Question 45161: Jill has $3.50 in nickels and dimes.If she has 50 coins, how many of each type of coin does she have?
Found 2 solutions by checkley71, maribeth_abella:
Answer by checkley71(8403)   (Show Source):
You can put this solution on YOUR website! x+y=50 & 5x+10y=350 then x=50-y & 5(50-y)+10y=350 or 250-5y+10y=350 or
5y=100 or y=20 then x=50-20 or x=30
proof 5*30+20*10=350 or 150=200=350 or 350=350
Answer by maribeth_abella(11) (Show Source):
You can put this solution on YOUR website! lets say the the nickels is n while the dimes is d.
first equation
n + d = the number of coins
n + d = 50
second equation
amount of a nickel ( n ) + amount of a dime ( d ) = the total amount
0.05 ( n ) + 0.10 ( d ) = $3.50
multiply to both side by 100 beacause we need to make the dollar to cent so if
you convert $1=100 cents.
100 [ 0.05 ( n ) + 0.10 ( d ) = $3.50 ] 100
5 ( n ) + 10 ( d ) = 350
5n + 10d = 350
so the second equation is 5n + 10d = 350
Now, we solve the two equations to get the no. of dimes:
n + d = 50
5n + 10d = 350
Then, we multiply the first equation by -5. The result would be like this:
-5n - 5d = -250
5n + 10d = 350
5d = 100
then, divide both side by the numerical co-efficient 5 .
d = 20
so, the number of dimes is 20.
now, the no. of nickels:
now, we have to substitute d to 20 and use equation 1(but you can use
equation 2 if you like) because we can get the number of nickels.
n + 20 = 50
then, transpose 20 to the other side.
n = 30
so, the no. of nickels is 30.
THEN, WE CAN CONCLUDE THAT THE NO> OF DIMES IS 20 AND THE NICKEL IS 30.
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