An investor invested a total of 3,200 in two mutual funds. One fund earned a 8% profit while the other earned a 6% profit. If the investor total profit was $232, how much was invested in each mutual fund.

Question

anonymous · Answer

let x and y represent the two mutual funds. the profit in fund x will be 8%x and in y will be 6%y. therefore
x + y = 3200 and 0.8x + 0.6y = 232. solve the two siml eqns

anonymous · Answer

solving by any method you get y = 1200 and x = 2000. which are the amount he/she is to invest in his/her funds

anonymous · Answer

can u explain it in a more understandable way plz

anonymous · Answer

ok, let x and y represent the two funds, such that x +y = 3,200 which is the total value of fund. but fund x earns a profit of 8% (ie 8% of x), and fund y earns a profit of 6% (ie 6% of x) and the total profit is 232, ie 0.08x + 0.06y = 232 
now solving    x  +  y = 3200
                   0.08x + 0.06y = 232 ie  8x + 6y = 23200
    by substution method
      make x the subject from x + y = 3200
    ie x = 3200 - y.
then plug this value in 8x + 6y = 23200 ie 8(3200 - y ) + 6y = 23200
  = 25600 -8y + 6y = 23200
  = - 2y =  - 2400
  =  y = 1200  not plug in x = 3200 - y ie x = 3200 - 1200 = 2000
 therefore the two funds worths 2000 and 1200. hope it will help u atleast