SOLUTION: Natasha invests $2100 in one account and $1400 in an account paying 2 % higher interest. At the end of one year she had earned $168 in interest. At what rates did she invest? $2

Algebra ->  Customizable Word Problem Solvers  -> Finance -> SOLUTION: Natasha invests $2100 in one account and $1400 in an account paying 2 % higher interest. At the end of one year she had earned $168 in interest. At what rates did she invest? $2      Log On

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Question 1105498: Natasha invests $2100 in one account and $1400 in an account paying 2 % higher interest. At the end of one year she had earned $168 in interest. At what rates did she invest?
$2100 invested at %
$1400 invested at %

Found 3 solutions by Alan3354, ikleyn, greenestamps:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Natasha invests $2100 in one account and $1400 in an account paying 2 % higher interest. At the end of one year she had earned $168 in interest. At what rates did she invest?
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Do NOT enter this -
$2100 invested at %
$1400 invested at %
Do NOT enter this -
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Answer by ikleyn(52782) About Me  (Show Source):
You can put this solution on YOUR website!
.
Interest + interest      = total interest


2100*x   + 1400*(x+0.02) = 168    <<<---=== where x is percentage for $2100 loan expressed as the decimal (like 3.5% = 0.0035 . . . )


2100x + 1400x + 28 = 168

3500x = 168 - 28 = 140  ====>  x = 140%2F3500 = 0.04 = 4%.


Answer.  $2100 was invested at 4%  and  $1400 was invested at 6%.


Check.   2100*0.04 + 1400*0.06 = 168 dollars.   ! Correct !


Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


You can use some logical reasoning to simplify the required calculations a bit.

The "extra" 2% on the $1400 produces $28 in interest: 1400%2A.02+=+28. The remaining $140 of interest is from the whole $3500 invested at the lower rate.

3500x+=+140
x+=+140%2F3500+=+.04

The lower rate was .04 = 4%; so the higher rate was 6%.

$2100 was invested at 4%; $1400 was invested at 6$.