Answered

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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%

Sagot :

pepe11
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: Option 'C' is correct.

Step-by-step 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,

[tex]\dfrac{x\times 2}{100}+\dfrac{(x+800)\times 4}{100}=92\\\\\dfrac{2}{100}(x+2(x+800))=92\\\\x+2x+1600=\dfrac{92\times 100}{2}\\\\3x+1600=4600\\\\3x=4600-1600\\\\3x=3000\\\\x=\dfrac{3000}{x}\\\\x=\$1000[/tex]

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

Hence, Option 'C' is correct.

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