# May Invested \$13,000, Part At 15% And Part At 10%. The Total Interest At The End Of The Year Is \$1,900, How Much Did She Invest At Each Interest Rate. How Do I Set Up And Solve This Type Of Problem?

To set up, add the two interest amounts from each rate.  Let x be the amount invested at 15%, and (1-x) be the remaining amount, since the two should add up to 100% which is represented as 1.00.

15%(13000)(x)  + 10%(13000)(1-x) = 1900

(.15)(13000)(x) + (.10)(13000)(1-x) = 1900

Now solve this before you scroll down...

1950x + 1300 - 1300x = 1900

650x = 600

x = 12/13 = 92.3%, or \$1800 interest

(1-x) = 1/13 = 7.7%, or \$100 interest
thanked the writer.
This can be solved in an algebraic equation or use of interpolation.

Algebraic - where X is the amount invested at 10%

(13,000-x)*15% +  X*10% = 1900
1,950 - .15X + .10X=1,900
.-15X+.10X=(1900-1950)
-.05X = -50
X= -50/-.05
X=1,000

12,000* 15% + 1,000*10% = 1,900

x = 1,000 or amount invested at 10%
leaving 12,000 as amount invested at 15%

Interpolation is also a way to solve for this but it an approximation that is very accurate.

If all was invested at 15% you would have \$1,950
If all was invested at 10% you would have \$1,300
This means the bulk by far has to be invested at 15% to reach \$1,900.
The effective interest rate to obtain \$1,900 would be 1,900/13,000 = 1.9/13 = 14.61%.

The difference betwee 15% and 10% is 5% the effective rate 14.61% is 4.61 of 5 of the way to 5%.   If we use this 4.61/5 * 13,000 = 4.61 * 2,600 = 10,400 + 1,600 = 12,000 the amount need to invest at 15%.

You might need to check my math as this was all in my old head.
thanked the writer. 