Linear equation percent

Question

Martha invested a total of $48,000 in two different funds. One fund earned 10.5% interest and the other earned 12% in one year. After that year her account showed $5520 in interest. How much did she have invested in each account?

Answer 1

Martha invested a total of $48,000 in two different funds. One fund earned 10.5% interest and the other earned 12% in one year. After that year her account showed $5520 in interest. How much did she have invested in each account?

 

Solu

 

This is linear equation

Martha invested in 10.5 % interest account = $x

Martha invested in 12 % interest account = 48000-x

 

Interest earn on 10.5 %  account = x*10.5%=0.105 x

Interest earn on 12 % account =(48000-x)*12% =(48000-x)*0.12 =5760-0.12x

 

 

total interest earn = Interest earn on 10.5 % account + Interest earn on 12 % account

                 5520 = 0.105x + 5760-0.12x

                 -240 =-0.015x

                 16000  =x

 

Martha invested in 10.5 % interest account = $x  = $16000
Martha invested in 12 % interest account = 48000-x =48000-16000=$32000

 

Answer 2

This is not a linear equation reference percent.

This is a compound interest question.

 

If we invest $x at 10.5%, we invest 48000-x at 12%

Since we are dealing with 1 year and annual compounding,

we can use simple interest  I = prt

 

5520 = .105x + .12(48000-x)

5520 = .105x + 5760 - .12x

-240 = -.015x

 

x = -240/(-.015)

x = $16,000 invested at 10.5%

48000 - 16000 = $32,000 invested at 12%