Question

Part of the proceeds from a garage sale was $455 worth of $5 bills and $20 bills. If there were 6 more $5 bills than 20$ bills find the number of each denomination

Answer 1

Answer:

$\large \boxed{\text{17 \ 20 bills and 23 \ 5 bills}}$

Step-by-step explanation:

               Let x = the number of $20 bills.

     Then x + 6 = the number of $5 bills  and

20x + 5(x +  6) = 455

$\begin{array}{rcl}20x + 5(x + 6) & = & 455\\20x + 5x + 30 & = & 455\\25x + 30 & = & 455\\25x & = & 425\\x & = &\dfrac{425}{25}\\\\x & = &\mathbf{17}\\\end{array}$

                    x = 17 = number of $20 bills

x + 6 = 17 + 6 = 23 = number of $5 bills

$\text{There were }\large \boxed{\textbf{17 \ 20 bills and 23 \ 5 bills}}$

Check:

$\begin{array}{rcl}20\times17+ 5(17 + 6) & = & 455\\340 + 5\times23& = & 455\\340 + 115 & = & 455\\455 & = & 455\\\end{array}$

OK.