The 15% of 8000 divided by 2

Question

If you have a total of $8000 invested in 2 accounts. In one account the annua interest rate is 5%. On the second account the annual interest rate is 10%. The total annual interest earned by the 2 accounts is $530. How much do invest in each account?

Answer 1

Hi Genesis,

let assume you invested in first account  x  and invested in second account y

so x+y = 8000---->eq1

also u have  0.05x interest rate for first account 

 0.1y    interest rate for second account

 0.05x + 0.1y = 530---->eq2 total interest earned

so we subtract eq2 from eq1 we get

-0.1x  - 0.1y = -800 (after multiply by -0.1)

0.05x + 0.1y = 530

------------------------

-0.05x = -270----> x = 270/0.05---> x = $5400 money invested in first account

and y = 8000 - x ----> y = 8000 - 5400 ---> y = $2600 money invested in second account

 

Best

Answer 2

Let Amount invested in account of interest rate 5% = x

Amount invested in account of interest rate 10%= y

 

x+y= 8000________________ equation 1

 

annual interest rate on 5% account is x* 5%=0.05x

annual interest rate on 10% account is y* 10%=0.10y

 

given that

 

The total annual interest earned by the 2 accounts is $530.

so

0.05 x+ 0.10y=530

 

multiply this equation with 100

 

5x+10y=53000 _________________ equation 2

 

x+y=8000 eq 1

 

y= 8000-x ___(subtract x from both sides)

 

substitute y= 800-x in equation 2

 

5x+ 10y=53000

5x+10(8000-x)=53000

5x+80000-10x=53000

-5x+80000=53000

-5x=53000-80000

-5x=-27000

 

x= 5400

 

y=8000-x

  =8000-5400=2600

 

So

Amount invested in account of interest rate 5% = x=5400
Amount invested in account of interest rate 10%= y=2600