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moll had some 10 cent coins and 50 cent coins in his savings bank . he had more 10 cent coins than 50 cent coins. the total value of the coins was $15.80.what was the total number of coins that moll had.

Guest Mar 13, 2017
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moll had some 10 cent coins and 50 cent coins in his savings bank . he had more 10 cent coins than 50 cent coins. the total value of the coins was $15.80.what was the total number of coins that moll had.

 

10T +50F = 1580           T > F

T     +5F   =  158

 

 

One solution would be T=158 and F=0

 

10(158 -  5a      ) +50(     b    ) =1580

 

158,  0

153,  1   =   154 coins

148,  2    =  150 coins

143,  3  =   146 coins

138,  4   =   144 coins

133,  5   =   138 coins

128,  6   =   134 coins

123,  7

118,  8

113,  9

108, 10

103,  11

98,    12

93,    13

88,    14

83,    15

78,    16

73,    17

68,    18

63,     19

58,     20  

53,     21

48,     22

43,     23

38,     24

33,     25         58 coins

------------------------------------

(154-58)/4 =  96/4  =  24

 

 

So there are many answers all of the form      

58+4k       where k is any integer and      \(0\le k \le 24\)

Melody  Mar 13, 2017

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