SOLUTION: Investment Problem A man invested Php 6,000, part of it at 3% and the rest at 4%. The total annual income from the two investments is Php 208. How much is invested at each of the

Question 1158770: Investment Problem
A man invested Php 6,000, part of it at 3% and the rest at 4%. The total annual income from the
two investments is Php 208. How much is invested at each of these rates?
Found 2 solutions by Boreal, greenestamps:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
x at 3%=.03x
6000-x at 4%=.04(6000-x)=240-.04x
those two add to 208
so 0.03x+240-0.04x=208
-0.01x=-32
x=3200 Php at 3%, earn 96
6000-x=2800 at 4%, earn 112
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

An alternative to the standard algebraic solution method which, in this example, makes the calculations very simple....

All 6000 at 3% would yield 180 interest; all at 4% would yield 240 interest.

Find where the actual interest of 208 lies between 180 and 240:
240-180 = 60; 208-180 = 28
208 is 28/60 of the way from 180 to 240

That means 28/60 of the total was invested at the higher rate.

28/60 of the total 6000 is 2800.

ANSWER: 2800 at 4%; the remaining 3200 at 3%.

CHECK: .04(2800)+.03(3200) = 112+96 = 208

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