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A man invests his savings in two accounts, one paying 6% and the other paying 10% simple interest per year. He puts twice as much in the lower-yielding account because it is less risky. His annual interest is $3872 dollars. How much did he invest at each rate? Your answer is: Amount invested at 6% equals $ Amount invested at 10% equals $

Respuesta :

Answer:

Amount invested at 6% equals $35200; Amount invested at 10% equals $17600

Step-by-step explanation:

For simple interest; A = P (1 + rt), where;

A= final amount, P=principal , r = rate , t = time in years

Let the amount invested in the 6% yielding account be 'x' and the amount invested in the 10% yielding account be 'y'.

From the question, the man invests twice as much in the lower yielding, Therefore;

⇒ x = 2y

Interest = A - P = Prt

Total interest = (Interest for 6% yielding account) + (Interest for 6% yielding account)

⇒ 3872 = ( [tex]x[/tex] X 0.06 x 1) + ( [tex]y[/tex] X 0.10 x 1)

⇒ 3872 = (2[tex]y[/tex] x 0.06 x 1) + ([tex]y[/tex] x 0.10 x 1)

⇒ 3872 = 0.12[tex]y[/tex] + 0.10[tex]y[/tex]

⇒ y = $17600

⇒ x = 2y = $ 35200

Answer: Amount invested at 6% equals $35200

Amount invested at 10% equals $17600

Step-by-step explanation:

Let x represent the amount invested at 6%.

Let y represent the amount invested at 10%.

He puts twice as much in the lower-yielding account because it is less risky. This means that

x = 2y

The formula for determining simple interest is expressed as

I = PRT/100

Where

P represents the principal

R represents the rate of investment

T represents the time in years.

Considering the amount invested at 6%,

I = (x × 6 × 1)/100 = 0.06x

Considering the amount invested at 10%,

I = (y × 10 × 1)/100 = 0.1y

If his annual interest is $3872, it means that

0.06x + 0.1y = 3872 - - - - - - - - - - - -1

Substituting x = 2y into equation 1, it becomes

0.06 × 2y + 0.1y = 3872

0.12y + 0.1y = 3872

0.22y = 3872

y = 3872/0.22

y = 17600

x = 2y = 2 × 17600

x = 35200

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