Compounded interest problem

[Guest]

I think I got this one figured out but again a pair of expert eyes would be great and extremely appreciated. Thanks!! Andrea

If I invest $1000 on May 18th in a savings account that pays 10% interest compounded every month, and I invest $1000 (on the same day) in an account which pays 10% interest that is compounded continuously:

(a)
Which will be worth more in one year?
Monthly
A = P(1 + r)^t
A = 1000(1 + .1)^12
A = $3138.43

Continuously
A = Pe^(rt)
A = 1000e^((.1)(1))
A = $1105.17

(b)
By how much?
$3138.43 – $1105.17 = $2033.26

 

[tkhunny]

Nope. In the "Monthly" version, r = 0.10/12. Try that.

Note: You should have been shocked by the results you produced. The question isn't even interesting if it is THAT different.

 

[Guest]

Is this correct now? The difference is very small. Thanks again. Andrea

Monthly
A = P(1 + r)t
A = 1000(1 + (.1/12))^12
A = 1000(1.01)^12
A = $1126.83

Continuously
A = Pe^(rt)
A = 1000e^((.1)(1))
A = $1105.17

(b)
By how much?
$1126.83 – $1105.17 = $21.66

 

[tkhunny]

0.1/12 is NOT 0.01

The ONLY difference is in the compounding methodology. I mean VERY close -- less than $1.00.

 

[Denis]

2 accounts; one is 10% compounded annually, other is 10% compounded semiannually:

Annual:
Jan 1                    1000.00
Dec 31  100.00 IN        1100.00 : 1000 * .10 = 100.00

Semi-annual:
Jan 1                    1000.00
Jun 30   50.00 IN        1050.00 : 1000 * .05 = 50.00
Dec 31   52.50 IN        1102.50 : 1050 * .05 = 52.50

Do you follow that, Andrea?

And: 1000 * (1.05)^2 = 1102.50 ; capish?