A sum of $40,000 earns interests at a rate of 10 percent per year compounded quarterly.?

Question

How long will it take to grow to $100,000? this is the rest of the part of it. Please answer in details with all of the working.

Answer 1

A = P(1 + r/n)^nt

100000 = 40000(1 + .1/4)^4t
2.5 = (1.025)^4t
log(2.5)/log(1.025) = 4t
t = log(2.5)/4log(1.025) ~ 9.277 years

Answer 2

There are two possibilities with respect to the "compounded quarterly" part. At first glance, 10% interest per year would mean 2.5% every quarter, but if you compound that quarterly, you actually get a rate of return of 10.381%. For a "true" 10% interest per year, the quarterly compounded interest would actually be 2.411%; you can figure this out as 1.10 ^ 0.25 = 1.02411

For whatever interest rate X compounded Y times, you have the following formula:

original amount * ((1+X) ^ Y) = final amount
(1+X) ^ Y = final amount / original amount

(1+X) ^ Y = 100000/40000
(1+X) ^ Y = 2.5

Now, solve for Y using logarithms:

Y = log 2.5 / log (1+X)

So, using 2.5%, you get:

log 2.5 / log 1.025 = 37.108

So you'd need 38 quarters to end up with more than $100,000. That's 9 and a half years.

Using 2.411%, you get:

log 2.5 / log 1.02411 = 38.461

So this slightly lower quarterly interest rate would cause you need 39 quarters -- 9 and three quarters years -- of interest to end up with more than $100,000.

I'd say let it sit for the full ten years, and take the extra couple thousand to buy yourself something nice. A solid 10% investment like that is hard to find! :)

Answer 3

P1=P0(1 = r/n)^ny, where,
P1 = 100,000
P0 = 40,000
r =10% = .1
n = 4 = number of quarters in a year
y = number of years
So, 100,000 = 40,000(1+.1/4)^4y
100,000/40,000 = 1.025^4y
2.5 = 1.0025^4y
ln 2.5 = ln 1.025^4y
ln 2.5 = 4y ln 1.025
4y = (ln 2.5)/(ln 1.0025)
y = (ln 2.5)/4(ln 1.0025)
y = .9163/(4* .02469)
y= .9163/.09876
y= 9.278 years