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.

Question 1186987: 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.

Found 2 solutions by Boreal, greenestamps:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
x at 4% for 0.04x interest
10000-x at 0.035 for 350-0.035x
those two add to 380
so 350+0.005x=380
0.005x=30
x=6000 at 4% (240)
10000-x=4000 at 3.5% (140)

Answer by greenestamps(13198) (Show Source): You can put this solution on YOUR website!

Here is a quick and easy way to solve any 2-part "mixture" problem like this if a formal algebraic solution is not required.

(1) P380 income from an investment of P10,000 is a rate of 3.8%; the two interest rates are 3.5% and 4%.
(2) 3.8% is 3/5 of the way from 3.5% to 4% (use a number line to see this, if it helps).
(3) That means 3/5 of the total was invested at the higher rate.

3/5 of P10,000 is P6000.

ANSWER: P6000 at 4%; the other P4000 at 3.5%

CHECK: .04(6000)+.035(4000) = 240+140 = 380

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