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If $x is invested at a constant annually compound interest r

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If $x is invested at a constant annually compound interest r

by Max@Math Revolution » Sun May 08, 2016 5:19 pm
If $x is invested at a constant annually compound interest rate of k percent, what is the ratio of the total amount including interest after 4n years to that after 3n years?

A. (1+k/100)^n
B. (1+k)^n
C. (1+kn/100)
D. (1+n/100)^k
E. (1+kn)

* A solution will be posted in two days.
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by 800_or_bust » Sun May 08, 2016 7:22 pm
Max@Math Revolution wrote:If $x is invested at a constant annually compound interest rate of k percent, what is the ratio of the total amount including interest after 4n years to that after 3n years?

A. (1+k/100)^n
B. (1+k)^n
C. (1+kn/100)
D. (1+n/100)^k
E. (1+kn)

* A solution will be posted in two days.
The formula for compounded interest is A = P(1+(r/n))^(nt), where P is the principal, r is the interest rate (expressed as a decimal), n is the number of times the interest is compounded each year, and t is the number of years.

Here, P=x, n=1, r=(k/100), and we're asked to find the ratio between t=4n and t=3n.

To determine the ratio, we need to set up a fraction with the expression for t=4n years as the numerator and the expression for t=3n years as the denominator.

[x(1+[k/100])^(4n)]/[x(1+[k/100])^(3n)], which simplifies to [1+(k/100)]^n, so the answer is A.
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by MartyMurray » Sun May 08, 2016 7:41 pm
800_or_bust wrote:To determine the ratio, we need to set up a fraction with the expression for t=4n years as the numerator and the expression for t=3n years as the denominator.

[x(1+[k/100])^(4n)]/[x(1+[k/100])^(3n)], which simplifies to [1+(k/100)]^n, so the answer is A.
Seems to be a lot of work.

Since you know that each year the previous year's ending amount is MULTIPLIED by (1+[k/100]), the ratio of the ending amount for 4n years to that of 3n years will be (1+[k/100]) raised to a power equal to the number of years that is the difference between 4n and 3n, which is n.

So, as you said, the answer is [1+(k/100]�.
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by 800_or_bust » Mon May 09, 2016 6:18 am
Marty Murray wrote:
800_or_bust wrote:To determine the ratio, we need to set up a fraction with the expression for t=4n years as the numerator and the expression for t=3n years as the denominator.

[x(1+[k/100])^(4n)]/[x(1+[k/100])^(3n)], which simplifies to [1+(k/100)]^n, so the answer is A.
Seems to be a lot of work.

Since you know that each year the previous year's ending amount is MULTIPLIED by (1+[k/100]), the ratio of the ending amount for 4n years to that of 3n years will be (1+[k/100]) raised to a power equal to the number of years that is the difference between 4n and 3n, which is n.

So, as you said, the answer is [1+(k/100]�.
True, I'm just glad I knew the compounding interest formula because I reviewed it the other day. A week ago, I would have had to derive it from scratch. Also, I just wanted the solution to be clear to somebody reading it. I can answer a lot of the questions on this forum without showing any work, although I still have a bad habit of making silly mistakes when I attempt to do everything in my head without at least jotting down some notes...
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by MartyMurray » Mon May 09, 2016 6:33 am
800_or_bust wrote:
Marty Murray wrote:
800_or_bust wrote:To determine the ratio, we need to set up a fraction with the expression for t=4n years as the numerator and the expression for t=3n years as the denominator.

[x(1+[k/100])^(4n)]/[x(1+[k/100])^(3n)], which simplifies to [1+(k/100)]^n, so the answer is A.
Seems to be a lot of work.

Since you know that each year the previous year's ending amount is MULTIPLIED by (1+[k/100]), the ratio of the ending amount for 4n years to that of 3n years will be (1+[k/100]) raised to a power equal to the number of years that is the difference between 4n and 3n, which is n.

So, as you said, the answer is [1+(k/100]�.
True, I'm just glad I knew the compounding interest formula because I reviewed it the other day. A week ago, I would have had to derive it from scratch. Also, I just wanted the solution to be clear to somebody reading it. I can answer a lot of the questions on this forum without showing any work, although I still have a bad habit of making silly mistakes when I attempt to do everything in my head without at least jotting down some notes...
Nice. I kind of figured that those things were going on.

Silly mistakes are doom aren't they. I resist doing much work though and resist writing anything even more.

You can definitely reduce silly mistakes even without using notes, partly by realizing that they can have an emotional basis. Maybe one of the most significant things of value that you can get from preparing for the GMAT is learning to think clearly and to not jump to conclusions and stuff. Having nice, solid, effective thought processes is so useful in all areas of life.
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by GMATGuruNY » Mon May 09, 2016 6:57 am
Max@Math Revolution wrote:If $x is invested at a constant annually compound interest rate of k percent, what is the ratio of the total amount including interest after 4n years to that after 3n years?

A. (1+k/100)^n
B. (1+k)^n
C. (1+kn/100)
D. (1+n/100)^k
E. (1+kn)
Let x = $1, k = 100%, and n = 2.
Since the amount in the account increases by 100% each year, it DOUBLES each year, as follows:
1, 2, 4, 8, 10, 12, 24, 48.
As the list above illustrates:
After 3n=6 years, the amount in the account = 12.
After 4n=8 years, the amount in the account = 48.
Thus:
(8th year)/(6th year) = 48/12 = 4. This is our target.

Now plug k=100 and n=2 into the answers to see which yields our target of 4.
Only A works:
(1+k/100)^n = (1 + 100/100)² = (1+1)² = 4.

The correct answer is A.
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by Max@Math Revolution » Mon May 09, 2016 7:45 pm
The balance after 4n years is (1+k/100)^4n. The balance after 3n is (1+k/100)^3n. Since it is asking for the ratio, we get (1+k/100)^4n:(1+k/100)^3n=(1+k/100)^4n/(1+k/100)^3n=(1+k/100)^4n-3n
=(1+k/100)^n. Hence, the correct answer is A.
 
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