To find the time required for an amount of 175 to grow to 189 at an interest rate of 16% per annum, compounded half-yearly, we can use the formula for compound interest:
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
Where:
- \( A \) is the amount of money accumulated after n years, including interest.
- \( P \) is the principal amount (the initial amount of money).
- \( r \) is the annual interest rate (decimal).
- \( n \) is the number of times that interest is compounded per year.
- \( t \) is the time in years.
Given:
- \( A = 189 \)
- \( P = 175 \)
- \( r = 16% = 0.16 \)
- Interest is compounded half-yearly, so \( n = 2 \).
We can substitute these values into the formula:
\[ 189 = 175 \left(1 + \frac{0.16}{2}\right)^{2t} \]
Calculating the term inside the parentheses:
\[ 1 + \frac{0.16}{2} = 1 + 0.08 = 1.08 \]
So our equation becomes:
\[ 189 = 175 (1.08)^{2t} \]
Next, we need to isolate \((1.08)^{2t}\):
\[ \frac{189}{175} = (1.08)^{2t} \]
Calculating the left-hand side:
\[ \frac{189}{175} = 1.08 \]
So we have:
\[ 1.08 = (1.08)^{2t} \]
Since the bases are the same, we can set the exponents equal to each other:
\[ 1 = 2t \]
Now solving for \(t\):
\[ t = \frac{1}{2} = 0.5 \text{ years} \]
Thus, the time required for 175 to amount to 189 at a rate of 16% per annum, compounded half-yearly, is 0.5 years or 6 months.
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