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Alexxx [7]
3 years ago
8

A man invests his savings in two accounts ,one paying6% 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 $7370dollars. How much did he invest at each rate?
Mathematics
1 answer:
AnnyKZ [126]3 years ago
6 0

Answer: he invested $46062.5 at 6% and $23031.25 at 10%

Step-by-step explanation:

Let x represent the amount which he invested in the account paying 6% interest.

Let y represent the amount which he invested in the account paying 10% interest.

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

Considering the account paying 6% interest,

P = $x

T = 1 year

R = 6℅

I = (x × 6 × 1)/100 = 0.06x

Considering the account paying 10% interest,

P = $y

T = 1 year

R = 10℅

I = (y × 10 × 1)/100 = 0.1y

His annual interest is $7370dollars. it means that

0.06x + 0.2y = 7370 - - - - - - - - - -1

Substituting x = 2y into equation 1, it becomes

0.06 × 2y + 0.2y = 7370

0.12y + 0.2y = 7370

0.32y = 7370

y = 7370/0.32

y = $23031.25

x = 2 × 23031.25

x = 46062.5

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