Question

The annual returns on three investments amounted to $21,600: 6% on a savings account, 8% on mutual funds, and 12% on bonds. The amount of the investment in bonds was twice the amount of the investment in the savings account, and the interest earned from the investment in bonds was equal to the dividends received from the investment in mutual funds. Find how much money was placed in each type of investment.

.06x+.08y+.12z=$21,600
z=2x
.12z=.08y

Answer 1

amount invested in saving account be x
therefore amount to be invested in bonds will be 2x
let us suppose amount invested in MF be y
interest earned by bond will be
\[Interest_{bonds} = 0.12 times 2x\]
Dividend received on MF will be
\[Dividend_{MF} = 0.08 times y\]

and these two are equal therefore
\[ 0.12 times 2x=0.08 times y \]
\[ \frac{x}{y}=\frac{8}{24} = \frac{1}{3}\]

Also, 0.06x+0.08y+0.12z=$21,600
0.06x+0.12z+0.12z=$21,600
0.06x+0.24z =$21,600
0.06x+0.24(2x) =$21,600
0.06x+0.48x =$21,600
0.54x =$21,600
x = $ 21,600/0.54
x = $ 40,000 is invested into Saving Deposit
z = $ 80,000 is invested into bonds
0.08 y = 0.12z
y = (3/2)2x = 3x = 3* $ 40,000 = $ 1,20,000 is invested into Mutual Fund.