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.
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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) (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%.
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