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A. Money is invested in a savings account at 12% simple interest. After one year there is $ 1680 in the account. How much was originally invested? 
  
B. The sum of three consecutive integers is 114. Find the integers
asked May 29, 2014 in ALGEBRA 1 by SanMar Rookie

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A) Rate of interest r = 12% = 0.12.

Balance of the account A = $1680.

Time period t = 1 year.

Formula for finding balance is A = P ( 1+ rt ).

1680 = P (1 + (0.12)(1))

1680 = P (1 + 0.12)

1680 = P (1.12)

P = 1680 / 1.12

P = 1500

$1500 was originally invested.

B) Let the consecutive integers be (a - 1), a, (a + 1).

Sum of three consecutive integers is 114.

Therefore (a - 1) + a + (a + 1) = 114

a - 1 + a + a + 1 = 114

3a + 0 = 114

3a = 114

a = 114/3 = 38.

a - 1 = 38 - 1 = 37

a + 1 = 38 + 1 = 39

Therefore the integers are 37, 38 and 39.

 

answered May 29, 2014 by joly Scholar

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