# Algebra Applications/Word problems

• Dec 21st 2009, 06:22 PM
jgjimcat
Algebra Applications/Word problems
I' am currently studying for the GRE and would really appreciate it if someone could help me with a word problem. If you could show work that would be awesome.

Problem:

Pat invested a total of \$3,000. Part of the money yields 10 percent interest per year, and the rest yields 8 percent interest per year. If the total yearly interest from this investment is \$256, how much did pat invest at 10 percent and how much at 8 percent?
• Dec 21st 2009, 06:31 PM
skeeter
Quote:

Originally Posted by jgjimcat
I' am currently studying for the GRE and would really appreciate it if someone could help me with a word problem. If you could show work that would be awesome.

Problem:

Pat invested a total of \$3,000. Part of the money yields 10 percent interest per year, and the rest yields 8 percent interest per year. If the total yearly interest from this investment is \$256, how much did pat invest at 10 percent and how much at 8 percent?

simple interest ...

I = Prt

t = 1 year

let x = amount invested at 10%

3000-x = amount invested at 8%

x(.10) + (3000-x)(.08) = 256

solve for x
• Dec 21st 2009, 06:53 PM
jgjimcat
Quote:

Originally Posted by skeeter
simple interest ...

I = Prt

t = 1 year

let x = amount invested at 10%

3000-x = amount invested at 8%

x(.10) + (3000-x)(.08) = 256

solve for x

Thank You Skeeter, I solved the problem, but could you explain to me what each variable stands for. The equation I=Prt
• Dec 21st 2009, 06:57 PM
skeeter
Quote:

Originally Posted by jgjimcat
Thank You Skeeter, I solved the problem, but could you explain to me what each variable stands for. The equation I=Prt

I = interest earned

P = principle (initial) investment

r = annual interest rate expressed as a decimal

t = time in years