Ms. Hagan invested twice as much money in an account that pays 7% interest as she did in an account that pays 6% in interest. Her total investment pays her $1,000 a year in interest. How much did she invest at each rate?

Answer: She invested $5000 in an account that pays 6% interest and $10000 in an account that pays 7% interest.

Step-by-step explanation:

Let P be the initial amount she invested in an account that pays 6% interest.

Then, amount invested in other account = 2P

Simple interest = Principal x rate x time

After one year, for the first account,

Interest = P(0.06)(1)  = 0.06P

For second account,

Interest = (2P)(0.07)(1)=0.14P

Total interest = [tex]0.06P+0.14P=1000[/tex]

[tex]\Rightarrow\ 0.20P=1000\\\\\Rightarrow\ P=\dfrac{1000}{0.20}\\\\\Rightarrow\ P=5000[/tex]

2P = 2(5000)=10000

Hence, She invested $5000 in an account that pays 6% interest and $10000 in an account that pays 7% interest.

At the interest rate of 6%, amount invested is; $5000

At interest rate of 7%, amount invested is; $10000

Let P be the initial amount that Ms. Hagan invested in the 6% interest account that

Since she invested twice as much in the 7% interest account as in the 6% interest account, then;

Initial amount invested in the 7% interest account = 2P

Formula for Simple interest is;

I = Principal × rate × time

Interest after 1 year for the 6% interest account is,

I_1 = P × 0.06 × 1

I_1 = 0.06P

Interest after 1 year for the 7% interest account is,

I_2 = 2P × 0.07 × 1

I_2 = 0.14P

Total investment pays her $1000 after a year. Thus;

0.06P + 0.14P = 1000

0.2P = 1000

P = 1000/0.2

P = $5000

Then amount initially invested in the 7% interest account = 2P = $10000

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