Worded algebraic question

Question

 

a person invested $6500 for 1 year, part at 7%, part at 10%, and the remainder at 14%.
the total annual income from these investments was $758. the amount of money invested at 14% was 700 more than invested at 7% and 10% combined. Find the amount invested at each rate. 

Answer 1

Hi Des,

 

There are 3 unknowns (amounts for the three parts)

 

We'll need 3 equations

 

X = money at 7%

 

y = money at 10%

 

z = money at 14%

 

x + y + z =6500  from the problem

 

z = 700 + x + y  from the problem,  substitute this for z

 

x + y + (700 + x + y) = 6500

 

2x + 2y + 700 = 6500

 

x + y =2900     

 

z = 700 + x + y     

 

z = 700 + 2900

 

z = 3600

 

$3600 invested at 14% = $504

 

Now we use the next 2 equations from above, x + y = 2900.

 

x = 2900 - y

 

This below is the interest earned by all three amounts

 

.07x + .10y + 504 = 758        substitute for x,  remember 504 was earned by the account at 14%

 

.07(2900 - y) +.10y + 504 = 758

 

203 - .07y + .10y = 254

 

203

 + .03y = 254

 

.03y = 51

 

y = 1700

 

1700 invested at 10%

 

x + y = 2900 

 

y = 1700

 

x = 1200

 

z = 3600

 

good luck with the rest of your work

 

Tracey

 

Answer 2

Let the three amounts be:

x, y, and x + y + 700

They total 6500

x + y + x + y + 700 = 6500

2x + 2y = 5800

x + y = 2900

y = 2900 - x

 

Asign the interest rates

 

0.07x + 0.10y + 0.14(x + y + 700) = 758

0.07x + 0.10(2900 - x) + 0.14(x + 2900 - x + 700) = 758

0.07x + 290 - 0.10x + 0.14(2900 + 700) = 758

0.07x - 0.10x + 290 + 504 = 758

-0.03x = -36

x = 1200

 

Can you determine the other two amounts and answer?