Question 176676
Let {{{n}}}= number of nickels
Let {{{d}}}= number of dimes
Let {{{q}}}= number of quarters
given:
(1) {{{n + d + q = 25}}}
(2) {{{d = 4q}}}
In pennies:
(3) {{{5n + 10d + 25q = 205}}}
There are 3 equations and 3 unknowns, so it's solvable
Multiply (1) by {{{5}}} and subtract from (3)
{{{5n + 10d + 25q = 205}}}
{{{-5n - 5d - 5q = -125}}}
--------------------------
{{{5d + 20q = 80}}}
And using (2),
{{{5*4q + 20q = 80}}}
{{{40q = 80}}}
{{{q = 2}}}
{{{d = 4q}}}
{{{d = 8}}}
(1) {{{n + d + q = 25}}}
{{{n + 8 + 2 = 25}}}
{{{n = 15}}}
There are 15 nickels, 8 dimes and 2 quarters
check:
(3) {{{5n + 10d + 25q = 205}}}
{{{5*15 + 10*8 + 25*2 = 205}}}
{{{75 + 80 + 50 = 205}}}
{{{205 = 205}}}
OK