above equation:
s/H = 1/(3 + 2r)

Answer: E.

We can let s = the amount of money, in dollars, Henry saves this year and t = his income, in dollars, this year; thus, the amount that he spends this year is (t - s). Since for each dollar that he saves this year he has 1 + r dollars available to spend next year, he has s(1 + r) dollars to spend next year. Furthermore, since this amount is half what he spends this year, we have:

s(1 + r) = (1/2)(t - s)

2s(1 + r) = t - s

2s + 2sr = t - s

2sr + 3s = t

s(2r + 3) = t

s = t/(2r + 3)

s = [1/(2r + 3)] * t

So, the amount he saves this year is 1/(2r + 3) of his income.

Answer: E
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I tried to explain this here: https://gmatclub.com/forum/this-year-he ... l#p1276031
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Somewhat similar questions:
https://gmatclub.com/forum/mary-decided ... 19290.html
https://gmatclub.com/forum/a-man-saves- ... 18796.html
https://gmatclub.com/forum/last-year-ca ... 67602.html
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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, \(E\)

Suppose he saves 50 dollars and spends 50 dollars the first year.

The next year he needs to spend 25 dollars. 50(1+r)=25

r=-.5

He spent 50 dollars and saved 50 dollars his first year so he saved 1/2 of his income. Only choice E gives 1/2 when]

we substitute r=-.5