Question 1041165
There are 3 kinds of coins and 3 equations, so I
already know it can be solved
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Let {{{ d }}} = number of dimes he has
Let {{{ q }}} = number of quarters he has
Let {{{ n }}} = number of nickels he has
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(1) {{{ 10d + 25q + 5n = 375 }}} ( in cents )
(2) {{{ d = 2q }}}
(3) {{{ n = q + 5 }}}
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Substitute (2) and (3) into (1)
(1) {{{ 10*(2q) + 25q + 5*( q+5 ) = 375 }}}
(1) {{{ 20q + 25q + 5q + 25 = 375 }}}
(1) {{{ 50q = 375 - 25 }}}
(1) {{{ 50q = 350 }}}
(1) {{{ q = 7 }}}
and
(2) {{{ d = 2q }}}
(2) {{{ d = 2*7 }}}
(2) {{{ d = 14 }}}
and
(3) {{{ n = q + 5 }}}
(3) {{{ n = 7 + 5 }}}
(3) {{{ n = 12 }}}
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He has 14 dimes, 7 quarters, and 12 nickels
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check:
(1) {{{ 10*14 + 25*7 + 5*12 = 375 }}}
(1) {{{ 140 + 175 + 60 = 375 }}}
(1) {{{ 375 = 375 }}}
OK