old coin shop

word problems for math?

You are 12 coins in a desk drawer of age, composed entirely of nickel coins dimes, and quarters, with a face value of $ 1.55. However, the pieces date from 1889 and worth more than face value. You been able to sell all the coins in a coin shop for a total of $ 65. Suppose you received $ 5 for every penny, $ 4 for every penny, and $ 7 per quarter. How many coins were quarters parties?

I. N + D + Q = 12 II. , 05 d N + .10 + .25 q = 1.55 III. 5N 4 D 7 q = 65 rewrite the equation I and n = 12-DQ write the equation II, 5n 10 D 25 Q = 155 and the equation of substitution in the equation becomes I II 5 (12-DQ) 10 D 25 Q = 155 simplifies 5d 20 q = 95 Now substitute the equation in the equation becomes I III 5 (12-DQ) 4 J 7 q = 65 is the simplified 1D-2 q = 5 solve However, these two new equations (I would remove all your fault number quarter required) 5d 20 q = 95 5 (1d-2 q = 5) is 5d 10 q = 25 then 30Q = 120 q = 4 4 quarters

Old Coin Collector