For a board game there is 3 kinds of bills. There are 11 more $2 bills than $1 dollar bills. There are 18 fewer $3 dollar bills than $1 bills. The total for the bill is $100. How many $1 bills are used?
100 = n+2(n -11)+3(n+18)
100= n-2n-22+3n +54
100 = 6n +76
24=6n 4=n
number of $1 --- n
number of $2 --- n+11
number of $3 --- n-18
now for the "value" :
1n + 2(n+11) + 3(n-18) = 100
n+2n+22 + 3n-54 = 100
6n = 132
n = 22
then n+11 = 33
n-18 = 4
so 22 $1 bills , 33 $2 bills and 4 $3 bills
check:
22 + 66 + 12 = 100
Yeahh
To solve this problem, we need to set up an equation using the given information:
Let's assume the number of $1 bills used is "n". According to the given information, there are 11 more $2 bills than $1 dollar bills, which means the number of $2 bills used will be (n + 11). Additionally, there are 18 fewer $3 dollar bills than $1 bills, so the number of $3 bills used will be (n - 18).
The sum of the values of each type of bill will equal the total amount of $100.
Therefore, the equation becomes:
1n + 2(n + 11) + 3(n - 18) = 100
Now, let's solve the equation step by step:
1n + 2n + 22 + 3n - 54 = 100 (Distribute)
6n - 32 = 100 (Combine like terms)
Next, we need to isolate the "n" variable:
6n = 100 + 32 (Add 32 to both sides)
6n = 132
Finally, solve for "n" by dividing both sides of the equation by 6:
n = 132/6
n = 22
Therefore, there are 22 $1 bills used in the board game.