John invests $x at the semi-annual constant compounded rate

Max@Math Revolution wrote:John invests $x at the semi-annual constant compounded rate of 2 percent and also does $10,000 at the quarterly constant compounded rate of 4 percent. If the interests are the same after 1 year, what is the value of x??
A. $20,001
B. $20,101
C. $20,201
D. $20,301
E. $20,401

* A solution will be posted in two days.

The general formula for compounding interest is A = P(1+r/n)^nt, where A is the final amount (interest and principal). Here, we just want to solve for interest, so we can simply subtract the principal, P, from both sides.

The new equation becomes:

I = P(1+r/n)^nt - P

Factoring a P from the right side...

I = P((1+r/n)^nt - 1)

Now, we're told that the two interest payments are equal, so setting up an equation that relates what we're given in the question prompt...

10,000[(1.01)^4 - 1] = x[(1.01)^2 - 1], where x = the principal amount we are solving for.

To make simplification a little easier, set y = 1.01 so we have...

10,000(y^4 - 1) = x(y^2 - 1)

10,000 / x = (y^2 - 1)/(y^4 - 1)

Reverse FOIL the denominator, and cancel like terms...

10,000 / x = 1 / (y^2 + 1)

Now let's evaluate the denominator on the right side of the equation (remember: y = 1.01)

So, we have: (101/100)(101/100) + 1 = 10201/10000 + 10000/10000 = 20201/10000.

Plugging this into our equation, we have...

10,000 / x = 1/(20201/10000)

Simplifying the right side of the equation...

10,000 / x = 10,000 / 20,201

Therefore x = 20,201.

Answer choice C

From x(1+2%/2)^2-x=10000(1+4%/4)^4-10000, we get x(1.1^2-1)=10000(1.1^4-1)=10000(1.1^2-1)(1.1^2+1). (1.1^2-1) can be canceled out in either sides. So we get x=10000(1.1^2+1)=20,201. Hence, the correct answer is C.