Question 341924
Let {{{n}}} = number of nickels
Let {{{d}}} = number of dimes
Let {{{q}}} = number of quarters
given:
(1) {{{d = n + 10}}}
(2){{{q = 2d}}}
(3){{{5n + 10d + 25q = 1575}}} (in cents)
By substitution:
From(1)
{{{n = d - 10}}}
(3){{{5*(d - 10) + 10d + 25q = 1575}}}
By substitution again:
(3){{{5*(d - 10) + 10d + 25*2d = 1575}}}
{{{5d - 50 + 10d + 50d = 1575}}}
{{{65d = 1625}}}
{{{d = 25}}}
From (1):
{{{n = d - 10}}}
{{{n = 25 - 10}}}
{{{n = 15}}}
From (2):
{{{q = 2d}}}
{{{q = 2*25}}}
{{{q = 50}}}
She has 15 nickels, 25 dimes, and 50 quarters
check:
(3){{{5n + 10d + 25q = 1575}}}
{{{5*15 + 10*25 + 25*50 = 1575}}}
{{{75 + 250 + 1250 = 1575}}}
{{{1575 = 1575}}}
OK