Question

11. A man invests his savings in two accounts, one paying 6 percent and the other paying 10 percent simple interest per year. He puts twice as much in the lower-yielding account because it is less risky. His annual interest is 198 dollars. How much did he invest at each rate?

Your answer is

total in the account paying 6 percent interest is   

total in the account paying 10 percent interest is

Answer 1

Solution-

Let in account two he invested x dollars and n account one , he invested 2x dollars

According to question,

Since ,account one paying 6 percent

So, Principal=$ 2x, Rate = 0.06, time = 1 year

Now, simple interestin dollars in one year is

SI = Principal ×rate ×time = 2x(0.06)(1) = 0.12x

Similarly,

Since the other paying 10 percent simple interest per year.

So, Principal =$ x , rate= 0.1 , time=1 year

So, Simple Interest here is

SI' = Principal×rate×time = x(0.1)(1) = 0.1x

Now, Since his annual interest is 198 dollars.

So, SI+ SI' = 198

Or 0.12x +0.1x = 198

Or 0.22x = 198

Or x = 198/0.22 =900 dollars

Hence, he invested x=$ 900 in account paying 10% interest and 2x= 9×00 =$ 1800 in the account paying 6% interest rate.

After one year total amount in account one (6% interest rate) is

= $(2x +SI)= $(2x +0.12x)= $(2.12x)

= $(2.12×900) = $ 1908.

After one year total amount in account two (10% interest rate) is

= $(x +SI) = $(x +0.1x)= $(1.1x) = $(1.1×900)

= $ 990.