Question

A and B together can do a piece of work in 8 days,B and C together in 10 days while C and A together in 6 days if they all work together the work will completed in...

$\rm \: Urgent \: Request...$
$\rm \: Help \: me \: out \: please...$



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Answer 1

(A + B)’s 1 day’s work = 1/8…….(i)

(B + C)’s 1 day’s work = 1/6…….(ii)

(C + A)’s 1 day’s work = 1/10…….(iii)

On adding all above equations

2(A + B + C)’s 1 day’s work = 1/8+1/6+1/10

= 15+20+12/120 = 47/120

∴ (A + B + C)’s 1 day’s work = 47/240

∴ A, B and C together will finish the work in

240/47 days

Answer 2

$\bf \: \underline{Given} :$ ↴

$\sf \: \rightarrow \: A \: and \: B \: together \: can \: do \: a \: piece \: of \: work \: in \: 8 \: days.$

$\sf \: \rightarrow \:B \: and \: C \: together \: in \: 10 \: days.$

$\sf \: \rightarrow C \: and \: A \: together \: in \: 6 \: days.$

$\bf \: \underline{To \: find} :$ ↴

$\sf \: \rightarrow How \: many \: days \: they \: will \: take \: if \: they \: work \: together.$

$\bf \: \green { \star} \: \underline{ So, \: Let's \: Start}\: \green{ \star}$

$\sf \: A \: and \: B's \:one \: day \: work = \dfrac{1}{8}$

$\sf \: \:B \: and \: C's \: one \: day \: work = \dfrac{1}{10}$

$\sf \: C \: and \: A's \: one \: day \: work = \dfrac{1}{6}$

$\bf \: Now,$

$\sf \: \: A, \: B \: and \: C's \: one \: day \: work = \dfrac{1}{8} + \dfrac{1}{10} + \dfrac{1}{6}$

$\sf \longrightarrow \: \dfrac{1}{8} + \dfrac{1}{10} + \dfrac{1}{6}$

$\sf \: Finding \: the \: common \: denominators.$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{2}}}&{\underline{\sf{\:\:8 - 10- 6 \:\:\:}}}\\ {\underline{\sf{2}}}&{\underline{\sf{\:\:4 - 5 - 3\:\:\:}}}\\ {\underline{\sf{2}}}&{\underline{\sf{\:\:2- 5- 3\:\:\:}}}\\ {\underline{\sf{3}}}& \underline{\sf{\:\:1 - 5 - 3 \:\:\:}} \\\underline{\sf{5}}&{\underline{\sf{\:\:1 - 5- 1\:\:\:}}} \\ \underline{\sf{}}&{\sf{\:\:1 - 1 - 1 \:\:\:}} \end{array}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

$\sf \: LCM = \underline{2 \times 2 \times2} \times 3 \times 5$

$\sf \: LCM = 8\times 3 \times 5 = 120$

$\longrightarrow \: \sf \dfrac{1}{8} \times \dfrac{15}{15} = \dfrac{15}{120}$

$\longrightarrow \: \sf \dfrac{1}{10} \times \dfrac{12}{12} = \dfrac{12}{120}$

$\longrightarrow \: \sf \dfrac{1}{6} \times \dfrac{20}{20} = \dfrac{20}{120}$

$\sf \: Now, \: we \: have \: to \: add : \dfrac{15}{120}, \: \dfrac{12}{120}, \: \dfrac{20}{120}$

$\sf \longrightarrow \: \dfrac{15}{120} + \: \dfrac{12}{120} + \: \dfrac{20}{120}$

$\sf \: Taking \: denominators \: as \: common.$

$\sf \longrightarrow \: \dfrac{15 + 12 + 20}{120}$

$\sf \: Adding \: the \: numerators.$

$\sf \longrightarrow \: \dfrac{47}{120}$

$\sf \: Thus \:time\: taken \: by \: A, \: B, \: C \: to \: finish \: the \: work = 1 \: \div \: \dfrac{47}{120}$

$\sf \: 1 \: \div \: \dfrac{47}{120} \: \: = \dfrac{120}{47}$

$\underline{ \boxed{ \pink{ \sf \: Answer = 5 \dfrac{4}{47} \: days}}}$

$\sf \: Thus \: A, \: B, \: C \: together \: will \: finish \: the \: work \: in \: 5 \dfrac{4}{47} \: days..$