Let the initial interest rate be \(r\%\).
Interest rate for the second year = \((1+10\%)r\% = 1.1r\%\)
Interest rate for the third year = \((1+20\%)1.1r\% = 1.32r\%\)
Then, \(30,000 + 33590.625 = 30,000(1+r\%)(1+1.1r\%)(1+1.32r\%)\)
\(\frac{63590.625}{30000} = \bigg(\frac{211.96}{100}\bigg) = (1+r\%)(1+1.1r\%)(1+1.32r\%)\)
Using options:
Option A: \((1+15\%)(1+16.5\%)(1+19.8\%) = \frac{160.5}{100}\)
Option B: \((1+20\%)(1+22\%)(1+26.4\%) = \frac{185.04}{100}\)
Option C: \((1+25\%)(1+27.5\%)(1+33\%) = \frac{211.96}{100}\)
Hence, option C is the correct answer.
A | B | C | D | Average |
---|---|---|---|---|
3 | 4 | 4 | ? | 4 |
3 | ? | 5 | ? | 4 |
? | 3 | 3 | ? | 4 |
? | ? | ? | ? | 4.25 |
4 | 4 | 4 | 4.25 |