Answer:
The money invested in the account of gain was x = $ 9,000
The money invested in the account of loss was y = $ 8,000
Explanation:
Solution:-
- Monica invested a total T = $ 17,000 in two of her accounts.
- Take the money invested in account A today = x
- Take the money invested in account B today = y
- The total money invested today is:
x + y = 17,000 ... Eq1
- After one year, the amount in each categorized by the following statement:
" The first account earned a rate of return of 14% (after a year) "
- The amount in account A after a year would be = ( 1 + gain/100 )*x
= ( 1 + 14/100)*x
= ( 1 + 0.14 )*x
= 1.14x
The account B:
" the second account suffered a 9% loss in the same time period "
- The amount in account B after a year would be = ( 1 - loss/100 )*y
= ( 1 - 9/100)*y
= ( 1 - 0.09 )*y
= 0.91*y
- The total amount of money in two accounts after one year would be:
Total money after a year = 1.14x + 0.91*y
- Given that the gain in one year was G = $540
G = Total money after one year - Total invested
540 = [ 1.14x + 0.91*y ] - 17,000
17,540 = 1.14x + 0.91*y .... Eq 2
- Solve the two equations Eq1 and Eq2 simultaneously:
x + y = 17,000
1.14x + 0.91*y = 17,540
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x = $ 9,000 , y = $ 8,000
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- The money invested in the account of gain was x = $ 9,000
- The money invested in the account of loss was y = $ 8,000