SOLUTION: in a jar of coins, all but 16 are nickels, all but 16 are quarters, all but 16 are dimes, and all but 18 are pennies. how many of each coin are in the jar?
Algebra ->
Customizable Word Problem Solvers
-> Coins
-> SOLUTION: in a jar of coins, all but 16 are nickels, all but 16 are quarters, all but 16 are dimes, and all but 18 are pennies. how many of each coin are in the jar?
Log On
Question 545801: in a jar of coins, all but 16 are nickels, all but 16 are quarters, all but 16 are dimes, and all but 18 are pennies. how many of each coin are in the jar? Answer by JBarnum(2146) (Show Source):
You can put this solution on YOUR website! total (x) amount = (n)ickels + (q)uarters + (d)imes + (p)ennies
*** <--- amount of nickels plus 16 coins = total amount of coins this means that the other 16 coins are made up of quarters dimes and pennies amount of quarters plus 16 coins = total amount of coins this means that the other 16 coins are made up of nickels dimes and pennies amount of dimes plus 16 coins = total amount of coins this means that the other 16 coins are made up of nickels quarters and pennies amount of pennies plus 18 coins = total amount of coins this means that the other 18 coins are made up of nickels quarters and dimes
.
Deducing from above set values of x against values of x to find what each amount is equal to
d+16=n+16=q+16
d=n=q so the amount of dimes/nickels/quarters are all the same amount
this means its like saying
*** <--- I chose (n)ickels as it was the first equation
*** <--- combining the 3 (***)<--- values I made this equation. if n=6 then:
.
check
Correct
