Question

A man invests a certain amount of money at 2% interest and $800 more than that amount in another account at 4% interest. At the end of one year, he earned $92 in interest. How much money was invested in each account?

A) $1,500 at 2%; $2,300 at 4%
B) $1,400 at 2%; $2,200 at 4%
C) $1,000 at 2%; $1,800 at 4%

Answer 1

x the first account interest = x/100*2 = 2x/100
(x + 800)/100 * 4 =( 4x + 3200)/100
2x/100 + (4x + 3200)/100 = 92
2x + 4x + 3200 = 9200
6x = 6000
x = 1000
C is the answer

Answer 2

Answer: Option 'C' is correct.

Explanation:

Since we have given that

Let the principal amount be 'x'.

First rate of interest = 2%

Second rate of interest = 4%

Number of years = 1

Amount of interest he earned = $92

According to question,

`(x* 2)/(100)+((x+800)* 4)/(100)=92\\\\(2)/(100)(x+2(x+800))=92\\\\x+2x+1600=(92* 100)/(2)\\\\3x+1600=4600\\\\3x=4600-1600\\\\3x=3000\\\\x=(3000)/(x)\\\\x=\$1000`

So, First investment of $1000 at 2% and Second investment of $1000+$800=$1800 at 4%.

Hence, Option 'C' is correct.