Question 1191699: A man invested P140,000. Part of his capital was invested at 11% and the rest at 9%. If the total income of the investments is P13,900, how much was invested at each rate?
Found 2 solutions by Theo, greenestamps:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! x = the amount invested at 11%.
y = the amount invested at 9%.
your equations are:
x + y = 140,000
.11x + .09y = 13,900
multiply both sides of the first equation by .11 and leave the second equation as it to get:
.11x + .11y = 15,400
.11x + 09y = 13,900
subtract the second equation from the first to get:
.02y = 1,500
solve for y to get:
y = 1,500 / .02 = 75,000
this means that x = 140,000 - 75,000 = 65,000.
this also means that .11 * x = .11 * 65,000 = 7,150.
this also means that .09 * y = .09 * 75,000 = 6,750.
total income is 7,150 + 6,750 = 13,900.
total invested is 65,000 + 75,000 = 140,000.
your solution is that 65,000 was invested at 11% and 75,000 was invested at 9%.
let me know if you have any questions or concerns.
theo
Answer by greenestamps(13198)  (Show Source):
You can put this solution on YOUR website!
Here is a non-traditional non-algebraic way to solve any two-part "mixture" problem like this.
All P140,000 invested at 9% would yield P12,600 interest; all at 11% would yield P15,400 interest; the actual interest was P13,900.
Consider the three interest amounts 12600, 13900, and 15400 on a number line and observe/calculate that 13900 is 1300/2800 = 13/28 of the way from 12600 to 15400.
That means 13/28 of the total was invested at the higher rate.
(13/28)(140000)=13(5000)=65000
ANSWER: P65000 at 11%; P75000 at 9%
CHECK: .11(65000)+.09(75000)=7150+6750=13900
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