Jill has $9.96 in nickels, pennies, and dollar bills in her bank. If she has 12 more pennies than nickels and 5 fewer dollar bills than nickels, how many of each does she have?

Show work.

Number of pennies:P Number of nickels:N Number of dollar bills:D P=N+12 N=D+5 1P+5N+100D=996 ^given Therefore P=(D+5)+12 P=D+17 Plug back into main equation (D+17)+5(D+5)+100D=996 Combine like terms 106D+42=996 106D=954 D=954/106=9
9 Dollar bills
N=D+5 N=(9)+5=14
14 Nickels
P=N+12 P=(14)+12=26
26 Pennies
26 1+14 5+9*100=996
Final answer: 9 Dollars 14 Nickels 26 Pennies

Answered by FirstSineOfMadness | 2024-06-10

Jill has $9.96 in nickels, pennies, and dollar bills. Let's denote the number of nickels as n , the number of pennies as p , and the number of dollar bills as d . Given that Jill has 12 more pennies than nickels and 5 less dollar bills than nickels, we can write the following equations:

p = n + 12
d = n - 5

The value of the coins and bills must add up to $9.96, or 996 pennies. Nickels are worth 5 cents each, pennies are worth 1 cent each, and dollar bills are worth 100 cents each. This gives us the value equation:
5n + p + 100d = 996
Substituting for p and d in terms of n from our previous adjustments, we get:
5n + (n + 12) + 100(n - 5) = 996
Now we can solve for n :

5n + n + 12 + 100n - 500 = 996
106n - 488 = 996
106n = 1484
n = 14

Now we can find the number of pennies p and dollar bills d :

p = n + 12 = 14 + 12 = 26
d = n - 5 = 14 - 5 = 9

Jill has 14 nickels, 26 pennies, and 9 dollar bills.

Answered by IanMckellen | 2024-06-19

Jill has 14 nickels, 26 pennies, and 9 dollar bills. The relationships between the amounts allow us to accurately calculate each type of currency using algebraic expressions. The total value confirms the solution is correct.
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Answered by FirstSineOfMadness | 2024-09-03