Question 1151971: One safe investment pays 7% per year, and a more risky investment pays 14% per year.
a. How much must be invested in each account if an investor of $100,000 would like a return of $9,100 per year?
b. Why might the investor use two accounts rather than put all the money in the 14% investment?
a. $__ is invested in the 7% account.
Found 3 solutions by Theo, MathLover1, greenestamps: Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! x = amount invested at 7%.
y = amount invested at 14%.
x + y = 100,000
.07x + .14y = 9100
solve for x in the first equaion to get:
x = 100,000 - y
replace x in the second equation with that to get:
.07 * (100,000 - y) + .14y = 9100
simplify to get:
.07 * 100,000 - .07y + .14y = 9100
simplify further and combine like terms to get:
7000 + .07y = 9100
subtract 7000 from both sides to get:
.07y = 2100
solve for y to get:
y = 2100 /.07 = 30,000
x = 100,000 - y = 70,000
.07 * 70,000 + .14 * 30,000 = 9100
this confirms the values for x and y are correct.
the solution is that 70,000 is invested in 7% account.
Answer by MathLover1(20849) (Show Source): Answer by greenestamps(13198) (Show Source):
You can put this solution on YOUR website!
Here is an easy alternative to the standard algebraic method (shown by the other tutors) for solving "mixture" problems like this.
(1) The $9,100 return on an investment of $100,000 is a rate of 9.1%.
(2) That 9.1% is 3/10 of the way from 7% to 14%.
(Think of starting at 7% on a number line and "walking" towards 14%, stopping when you get to 9.1%. What fraction of the distance have you gone? 7% to 14% is a difference of 7%; 7% to 9.1% is a difference of 2.1%. The fraction of the distance you have gone is 2.1/7 = 3/10.)
(3) Since 9.1% is 3/10 of the way from 7% to 14%, 3/10 of the total was invested at the higher rate.
ANSWER: 3/10 of $100,000, or $30,000, at 14%; the other 7/10, or $70,000, at 7%.
CHECK:
.14(30,000)+.07(70,000) = 4200+4900 = 9100
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