Pembahasan Ujian PAI: A60 – No. 17 – Juni 2014

PEMBAHASAN

Asumsi P adalah premi tunggal neto

Diketahui:
\({{\ddot a}_{65}} = 9,90\)
\({A_{35:\left. {\overline {\, {30} \,}}\! \right| }} = 0,21\)
\(A_{35:\left. {\overline {\, {30} \,}}\! \right| }^1 = 0,07\)

Sehingga didapatkan:
\(_{30}{E_{35}} = {A_{35:\left. {\overline {\, {30} \,}}\! \right| }} – A_{35:\left. {\overline {\, {30} \,}}\! \right| }^1\)
\(_{30}{E_{35}} = 0,21 – 0,07 = 0,14\)

Maka:
\(P = \left. {_{30}} \right|{{\ddot a}_{35}} + P \cdot A_{35:\left. {\overline {\, {30} \,}}\! \right| }^1\)
\(P{ = _{30}}{E_{35}} \cdot {{\ddot a}_{65}} + P \cdot A_{35:\left. {\overline {\, {30} \,}}\! \right| }^1\)
\(P – P \cdot A_{35:\left. {\overline {\, {30} \,}}\! \right| }^1{ = _{30}}{E_{35}} \cdot {{\ddot a}_{65}}\)
\(P(1 – A_{35:\left. {\overline {\, {30} \,}}\! \right| }^1){ = _{30}}{E_{35}} \cdot {{\ddot a}_{65}}\)
\(P = \frac{{_{30}{E_{35}} \cdot {{\ddot a}_{65}}}}{{(1 – A_{35:\left. {\overline {\, {30} \,}}\! \right| }^1)}}\)
\(P = \frac{{(0,14) \cdot (9,90)}}{{(1 – 0,07)}}\)
\(P = \frac{{1,386}}{{0,93}} = 1,49032\)

Jawaban: B. 1,49032