(MC) If the compound interest on $1000 for two years at 9% p.a., payable half-yearly is $x, find x.
A) 1000x9/100x2
B) 1000(1+9/100)⁴
C) 1000(1+4.5/100)⁴
D) 1000(1+9/100)²-1000
E) 1000(1+4.5/100)⁴-1000
To find the compound interest on $1000 for two years at 9% p.a., payable half-yearly, we can use the formula:
A = P(1 + r/n)^(nt)
Where:
A = the final amount after interest
P = the principal amount (the initial amount)
r = annual interest rate (as a decimal)
n = number of times that interest is compounded per year
t = number of years
In this case, the principal amount (P) is $1000, the annual interest rate (r) is 9% (0.09 as a decimal), and the interest is compounded half-yearly, so n = 2 (twice a year). The time period (t) is 2 years.
Plugging in these values, we have:
A = 1000(1 + 0.09/2)^(2*2)
= 1000(1 + 0.045)^(4)
= 1000(1.045)^4
Therefore, the compound interest is equal to the final amount minus the principal amount, so the answer is:
E) 1000(1+4.5/100)⁴-1000
To find the compound interest on $1000 for two years at 9% p.a., payable half-yearly, we need to use the compound interest formula:
A = P(1 + r/n)^(nt)
where:
A = final amount
P = principal amount (initial investment)
r = annual interest rate (in decimal form)
n = number of times interest is compounded per year
t = number of years
In this case, the principal amount is $1000, the annual interest rate is 9% (or 0.09), interest is compounded semi-annually (n = 2), and the time period is 2 years (t = 2).
Plugging in the values into the formula, we get:
A = 1000(1 + 0.09/2)^(2*2)
Simplifying further:
A = 1000(1 + 0.045)^(4)
Now, let's calculate this:
A = 1000(1.045)^(4)
A = 1000(1.1932025)
A ≈ $1193.20
The final amount after two years is approximately $1193.20.
To find the compound interest, we subtract the principal amount from the final amount:
Compound Interest (x) = A - P
x = 1193.20 - 1000
x = $193.20
Therefore, the compound interest on $1000 for two years at 9% p.a., payable half-yearly is $193.20.
Looking at the answer choices, the correct answer is option E) 1000(1 + 4.5/100)⁴ - 1000.