Math money problem

Question

Last year, Chau had $30,000 to invest. He invested some of it in an account that paid 6% simple interest per year, and he invested the rest in an account that paid 8% simple interest per year. After one year, he received a total of $2260 in interest. How much did he invest in each account?

Answer 1

a + b = 30,000

0.06a + 0.08b = 2260

 

a = 30,000 - b

 

0.06(30,000 - b) + 0.08b = 2260

1800 - 0.06b + 0.08b = 2260

0.02b = 460

b = 23,000

 

a = 30,000 - 23,000

a = 7000

 

 

Answer 2

Total investment:  $30,000

Put $x in 6% plan

Put $(30000-x) in 8% plan

 

Total interest earned, $2,260, is sum of interest earned in the two accounts

 

I = prt

I = 2260

p = x and 30000-x

r = 0.06 and 0.08

t = 1

 

2260 = 0.06x + 0.08(30000-x)

2260 = 0.06x + 2400 - 0.08x

2260 - 2400 = -0.02x

-140 = -0.02x

 

x = $7000 invested at 6%

30000-x = $23000 invested at 8%

 

Check:  .06(7000) + .08(23000) = 420 + 1840 = $2260

Answer 3

x+Y = 30000

 

0.06X + 0.08Y = 2260

 

0.06X + 0.08(30000 - X) = 2260

0.06X + 2400 - 0.08X = 2260

 

-0.02X = 140

 

2X = 14000

 

X=7000

 

7000 and 23000