Original Amount

[Guest]

A man adds annually to his capital, 1/3 of it, but deducts from the capital, at the end of each year, $5000 for expenses. At the end of the 3rd year, after deducting the final $5000, he has twice the original capital. Determine original amount of capital.


My partial solution:

Let C = original amount of capital.

Has $15,000 expenses for the 3 years.

2C = capital at end of 3rd year.

How do I proceed from this point ?

[Guest]

Let [tex]C[/tex] be the amount of capital he starts with.

Every year he adds 1/3 of his capital (so effectively multiplies it by 4/3) and subtracts 5000.

After year one he has
[tex]\frac{4}{3}C - 5000[/tex]

After year two he has
[tex]\frac{4}{3}\left(\frac{4}{3}C - 5000\right) - 5000[/tex]
which simplifies to
[tex]\frac{16}{9}C - \frac{35000}{3}[/tex]

After year three he has
[tex]\frac{4}{3}\left(\frac{16}{9}C - \frac{35000}{3}\right) - 5000[/tex]
which simplifies to
[tex]\frac{64}{27}C - \frac{185000}{9}[/tex]

We are told that at the end of year three the amount is equivalent to [tex]2C[/tex].
So this means
[tex]\frac{64}{27}C - \frac{185000}{9} = 2C[/tex]

Solving this gives [tex]C = 55500[/tex] dollars.

Hope this helped,

R. Baber.

[Guest]

I don't understand multiplying by 4/3 and how the 35000 and 185000 were obtained. Thanks.

[Guest]

Suppose the amount of capital is [tex]x[/tex] the amount added is 1/3 of this, i.e. [tex]x/3[/tex]. So in total the capital has increased to [tex]x+x/3 = \frac{4}{3} x[/tex], so the overall effect is to multiply by 4/3.

After year two he has
[tex]\frac{4}{3}\left(\frac{4}{3}C-5000\right)-5000[/tex]
removing the brackets gives
[tex]\frac{4}{3}\times\frac{4}{3}C-\frac{4}{3}\times 5000-5000[/tex]
which becomes
[tex]\frac{16}{9}C-\frac{20000}{3}-5000[/tex]
which is the same as
[tex]\frac{16}{9}C-\frac{35000}{3}[/tex]

The 185000 arises in the same way by multiplying -35000/3 by 4/3 and subtracting 5000.

Hope this helped,

R. Baber.

[Guest]

4/3 * 4/3C - 4/3 * 5000 - 5000

16/9C - 20000 / 3 - 5000 - I understand.

16/9C - 35000 / 3 - I still don't understand this step.

[Guest]

When adding fractions the rule is first multiply the numerators and denominators so that the two fractions have a common denominator, then add the numerators.

For example [tex]\frac{3}{4}+\frac{5}{6}[/tex]:
We can choose 12 as a common denominator as both denominators (4 and 6) divide into 12.
To get a denominator of 12 in the first fraction we multiply the numerator and denominator by 3.
To get a denominator of 12 in the second fraction we multiply the numerator and denominator by 2.
So
[tex]\frac{3}{4}+\frac{5}{6} = \frac{3\times 3}{4\times 3}+\frac{5\times 2}{6\times 2} = \frac{9}{12}+\frac{10}{12} = \frac{9+10}{12} = \frac{19}{12}[/tex]

In the question we are interested in we have -20000/3-5000 this is the same as -(20000/3+5000) (the rule for a negative number minus a positive number, (or equivalently adding two negative numbers) is add the numbers as if they were both positive and stick a minus sign in front). Put another way: [tex]-a-b[/tex] means taking away [tex]a[/tex] then taking away [tex]b[/tex] which is the same as first finding out the total then subtracting it all in one go, i.e. [tex]-(a+b)[/tex].

Although 20000/3 is a fraction 5000 isn't, but it can easily be changed into one by remembering 5000 = 5000/1.
So
[tex]\frac{20000}{3}+5000 = \frac{20000}{3}+\frac{5000}{1} = \frac{20000\times 1}{3\times 1}+\frac{5000\times 3}{1\times 3} = \frac{20000}{3}+\frac{15000}{3} = \frac{20000+15000}{3} = \frac{35000}{3}[/tex]
and because they were both originally negative we get a -35000/3 term in our expression.

Hope this helped,

R. Baber.

[Guest]

That helped again. Thanks.