Question 187158
Let {{{n}}} = number of nickels
Let {{{d}}}= number of dimes
Let {{{q}}}= number of quarters
given:
(1) {{{5n + 10d + 25q = 225}}}
(2) {{{5*2n + 10*(1/2)*d + 25q = 250}}}
(3) {{{n + d + q = 27}}}
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There are 3 unknowns and 3 equations, so it's solvable
(2) {{{5*2n + 10*(1/2)*d + 25q = 250}}}
(2) {{{10n + 5d + 25q = 250}}}
Subtract (1) from (2)
(2) {{{10n + 5d + 25q = 250}}}
(1) {{{-5n - 10d - 25q = -225}}}
(4) {{{5n - 5d = 25}}}
(4) {{{n - d = 5}}}
(4) {{{d = n - 5}}}
And, from (3),
(3) {{{n + d + q = 27}}}
(3) {{{q = -n - d + 27}}}
From(1)
(1) {{{5n + 10d + 25q = 225}}}
(1) {{{n + 2d + 5q = 45}}}
(1) {{{n + 2*(n - 5) + 5*(-n -d + 27) = 45}}}
(1) {{{n + 2n - 10 - 5n - 5d + 135 = 45}}}
(1) {{{-2n - 5d = -80}}}
(1) {{{-2n - 5*(n - 5) = -80}}}
(1) {{{-2n -5n + 25 = -80}}}
(1) {{{-7n = -105}}}
(1) {{{ n = 15}}}
and
(4) {{{d = n - 5}}}
(4) {{{d = 10}}}
and
(3) {{{q = -n - d + 27}}}
(3) {{{q = -15 - 10 + 27}}}
(3) {{{q = -25 + 27}}}
(3) {{{q = 2}}}
Harry has 15 nickels, 10 dimes, and 2 quarters
check:
(1) {{{5n + 10d + 25q = 225}}}
(1) {{{5*15 + 10*10 + 25*2 = 225}}}
(1) {{{75 + 100 + 50 = 225}}}
(1) {{{225 = 225}}}
OK