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| Question 1207481:  Linda Smaoke invests a total of $10,000 in two savings accounts. One account pays 6% interest, and the other,7%. Find the amount placed in each account if the accounts receive a total of $660 in interest after 1 year.
 Found 3 solutions by  Msexcel , ikleyn, greenestamps:
 Answer by Msexcel (1)
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You can put this solution on YOUR website! x + y = 10000
 .06x + .07y = 660
 From equation (1)
 x = 10,000 - y
 Substitute into (2)
 .06x + .07y = 660
 .06(10,000 - y) + .07y = 660
 600 - .06y + .07y = 660
 - .06y + .07y = 660 - 600
 .01y = 60
 y = 60/.01
 y = 600
 Substitute the value of y into equation (1)
 x + y = 10,000
 x + 600 = 10,000
 x = 10,000 - 600
 x = 9400
 Therefore, $9,400 was placed in the account which pays 6% interest rate and $600 was placed in the account which pays 7% interest rate
Answer by ikleyn(52879)
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You can put this solution on YOUR website! . 
 
Let the 7% investment be x dollars;
 
then the 6% investment is (10000-x) dollars.
Partial interests from separate accounts are 0.07x and 0.06*(10000-x) dollars, respectively.
The total annual interest equation is
    0.07x + 0.06*(10000-x) = 660  dollars.
Simplify this equation and find x
    0.07x + 600 - 0.06x = 660
    0.07x - 0.06x = 660 - 600
        0.01x     =    60
            x     =    60/0.01 = 6000  dollars.
So, the 7% investment is $6000;  the 6% investment is the rest,  $10000 - $6000 = $4000.
        Thus you get the answer.
CHECK.  0.07*6000 + 0.06*4000 = 420 + 240 = 660  dollars.  ! precisely correct !
Solved.
 
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 The other person made an arithmetic error on the way and did not fix it.
 
 This led him/her to wrong answer.
 
 
 Let this example teaches you on how important is to make a full-size check at the end.
 
 Without such full-size check the solution is not complete !
 
 And can not be considered/recognized/accepted as an error-free !
 
 
 
Answer by greenestamps(13209)
      (Show Source): 
You can put this solution on YOUR website! 
 Here is a quick and easy non-algebraic method for solving any 2-part "mixture" problem like this.
 
 $10,000 all invested at 6% would yield $600 interest; all at 7% would yield $700 interest.
 
 Use any arithmetic method you want (perhaps a number line) to observe/calculate that $660 is 60/100 = 3/5 of the way from $600 to $700.
 
 That means 3/5 of the total was invested at the higher rate.
 
 3/5 of $10,000 is $6000.
 
 ANSWER: $6000 was invested at 7%, the other $4000 at 6%
 
 CHECK: .07($6000)+.06($4000) = $420+$240 = $660
 
 
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