Business, 22.10.2020 19:01 williamsgregg5214

Fund A is invested at an effective annual interest rate of 3%. Fund B is invested at an effective annual interest rate of 2.5%. At the end of 20 years, the total in the two funds is 10,000. At the end of 31 years, the amount in Fund A is twice the amount in Fund B. Calculate the total in the two funds at the end of 10 years.

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Answer from: stmy1969

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:

$\to P_A1.03^{31} = 2(10000 - PA1.03^{20})1.025^{11}\\\\\to P_A(1.03^{31} + 2 \times 1.03^{20} \times 1.025^{11})\\\\ = 20000 \times 1.025^{11}\\\\\to P_A = \frac{(20000 \times 1.025^{11})}{(1.0331 + 2 \times 1.03^{20} \times 1.025^{11})}$

$P_B = \frac{(10000 - P_A1.03^{20})}{1.025^{20}}$ , So putting in this equation the above P_A value

$P_B = \frac{[ {10000 - {(20000 \times 1.025^{11})}{(1.03^{31} + 2 \times 1.03^{20} \times 1.025^{11})}1.03^{20}]}}{1.025^{20}}$

We want to find $P_A1.03^{10} + P_B1.025^{10}$ so we have the last $P_A \ \& \ P_B$ values (in bold)-

$\to \frac{[(20000 \times 1.025^{11})}{(1.03^{31} + 2 \times 1.03^{20} \times 1.025^{11})] \times 1.03^{10}}$ $+ \frac{(10000 \times 1.03^{11})}{(1.03^{11} \times1.025^{20}+2 \times 1.025^{31}) \times 1.025^{10}}\\\\$

$=( \frac{26241.73}{7.24}) \times 1.03^{10} + (\frac{13842.34}{6.57}) \times 1.025^{10}\\\\= 4871.09 + 2697.01\\\\ = 7568$

Answer from: Quest

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explanation:

Answer from: Quest

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