Pembahasan Ujian PAI: A50 – No. 17 – November 2018

\({m_x} = \frac{{{d_x}}}{{{L_x}}}\)
\(= \frac{{{l_x} – {l_{x + 1}}}}{{{L_x}}}\)

\({}_t{p_{x + s}} = \frac{{{l_{x + s + t}}}}{{{l_{x + s}}}}\)
\(= \frac{{{l_{x + t}}{{\left( {{p_{x + t}}} \right)}^s}}}{{{l_x}{{\left( {{p_x}} \right)}^s}}}\)

\({q_{x + 1}} = 1 – {p_{x + 1}}\)
\(= 1 – \frac{{{l_{x + 2}}}}{{{l_{x + 1}}}}\)

\(0,25 = \frac{{{l_{x + 1}} – {l_{x + 2}}}}{{{l_{x + 1}}}}\)
\(0,25 \cdot {l_{x + 1}} = {l_{x + 1}} – {l_{x + 2}}\)
\({l_{x + 2}} = {l_{x + 1}} – 0,25 \cdot {l_{x + 1}}\)
\(= 0,75 \cdot {l_{x + 1}}\)
\(= 0,75 \cdot 8.100 = 6.075\)

\({m_x} = \frac{{{d_x}}}{{{L_x}}} = \frac{{{l_x} – {l_{x + 1}}}}{{{L_x}}}{\rm{ }}\) sehingga
\({m_{x + 2}} = \frac{{{l_{x + 2}} – {l_{x + 3}}}}{{{L_{x + 2}}}}\)
\(0,3645 = \frac{{6.075 – {l_{x + 3}}}}{{6.000}}\)
\({l_{x + 3}} = 6.075 – 0,3645 \cdot 6.000\)
\(= 3.888\)

\({}_2{p_{x + 0,5}} = \frac{{{l_{x + 0,5 + 2}}}}{{{l_{x + 0,5}}}}\)

\(= \frac{{{l_{x + 2}}{{\left( {{p_{x + 2}}} \right)}^{\frac{1}{2}}}}}{{{l_x}{{\left( {{p_x}} \right)}^{\frac{1}{2}}}}}\)

\(= \frac{{{l_{x + 2}}{{\left( {\frac{{{l_{x + 3}}}}{{{l_{x + 2}}}}} \right)}^{\frac{1}{2}}}}}{{{l_x}{{\left( {\frac{{{l_{x + 1}}}}{{{l_x}}}} \right)}^{\frac{1}{2}}}}}\)

\(= \frac{{6.075{{\left( {\frac{{3.888}}{{6.075}}} \right)}^{\frac{1}{2}}}}}{{10.000{{\left( {\frac{{8.100}}{{10.000}}} \right)}^{\frac{1}{2}}}}} = \frac{{4860}}{{9000}} = 0,54\)