Question

A and B can do a piece of work in 30 days B and C in 24 days B and C in 40 days how long will it take them to do the work together in what time can each finish it working alone​

Answer 1

Answer:

They can finish it alone as A in 40 days , B in 60 days and C in 120 days.

Answer 2

$\huge{\blue{\bold{\underline{\underline{Answer :}}}}}$

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$\large{\green{\underline \bold{\tt{Given :-}}}}$

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• A and B can do a piece of work in 30 days

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• B and C in 24 days

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• C and A in 40 days

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$\large{\red{\underline \bold{\tt{To \: Find :-}}}}$

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• Time required when all of them do the work together

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• Time in which each can finish the work alone

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$\large{\orange{\underline{\tt{Solution :-}}}}$

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• Let A , B , C can complete the work in A , B , C days respectively

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$\underline{\bold{\texttt{One day work of A :}}}$

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➠ $\sf \dfrac { 1 } { A }$

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$\underline{\bold{\texttt{One day work of B :}}}$

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➠ $\sf \dfrac { 1 } { B}$

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$\underline{\bold{\texttt{One day work of C :}}}$

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➠ $\sf \dfrac { 1 } { C}$

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A and B can do the work in 30 days

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➠ $\sf \dfrac { 1 } { A } + \dfrac { 1 } { B } = \dfrac { 1 } { 30 }$ ------- (1)

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B and C can do it in 24 days

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➠ $\sf \dfrac { 1 } { B} + \dfrac { 1 } { C} = \dfrac { 1 } { 24 }$ ------ (2)

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C and A required 40 days

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➠ $\sf \dfrac { 1 } { C} + \dfrac { 1 } { A} = \dfrac { 1 } { 40 }$ ------ (3)

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⟮ Adding equation (1) , (2) & (3) ⟯

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➜ $\footnotesize{ \sf \dfrac { 1 } { A } + \dfrac { 1 } { B } + \dfrac { 1 } { B} + \dfrac { 1 } { C} + \dfrac { 1 } { C} + \dfrac { 1 } { A} = \dfrac { 1 } { 30 } + \dfrac { 1 } { 24 } + \dfrac { 1 } { 40 }}$

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➜ $\sf \dfrac { 2 } { A } + \dfrac { 2 } { B } + \dfrac { 2} { C } = \dfrac { 5 + 3 + 4 } { 120 }$

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➜ $\sf 2 (\dfrac { 1 } { A } + \dfrac { 1 } { B } + \dfrac { 1 } { C }) = \dfrac { 12 } { 120 }$

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➜ $\sf 2(\dfrac { 1 } { A } + \dfrac { 1 } { B } + \dfrac { 1 } { C }) = \dfrac { 1 } { 10 }$

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➜ $\sf \dfrac { 1 } { A } + \dfrac { 1 } { B } + \dfrac { 1 } { C } = \dfrac { 1 } { 20}$ ------- (4)

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➨ A + B + C = 20

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Hence working together they can finish the work in 20 days

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Now ,

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From (1)

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➠ $\sf \dfrac { 1 } { A } = \dfrac { 1 } { 30 } - \dfrac { 1 } { B }$ ----- (5)

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From (2)

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➠ $\sf \dfrac { 1 } { C} = \dfrac { 1 } { 24 } - \dfrac { 1 } { B}$ ------ (6)

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⟮ Putting the values got from (5) & (6) to (4) ⟯

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➜ $\sf \dfrac { 1 } { A } + \dfrac { 1 } { B } + \dfrac { 1 } { C } = \dfrac { 1 } { 20}$

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➜ $\sf \dfrac { 1 } { 30 } - \dfrac { 1 } { B } + \dfrac { 1 } { B } + \dfrac { 1 } { 24 } - \dfrac { 1 } { B} = \dfrac { 1 } { 20}$

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➜ $\sf \dfrac { 1 } { 30 } + \dfrac { 1 } { 24 } - \dfrac { 1 } { B } = \dfrac { 1 } { 20 }$

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➜ $\sf \dfrac { 1 } { 30 } + \dfrac { 1 } { 24 } -\dfrac { 1 } { 20 } = \dfrac { 1 } { B }$

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➜ $\sf \dfrac { 1 } { B } = \dfrac { 4 + 5 - 6 } { 120 }$

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➜ $\sf \dfrac { 1 } { B } = \dfrac { 3} { 120 }$

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➜ $\sf \dfrac { 1 } { B } = \dfrac { 1} { 40 }$

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➨ B = 40 days ------- (7)

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• Hence B can finish the work alone in 40 days

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⟮ Putting B = 40 from (7) to (1) ⟯

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➜ $\sf \dfrac { 1 } { A } + \dfrac { 1 } { B } = \dfrac { 1 } { 30 }$

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➜ $\sf \dfrac { 1 } { A } + \dfrac { 1 } { 40} = \dfrac { 1 } { 30 }$

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➜ $\sf \dfrac { 1 } { A } = \dfrac { 1 } { 30 } - \dfrac { 1 } { 40}$

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➜ $\sf \dfrac { 1 } { A } = \dfrac { 4 - 3 } { 120 }$

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➜ $\sf \dfrac { 1 } { A } = \dfrac { 1} { 120 }$

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➨ A = 120 days -------- (8)

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• Hence A can finish the work alone in 120 days

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⟮ Putting the values of A & B from (7) & (8) to (4) ⟯

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➜ $\sf \dfrac { 1 } { A } + \dfrac { 1 } { B } + \dfrac { 1 } { C } = \dfrac { 1 } { 20}$

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➜ $\sf \dfrac { 1 } { 120} + \dfrac { 1 } { 40} + \dfrac { 1 } { C} = \dfrac { 1 } { 20}$

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➜ $\sf \dfrac { 1 } { C} = \dfrac { 1 } { 20} - \dfrac { 1 } { 120} - \dfrac { 1 } { 40}$

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➜ $\sf \dfrac { 1 } { C } = \dfrac { 6 - 1 - 3 }{ 120 }$

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➜ $\sf \dfrac { 1 } { C } = \dfrac { 2}{ 120 }$

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➜ $\sf \dfrac { 1 } { C } = \dfrac { 1}{ 60 }$

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➨ C = 60 days

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• Hence C alone can finish the work in 60 days