Question 1162083: Diane has $1.85 in dimes and nickels. She has a total of 23 coins. How many of each kind does she have? Found 2 solutions by Boreal, Edwin McCravy:Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! x dimes
23-x nickels
.10x+(.05)(23-x)=1.85
.10x+1.15-0.05x=1.85
0.05x=0.70
x=14 dimes ($1.40)
23-x=9 nickels ($0.45)
You can put this solution on YOUR website! Diane has $1.85 in dimes and nickels. She has a total of 23 coins. How many
of each kind does she have?
Without using algebra:
If all 23 coins had been nickels, the amount of money would have been only
23×$0.05 or $1.15. Therefore the extra $1.85-$1.15 = $0.70 must have come
from the dimes, each of which contributed an extra $0.05 per coin. So we
divide $0.70 by $0.05, getting 14. So there were 14 dimes and 23-14=9
nickels.
Answer: 14 dimes and 9 nickels.
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Using algebra:
coins | number | value per coin | value of all coins |
--------------------------------------------------------
dimes | x | $0.10 | $0.10x |
nickels| y | $0.05 | $0.05y |
--------------------------------------------------------
| 23 | | $1.85
Clear the second equation of decimals by multiplying through by 100
and dropping the "$"
Divide the second equation through by 5
Solve the first equation for y:
y = 23-x
Substitute into the second equation
2x+(23-x) = 37
2x+23-x = 37
x = 14
Substitute in
y = 23-x
y = 23-14
y = 9
Answer: 14 dimes and 9 nickels.
[Easier without algebra!!]
Edwin
