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math question; grade 10?
briefly:
there are nickels dimes and quarters, a total of 49 coins. total value $5.20. if there are 5 more dimes than all the nickel and quarters combined then how many of each type of coin are there?
1 Answer
- Anonymous1 decade agoFavorite Answer
This, took a while, but I finally did it. You can form these 3 equations from the data supplied....
x = nickels, y = dimes, z = quarters....
x + y + z = 49
5x + 10y + 25z = 520
x + z = y - 5
rearrange the 1st eqn. to get.... z = 49 - x - y
then plug that into the 3rd eqn, getting..... x + 49 - x - y = y - 5
simplify that to.... 2y = 54...... y = 27
Plug y = 27 into the 1st eqn...... x + z = 22
Plug y = 27 into the 2nd eqn....... x + 5z = 50
Then minus the 1st eqn from the 2nd one...... 4z = 28......therefore z = 7
plug z = 7, and y = 27 into the 1st eqn...... x = 49 - 27 - 7 = 15
Therefore there are...
15 nickels
27 dimes
7 quarters