Question

Invest $6,300 in two different accounts the first account paid 11% the second account paid 6% in interest at the end of the year he had earned $643 in interest how much was in each account

Answer 1

Answer: he invested $5300 at 11% and $1000 at 6%

Step-by-step explanation:

Let x represent the amount which he invested in the first account paying 11% interest.

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

He Invest $6,300 in two different accounts the first account paid 11% the second account paid 6% in interest. This means that

x + y = 6300

The formula for determining simple interest is expressed as

I = PRT/100

Considering the first account paying 11% interest,

P = $x

T = 1 year

R = 11℅

I = (x × 11 × 1)/100 = 0.11x

Considering the second account paying 6% interest,

P = $y

T = 1 year

R = 6℅

I = (y × 6 × 1)/100 = 0.06y

At the end of the year, he had earned $643 in interest , it means that

0.11x + 0.06y = 643 - - - - - - - - - -1

Substituting x = 6300 - y into equation 1, it becomes

0.11(6300 - y) + 0.06y = 643

693 - 0.11y + 0.06y = 643

- 0.11y + 0.06y = 643 - 693

- 0.05y = - 50

y = - 50/ - 0.05

y = 1000

x = 6300 - y = 6300 - 1000

x = 5300