Question 1109172
Let the rate, {{{ R }}}  = [ jobs done ] / [ time in min ]
Let the rates of the 3 copiers = {{{ R[a] }}}, {{{ R[b] }}}, {{{ R[c] }}}
You can add the rates to get their rate working together.
Also: {{{ R }}} x time working = fraction of the job completed.
(1) {{{ R[a] + R[b] + R[c] = 1/50 }}}
(2) {{{ R[a]*20 + R[b]*50 = 1/2 }}}
(3) {{{ R[b]*30 + R[c]*80 = 3/5 }}}
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(2) {{{ 20*R[a] = -50*R[b] + 1/2 }}}
(3) {{{ 80*R[c] = -30*R[b] + 3/5 }}}
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(2) {{{ R[a] = -(50/20)*R[b] + 1/40 }}}
(3) {{{ R[c] = -(30/80)*R[b] + 3/400 }}}
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(1) {{{ -(5/2)*R[b] + 10/400 + R[b] - (3/8)*R[b] + 3/400 = 8/400 }}}
(1) {{{ -(20/8)*R[b] + 13/400 + R[b] - (3/8)*R[b] = 8/400 }}}  
(1) {{{ (8/8)*R[b] - (23/8)*R[b] = -5/400 }}}
(1) {{{ -(15/8)*R[b] = -5/400 }}}
(1) {{{ R[b] = ( 8/15 )*( 5/400 ) }}}
(1) {{{ R[b] = ( 1/15 )*( 5/50 ) }}}
(1) {{{ R[b] = 1/150 }}}
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(2) {{{ R[a] = -(50/20)*R[b] + 1/40 }}}
(2) {{{ R[a] = -(5/2)*(1/150) + 1.40 }}}
(2) {{{ R[a] = -1/60 + 1/40 }}}
(2) {{{ R[a] = -2/120 + 3/120 }}}
(2) {{{ R[a] = 1/120 }}}
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(3) {{{ R[c] = -(30/80)*R[b] + 3/400 }}}
(3) {{{ R[c] = -(3/8)*R[b] + 3/400 }}}
(3) {{{ R[c] = -(3/8)*(1/150) + 3/400 }}}
(3) {{{ R[c] = -3/1200 + 3/400 }}}
(3) {{{ R[c] = -1/400 + 3/400 }}}
(3) {{{ R[c] = 1/200 }}}
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The fastest rate is copier A which is
1 job / 120 min
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check answer:
(1) {{{ R[a] + R[b] + R[c] = 1/50 }}}
(1) {{{ 1/120 + 1/150 + 1/200 = 1/50 }}}
(1) {{{ 5/600 + 4/600 + 3/600 = 12/600 }}}
(1) {{{ 12/600 = 12/600 }}}
OK
You can check (2) and (3)
Find a way to plug the numbers into the calculator, I guess.
Check my math & get 2nd opinion if needed