Investment problem

Question

Joy invests a total of $10,500 in two accounts. The first account earned a rate of return of 5% (after a year). However, the second account suffered a 2% loss in the same time period. At the end of one year, the total amount of money gained was $315.00. How much was invested into each account?

Answer 1

Solution:

The present worth of Joy's in the two account is P = $10,500. The rate of return of the first account annually annually is 5% or, i1 = 0.05, and the return period n is 1 year

The 2nd account suffered a loss of 2%, i2 = 0.02, and the return period is n 1 year also

The future worth of both accounts F

F = P(1+i)ⁿ

The total present worth of both accounts P = P1 + P2 = $10,500 ----- Eq 1

The future worth of both accounts is F = P + Money gained = $10,500 + $315 = $10,815

the future worth of both accounts can be written as such

F = P1(1+i1)ⁿ + P2(1-i2)ⁿ , substituting $10,815 = P1(1+0.05)¹ + P2(1-0.02)¹

$10,815 = 1.05 P1 + 0.98 P2 ----- Eq 2

solve for two equations two unknowns to get P1 and P2

P1 + P2 = $10,500 ----- Eq 1

1,05P1+0.98P2 = $10,815----Eq 2

Multiply Eq 1 by -1.05

you get -1.05 P1 - 1.05 P2 = -11025 Eq 3

1.05 P1 + 0.98 P2 = 10815 Eq 2 Add Eq 1 and Eq 3 together

-0.07 P2 = -210, and P2 = $3000 Substitute P2 in Eq 1

P1 + 3000 = 10500, and P1 = 10500 - 3000 = $7500

Answer 2

0.05x - 0.02(10500 - x) = 315

Can you solve for x and answer?