Suppose $6,500 is divided into two bank accounts. one account pays 10% simple interest per year and the other pays 4.6%. after two years there is a total of $1000 in interest between the two accounts. how much was invested into the bank account that pays 4.6% simple interest (rounded to the nearest cent) ?

Question

Douglas Parrish

Mathematics

13 January, 09:14

Suppose $6,500 is divided into two bank accounts. one account pays 10% simple interest per year and the other pays 4.6%. after two years there is a total of $1000 in interest between the two accounts. how much was invested into the bank account that pays 4.6% simple interest (rounded to the nearest cent) ?

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Answers (1)

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Yasmin Fletcher · Answer 1

Interest1+interest2=1000

case 1

principal = x

rate of interest=10/100

time = 2 years

I=P*T*R/100

(10*x*2) / 100

20x/100

case 2

principal = 6500 - x

rate of interest = 4.6/100

time = 2 years

I=P*T*R/100

(6500-x*2*4.6) / 100

(6500-x) * 9.2/100

61800-9.2x/100

interest1+interest2=1000

20x/100+61800-9.2x/100 = 1000

61800-9.2x+20x/100=1000

61800+10.8x=100000

x = (100000 - 61800) / 10.8

x = 38200/10.8

x=$ 3,537.04

6500 - x = 6500-3,537.04

=$2962.96

if rounded to the nearest cent the answer is

x=$3,537.00

6500-x=$2963.00