Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Matematika Keuangan |
| Periode Ujian |
: |
Mei 2018 |
| Nomor Soal |
: |
21 |
SOAL
Dari persamaan-persamaan berikut, yang manakah yang benar.
- \({{\ddot a}_{\left. {\overline {\, n \,}}\! \right| }} = 1 + {a_{\left. {\overline {\, n \,}}\! \right| }}\)
- \({{\ddot s}_{\left. {\overline {\, n \,}}\! \right| }} = (1 + i){s_{\left. {\overline {\, n \,}}\! \right| }}\)
- \(\frac{1}{{{a_{\left. {\overline {\, n \,}}\! \right| }}}} = \frac{1}{{{s_{\left. {\overline {\, n \,}}\! \right| }}}} + i\)
- Hanya 1
- Hanya 2
- 1 dan 2
- 1 dan 3
- 2 dan 3
| Diketahui |
- \({{\ddot a}_{\left. {\overline {\, n \,}}\! \right| }} = 1 + {a_{\left. {\overline {\, n \,}}\! \right| }}\)
- \({{\ddot s}_{\left. {\overline {\, n \,}}\! \right| }} = (1 + i){s_{\left. {\overline {\, n \,}}\! \right| }}\)
- \(\frac{1}{{{a_{\left. {\overline {\, n \,}}\! \right| }}}} = \frac{1}{{{s_{\left. {\overline {\, n \,}}\! \right| }}}} + i\)
|
| Rumus yang digunakan |
\({a_{\left. {\overline {\, n \,}}\! \right| }} = \frac{{1 – {v^n}}}{i}\)
\({{\ddot a}_{\left. {\overline {\, n \,}}\! \right| }} = \frac{{1 – {v^n}}}{d}\)
\({s_{\left. {\overline {\, n \,}}\! \right| }} = \frac{{{{(1 + i)}^n} – 1}}{i}\)
\({{\ddot S}_{\left. {\overline {\, n \,}}\! \right| }} = \frac{{{{(1 + i)}^n} – 1}}{d}\)
\(d = iv = 1 – v\) |
| Proses Pengerjaan |
- \({{\ddot a}_{\left. {\overline {\, n \,}}\! \right| }} = 1 + {a_{\left. {\overline {\, n \,}}\! \right| }}\)
\(\Leftrightarrow \frac{{1 – {v^n}}}{d} = 1 + \frac{{1 – {v^n}}}{i}\)
\(\Leftrightarrow \frac{{1 – {v^n}}}{d} = \frac{i}{i} + \frac{{1 – {v^n}}}{i}\)
\(\Leftrightarrow \frac{{1 – {v^n}}}{d} = \frac{{(1 + i) – {v^n}}}{i}\)
\(\Leftrightarrow \frac{{1 – {v^n}}}{{iv}} = \frac{{{v^{ – 1}} – {v^n}}}{i}\)
\(\Leftrightarrow \frac{{{v^{ – 1}} – {v^{n – 1}}}}{i} \ne \frac{{{v^{ – 1}} – {v^n}}}{i}\)
\({\rm{Pernyataan (1) salah}}\)
- \({{\ddot s}_{\left. {\overline {\, n \,}}\! \right| }} = (1 + i){s_{\left. {\overline {\, n \,}}\! \right| }}\)
\(\Leftrightarrow \frac{{{{(1 + i)}^n} – 1}}{d} = (1 + i).\frac{{{{(1 + i)}^n} – 1}}{i}\)
\(\Leftrightarrow \frac{{{v^{ – n}} – 1}}{{iv}} = {v^{ – 1}}.\frac{{{v^{ – n}} – 1}}{i}\)
\(\Leftrightarrow \frac{{{v^{ – n}} – 1}}{i} = \frac{{{v^{ – n}} – 1}}{i}\)
\({\rm{Pernyataan (2) benar}}\)
- \(\frac{1}{{{a_{\left. {\overline {\, n \,}}\! \right| }}}} = \frac{1}{{{s_{\left. {\overline {\, n \,}}\! \right| }}}} + i\)
\(\Leftrightarrow \frac{1}{{\frac{{1 – {v^n}}}{i}}} = \frac{1}{{\frac{{{{(1 + i)}^n} – 1}}{i}}} + i\)
\(\Leftrightarrow \frac{i}{{1 – {v^n}}} = \frac{i}{{{{(1 + i)}^n} – 1}} + i\)
\(\Leftrightarrow \frac{i}{{1 – {v^n}}} = \frac{i}{{{v^{ – n}} – 1}} + \frac{{i{v^{ – n}} – i}}{{{v^{ – n}} – 1}}\)
\(\Leftrightarrow \frac{i}{{1 – {v^n}}} = \frac{{i{v^{ – n}}}}{{{v^{ – n}} – 1}}\)
\(\Leftrightarrow i{v^{ – n}} – i = i{v^{ – n}} – i\)
\({\rm{Pernyataan (3) benar}}\)
|
| Jawaban |
e. Pernyataan 2 dan 3 benar |
