Saturday

April 18, 2015

April 18, 2015

Posted by **Me** on Thursday, March 25, 2010 at 12:57pm.

- Math -
**Tiffany**, Thursday, March 25, 2010 at 1:11pmAll you have to do is multiply parts of the 2500 by .08 and .12 then find which amount is the same.

EXP 1250 * .08 = 100 1250 * .12 = 150 since those don't match go to 1300*.08 and 1200*.12 and so on untill you find the same answer

- Math -
**Reiny**, Thursday, March 25, 2010 at 1:58pmLet the amount invested at 8% be x

then the amount invested at 12% is 2500-x

.08x = .12(2500-x)

times a 100

8x = 12(2500-x)

8x = 30000 - 12x

20x = 30000

x = 1500

So John investe $1500 at 8% and $1000 at 12%

Check: .08(1500) = 120 , and .12(1000) = 120

- Math -
**Tiffany**, Thursday, March 25, 2010 at 2:13pmwould you really have to multiply .08 and .12 by 100 or could you leave it the way it is?

- Math -
**Reiny**, Thursday, March 25, 2010 at 2:38pmYou could just work with the decimals.

I just like to work with whole numbers rather than decimals or fractions.

That is because calculators did not exist for most of my teaching career and we had to do calculations in our heads or with pencil and paper.

- Math -
**Tiffany**, Thursday, March 25, 2010 at 3:24pmI see well think you for showing me an easier and faster way to figure that problem out. I was taught the long way on most math problems.

**Answer this Question**

**Related Questions**

math - 8.Mary invested RM3000 in two accounts, some at 12% per annum and the ...

Help Please Math - A principal of $2500 is invested,part at 8% per annum and the...

math - If you invest $2500 for one year, part at 8% and part at 12%.....how much...

Math - I hate word problems! Can anyone just change this Following word prob to...

math - you invested $4000 in two accounts paying 2% and 9% annual interest, ...

Math - An investment advisor invested a total of $12,000, part at 2.5% annual ...

math(urgent) - Mary invested $30,000 in two accounts some at 12% per annum and ...

Math - Larry has an annual return of $213.00 from $3000.00 invested at simple ...

intermediate algebra - need the equation for this problem. Mike invested $706 ...

algebra - Select the set of equations that represents the following situation: ...