Question 1150545
Let {{{ d[1] }}} = Hillary's head start in km
Let {{{ d }}} = the distance between city A and city B
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Start a stop watch when Charlene leaves city A
Hillary's equation:
(1) {{{ d - d[1] = 120*( 50/60 ) }}}
Charlene's equation:
(2) {{{ d = 144*( 50/60 ) }}}
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By substitution:
(1) {{{ 144*( 50/60 ) - d[1] = 120*( 50/60 ) }}}
Multiply both sides by {{{ 60/50 }}}
(1) {{{ 144 - ( 60/50 )*d[1] = 120 }}}
(1) {{{ ( 6/5 )*d[1] = 24 }}}
(1) {{{ d[1] = ( 5/6 )*24 }}}
(1) {{{ d[1] = 20 }}} km
Plug this result back into (1)
(1) {{{ d - d[1] = 120*( 50/60 ) }}}
(1) {{{ d - 20 = 120*( 50/60 ) }}}
(1) {{{ d = 100 + 20 }}}
(1) {{{ d = 120 }}}
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Charlene took 50 minutes after the time that
Hillary drove {{{ d[1] }}} km
{{{ 20 = 120*t }}}
{{{ t = 10/60 }}}
Charlene left city A at 10:10 AM
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check:
(1) {{{ d - d[1] = 120*( 50/60 ) }}}
(1) {{{ 120 - 20 = 120*( 50/60 ) }}}
(1) {{{ 100 = 100 }}}
and
(2) {{{ d = 144*( 50/60 ) }}}
(2) {{{ 120 = 144*( 50/60 ) }}}
(2) {{{ 120 = 120 }}}
OK