Question 1154071
Let {{{ a }}} = number of $20 bills
Let {{{ b }}} = number of $50 bills
Let {{{ c }}} = number of $100 bills
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(1) {{{ 20a + 50b + 100c = 1560 }}}
(2) {{{ c = 2b }}}
(3) {{{ b = 2a }}}
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There are 3 equations & 3 unknowns, so
it's solvable
(3) {{{ a = (1/2)*b }}}
and
(1) {{{ 20a + 50b + 100c = 1560 }}}
(1) {{{ 20*(1/2)*b + 50b + 100*2b = 1560 }}}
(1) {{{ 10b + 50b + 200b = 1560 }}}
(1) {{{ 260b = 1560 }}}
(1) {{{ b = 6 }}}
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(3) {{{ a = (1/2)*b }}}
(3) {{{ a = (1/2)*6 }}}
(3) {{{ a = 3 }}}
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(2) {{{ c = 2b }}}
(2) {{{ c = 2*6 }}}
(2) {{{ c = 12 }}}
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{{{ a + b + c = 3 + 6 + 12 }}}
{{{ a + b + c = 21 }}}
Philip has 21 bills
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check:
(1) {{{ 20a + 50b + 100c = 1560 }}}
(1) {{{ 20*3 + 50*6 + 100*12 = 1560 }}}
(1) {{{ 60 + 300 + 1200 = 1560 }}}
(1) {{{ 1560 = 1560 }}}
OK