Respuesta :
Answer:
The answer is "7568"
Explanation:
It took the funds through account the interest becomes compound. s Lets people, therefore, consider the primary in funds A as PA & the main sum of finances B as PB, the details presented as follows:
We get , and we'll have the P_B value throughout the second equation:
[tex]\to P_A1.03^{31} = 2(10000 - PA1.03^{20})1.025^{11}\\\\\to P_A(1.03^{31} + 2 \times 1.03^{20} \times 1.025^{11})\\\\ = 20000 \times 1.025^{11}\\\\\to P_A = \frac{(20000 \times 1.025^{11})}{(1.0331 + 2 \times 1.03^{20} \times 1.025^{11})}[/tex]
[tex]P_B = \frac{(10000 - P_A1.03^{20})}{1.025^{20}}[/tex], So putting in this equation the above [tex]P_A[/tex] value
[tex]P_B = \frac{[ {10000 - {(20000 \times 1.025^{11})}{(1.03^{31} + 2 \times 1.03^{20} \times 1.025^{11})}1.03^{20}]}}{1.025^{20}}[/tex]
We want to find [tex]P_A1.03^{10} + P_B1.025^{10}[/tex] so we have the last [tex]P_A \ \& \ P_B[/tex] values (in bold)-
[tex]\to \frac{[(20000 \times 1.025^{11})}{(1.03^{31} + 2 \times 1.03^{20} \times 1.025^{11})] \times 1.03^{10}}[/tex] [tex]+ \frac{(10000 \times 1.03^{11})}{(1.03^{11} \times1.025^{20}+2 \times 1.025^{31}) \times 1.025^{10}}\\\\[/tex]
[tex]=( \frac{26241.73}{7.24}) \times 1.03^{10} + (\frac{13842.34}{6.57}) \times 1.025^{10}\\\\= 4871.09 + 2697.01\\\\ = 7568[/tex]
