Question

A women invests 37,000, part at 8% and the rest at 91/2 annual interest. If the 91/2% investment provides 627.50 more income than the 8% investment, how much is invested at each rate?

Answer 1

Answer: the amount invested at 8% is $16500

the amount invested at 9.5% is $20500

Step-by-step explanation:

Let x represent the amount invested at 8%.

Let y represent the amount invested at 9.5%.

A women invests 37,000, part at 8% and the rest at 91/2 annual interest. It means that

x + y = 37000

The formula for simple interest is expressed as

I = PRT/100

Where

P represents the principal

R represents interest rate

T represents time in years

I = interest after t years

Considering the 8% investment,

T =1 year

P = $x

R = 8%

Therefore

I = (x × 8 × 1)/100

I = 0.08x

Considering the 9.5% investment,

T =1 year

P = $y

R = 9.5%

Therefore

I = (y × 9.5 × 1)/100

I = 0.095y

If the 91/2% investment provides 627.50 more income than the 8% investment, it means that

0.095x - 0.08x = 627.5 - - - - - - -1

Substituting x = 37000 - y into equation 1, it becomes

0.095y - 0.08(37000 - y) = 627.5

0.095y - 2960 + 0.08y = 627.5

0.095y + 0.08y = 627.5 + 2960

0.175y = 3587.5

y = 3587.5/0.175

y = 20500

x = 37000 - y = 37000 - 20500

x = 16500