SOLUTION: A part of P10,000 is invested at 4% and the remainder at 3. ½%. The annual income from both is P380. Find the amount of each investment.

Algebra ->  Customizable Word Problem Solvers  -> Age -> SOLUTION: A part of P10,000 is invested at 4% and the remainder at 3. ½%. The annual income from both is P380. Find the amount of each investment.      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 1186993: A part of P10,000 is invested at 4% and the remainder at 3. ½%. The annual
income from both is P380. Find the amount of each investment.
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
x equals the investment at 4%.
y equals the investment at 3.5%.

you have two equations that need to be solved simultaneously.

they are:

x + y = 10000
.04x + .035y = 380

multiply both sides of the first eqution by .04 and leave the second equation as is to get:

.04x + .04y = 400
.04x + .035y = 380

subtract the second equation from the first to get:

.005y = 20

solve for y to get:

y = 20/.005 = 4000

that makes x equal to 10000 - 4000 = 6000.

.04*6000 + .035*4000 = 240 + 140 = 380.

this confirms the solution is correct.

your solution is that 6000 is invested at 4% and 4000 is invested at 3.5%.