Question 697851
Let {{{ a }}} = number of 5 c coins
Let {{{ b }}} = number of 10 c coins
Let {{{ c }}} = number of 25 c coins
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given:
(1) {{{ 5a + 10b + 25c = 490 }}}
(2) {{{ b = 3a }}}
(3) {{{ c = b + 2 }}}
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(2) {{{ a = b/3 }}}
Substitute (2) and (3) into (1)
(1) {{{ 5*(b/3) + 10b + 25*( b + 2 ) = 490 }}}
(1) {{{ 5*( b/3 ) + 10b + 25b + 50 = 490 }}}
(1) {{{ (5/3)*b + 35b = 440 }}}
Multiply both sides by {{{ 3 }}}
(1) {{{ 5b + 105b = 1320 }}}
(1) {{{ 110b = 1320 }}}
(1) {{{ b = 12 }}}
and, since
(2) {{{ a = b/3 }}}
(2) {{{ a = 12/3 }}}
(2) {{{ a = 4 }}}
and
(3) {{{ c = b + 2 }}}
(3) {{{ c = 12 + 2 }}}
(3) {{{ c = 14 }}}
There are 4 of the 5 c coins
There are 12 of the 10 c coins
There are 14 of the 25 c coins
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check answer:
(1) {{{ 5a + 10b + 25c = 490 }}}
(1) {{{ 5*4 + 10*12 + 25*14 = 490 }}}
(1) {{{ 20 + 120 + 350 = 490 }}}
(1) {{{ 490 = 490 }}}
OK