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# Original Amount

Fri 26 Jun, 2015 08:23 pm
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.

What is known:

A capital is added to annually by 1/3.

\$5000 is deducted for expenses at the end of each year.

The final \$5ooo is deducted at the end of the 3rd year and the amount of the capital is twice the original amount.

Want to know:

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 ?

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ekename

1
Sat 27 Jun, 2015 12:33 am
@Randy Dandy,
Quote:
Want to know:
Original amount of capital.

55,500

Quote:
2C = capital at end of 3rd year.

How do I proceed from this point ?

Form an equation and solve for C.

2C = an expression which describes the way that the money has accumulated over the 3 year period after annual additions and subtractions.

C is the amount at the beginning

(4/3 times C minus 5,000) is the accumulated amount at the end of year 1

[4/3 times (4/3 times C minus 5,000) minus 5,000] is the amount at end year 2

{4/3 times [4/3 times (4/3 times C minus 5,000) minus 5,000] minus 5,000 } is the amount at end year 3 = 2 times C
Randy Dandy

0
Sat 27 Jun, 2015 08:10 am
@ekename,
What does the 4/3 represent ?
ekename

1
Sat 27 Jun, 2015 10:08 pm
@Randy Dandy,
The number 4 divided by the number 3.

It's a convenient way off representing an increase of one third.

Solve for C:

2C = {(4/3)x [(4/3)x ((4/3) x C - 5,000) - 5,000] - 5,000 }
Randy Dandy

1
Sun 28 Jun, 2015 11:23 am
@ekename,
I still can't solve it.
engineer

1
Sun 28 Jun, 2015 11:43 am
@Randy Dandy,
Try for one year first. If the man had X dollars, how much do you think he would have after one year?
Randy Dandy

1
Sun 28 Jun, 2015 11:52 am
@engineer,
4/3X - 5000 ??
engineer

1
Sun 28 Jun, 2015 11:59 am
@Randy Dandy,
Yes! Now, how much would he have after two years?
Randy Dandy

1
Sun 28 Jun, 2015 12:19 pm
@engineer,
4/3X - 5000 * 4/3X - 5000 ??
engineer

1
Sun 28 Jun, 2015 12:28 pm
@Randy Dandy,
Not quite. The X for the second year is the amount after the first year. You have to multiply the amount you have after the first year by 4/3 and then substract 5000. Try to write the equation again.
Randy Dandy

1
Sun 28 Jun, 2015 12:54 pm
@engineer,
4/3 - 5000 * 4/3X - 5000 ?
engineer

1
Sun 28 Jun, 2015 02:28 pm
@Randy Dandy,
Take the amount after one year (which you calculated) and make that the new X to find the amount in the next year. Don't randomly try to arrange 4/3, X and 5000. Think about what is happening.

Try this. Let's say the amount is \$50,000. What is the amount after year one in dollars?
Randy Dandy

1
Sun 28 Jun, 2015 03:50 pm
@engineer,
4/3 * 50000 = 200, 000 / 3 = 6666 - 5000 = 1666
engineer

1
Sun 28 Jun, 2015 04:13 pm
@Randy Dandy,
You dropped a zero off that division effort, it should be 61,667. Now for step two. How much would he have after two years?
Randy Dandy

1
Sun 28 Jun, 2015 05:34 pm
@engineer,
61667 * 4 = 246668 / 3 = 82223 - 5000 = 77223
ekename

1
Sun 28 Jun, 2015 11:08 pm
@Randy Dandy,
Quote:
I still can't solve it.

2C = {(4/3)x [(4/3)x ((4/3) x C - 5,000) - 5,000] - 5,000 }

re-arrange by adding 5000 to both sides of the equation

2C +5000 = (4/3)x [(4/3)x ((4/3) x C - 5,000) - 5,000]

simplify by multiplying both sides of the equation by 3/4

(3/4) x (2C +5000) = [(4/3)x ((4/3) x C - 5,000) - 5,000]

re-arrange by adding 5000 to both sides of the equation

(3/4) x (2C +5000) +5000 = (4/3)x ((4/3) x C - 5,000)

simplify by expanding the terms in brackets using multiplication

(6/4)C +3750 +5000 = (16/9) C - 20,000/3 ( on the next line 8,750 becomes 26250/3)

simplify by collecting like terms (numbers on one side and the unknown C on the other side

(26,250/3) + (20,000/3) = [(16/9) - (6/4)]C = [(32/18) - (27/18)] C = (5/18)C

(46250/3) = (5/18)C

(46250/3)x(18/5) = C

C = 55,500
engineer

1
Mon 29 Jun, 2015 06:11 am
@Randy Dandy,
Randy Dandy wrote:

61667 * 4 = 246668 / 3 = 82223 - 5000 = 77223

Good. You see how you are using the formula over and over.

After 1 year: X * 4/3 -5000
After 2 years: (X * 4/3 -5000) * 4/3 -5000

See how the value of X in the second year equation is the total value from the first year equation? What will the third year equation look like?
Randy Dandy

1
Mon 29 Jun, 2015 09:51 am
@engineer,
X * 4/3 - 5000 * 4/3 - 5000 * 4/3X - 5000 ?
engineer

1
Mon 29 Jun, 2015 11:01 am
@Randy Dandy,
This will be correct if you add in two sets of parentheses and remove that second X. Can you see where they go? Look at my example.
Randy Dandy

1
Mon 29 Jun, 2015 12:28 pm
@engineer,
X * 4/3 - 5000 *( 4/3 - 5000) *( 4/3 - 5000) ?
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