Donald plans to invest x dollars in a savings account that

Compound interest formula

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

given, n= 4 (quaterly);r =.08

the approach is substitution,

our interest requirement is 100$ after 6 months, 2 compounding period. interest per compounding period is 2%

lets take 1500, after 3 months interest accumulated is 30$, total amount is 1530
after 6 months, interest is 30.6$ and total is 1560.6$, so not 1500

1500 & 1750 have a difference of 250$ only , but the expected interest different is around 40$ hence you can straightaway rule out 1750

2000 is again can be ruled out as approx 4% interest yeilds only 80$

2500$ is a good bet, first 3 months it earns 50$ as interest, next 3 months it will earn 51$ as interest.
hence answer is D

E.

For CI:
Final amount = Principal(1+ r/n)^nt
Let principal = x, then final amount = x+100

x+100 = x[1+ 0.02]^2
x+100 = 1.04x
0.04x=100
x=2500

so if 2500 is invested CI will be exactly 100. To earn more interest more principal should be invested. So E.

Thank you veyr much here is the book explanation

The formula for calculating compound interest is A = P(1 + r/n)nt where the variables represent the following:
A = amount of money accumulated after t years (principal + interest)
P = principal investment
r = interest rate (annual)
n = number of times per year interest is compounded
t = number of years
In this case, x represents the unknown principal, r = 8%, n = 4 since the compounding is done quarterly, and t = .5 since the time frame in question is half a year (6 months).
You can solve this problem without using compound interest. 8% interest over half a year, however that interest is compounded, is approximately 4% interest. So, to compute the principal, it's actually a very simple calculation:
100 = .04x
2500 = x
The correct answer is D.

Unfortunately I dont agree with the OE if I am understanding the question correctly.

If we invest 2500, then CI will ONLY be 100.
Question asks "to earn over $100" how much amount is to be invested. Considering rate of interest, number of years and everything else to be the same, the only way to earn CI>100 is to increase the Principal Amount because CI is directly proportional to Principal Amount.

On test day I would have chosen E for sure. What is the source though?

For P=2500, first 3 months you earn 50$ as interest, next 3 months it will earn 51$ as interest. So total 101$. I guess D should be fine

OK. I got the problem, the reason is that I rounded off 1.02^2 in the above calculation. Precisely it is 1.0404 and then we get x= 2475. So anything >2475 will give CI >100
The .xx04 made the difference Hence, either I should be very precise and not round off for CI problems or solve via back tracking.

Donald plans to invest x dollars in a savings account that pays interest at an annual rate of 8% compounded quarterly. Approximately what amount is the minimum that Donald will need to invest to earn over $100 in interest within 6 months?

A. $1500
B. $1750
C. $2000
D. $2500
E. $3000

The formula for calculating compound interest is A = P(1 + r/n)nt where the variables represent the following:
A = amount of money accumulated after t years (principal + interest)
P = principal investment
r = interest rate (annual)
n = number of times per year interest is compounded
t = number of years

IMO D


Solution:
r = Rate = 8% compounded quarterly.. = 2% per quarter
t = 6 motnhs = 2 quarters
So A = P (1+0.02)^2
=> A = P * 1.0404

Subsitute, P for 1500, 1700, 2000.. The A-P should be greater than 100 (since Interest = Amount - principal).
For 2500, interest is 101...

Hence, D

A = P(1 + r/n)nt here t =1/2 and n=4
=> A = P(1 + 8 %/4)4*(1/2)
=> A = P(1 + 2%)^2
=> A = P ( 1+ (2%)^2 + 4%) ,(2%)^2 is very small and can be ignored

=> A = P ( 1+4%) => A-P = P*4%
Also A-P = 100 => P = 100/4% = 100*100/4 = 2500

what wrong am I doing, can someone point out?

{x(1+.02)^2 } -x > 100

taking x common on LHS
(x)(1.02)^2 -1>100

(x)(1.02)^2 >101

(x) 1.04 >101

x= 101\1.04

I think i am doing some silly calculation mistake here, can someone point me out pl.

Hi Nikhil,

why don't use the simple formula:

CI = P(R/N)^NT

P= x
CI = 100
N= 4
T = 1/2 = 0.5

=> P = 100/ (.08/4)^(4*0.5)
=> P = 100/ (.02)^2
=> p = 100/ (4 * 10^-4)
=> P = 10^6/4
=> P = 25,000

Damn..when will I stop making these silly mistakes..

I think you have quoted the incorrect formula for CI.

Hi Bunnel,

Would it be correct to do the following.

I used the simple interest formula.
Since CI will give a greater amount than SI. an the interest is compounded quaterly. The rate for a quater would be 2%
Since the duration is 6 months there would be two compounding periods. $50 per period.

so... 50= (X. 2.1)/ 100

X would be 2500.... This is SI so CI would be obviously greater.

IS this method correct.?

Yes, 50= x*0.02 is correct.

Check other Compound Interest Problems in our Special Questions Directory.

Hope it helps.

Answer = D = 2500

For small amount, small duration (in 6 months), with quarterly compounding, there would be hardly major difference between SI & CI.

We can safely use the SI formula after proper adjustments

SI = 100

Principal = p (Required to be calculated)

RoI = 8% PA = 4% for 6 months

\(100 = p * \frac{4}{100}\)

\(p = \frac{100*100}{4} = 2500\)

if 4% is 100 then 100% would be 2500 -->> how did you get to 2500 exactly