can someone solve this problem ?

Question

Fran receives ?$153 per year in simple interest from investing ?$1900. Part is invested at? 7%, part at? 8% and part at? 9%. There is ?$100 more invested at? 9% than at? 8%. Find the amount invested at each rate.

Answer 1

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Answer 2

Hi Fadi.

 

The simple interest equation looks like this:

 

I = Prt

 

I = interest paid out

P = amount invested

r = interest rate

t = time

 

You have three separate accounts: one paid at 7%, one paid at 8%, and one paid at 9%. I'm going to call some variables:

 

S = amount of principal invested at 7%

E = amount of principal invested at 8%

N = amount of principal invested at 9%

 

From your first statement you know that

 

S + E + N = 1900

 

since the sum of all your investments is $1900.

 

Also from the first statement you know that

 

I_s + I_e + I_n = 153

 

since the sum of all the interest paid out is $153

 

Finally, you know that

 

N = E + 100

 

since you invest $100 more at 9% than you do at 8%, so amount at 8% plus 100 equals amount at 9%.

 

So you have three equations we can simplify:

 

S + E + N = 1900

I_s + I_e + I_n = 153

N = E + 100

 

Since I = Prt:

 

I_s = S*.07*1

I_e = E*.08*1

I_n = N*.09*1

 

So

 

.07S + .08E + .09N = 153

 

And since N = E + 100, we'll have:

 

S + E + (E + 100) = 1900

.07S + .08E + .09(E + 100) = 153

 

S + 2E + 100 = 1900

.07S + .17E + 9 = 153

 

S + 2E  = 1800

.07S + .17E = 144

 

Multiply the bottom row by 100 to clear out the decimals.

 

S + 2E = 1800

7S + 17E = 14400

 

Multiply the top row by -7.

 

-7S - 14E = -12600

7S + 17E = 14400

 

Add both rows to eliminate S

 

3E = 1800

 

E = 600

 

Now we know we've got $600 invested at 8%.

 

Since N = E + 100

 

N = 600 + 100 = 700, so we've got $700 invested at 9%.

 

We've got $1900 total invested.

 

S + 600 + 700 = 1900

 

S + 1300 = 1900

 

S = 600

 

So we've got $600 invested at 7%.

 

In summary:

 

$600 at 7%

$600 at 8%

$700 at 9%

 

Hope this helps.