Math, asked by mace37, 7 months ago

a and b can do a peice of work in 30 days . b and c in 24 days , c and a in 40 days.how long will it take to complete altogether

Answers

Answered by Nemha
2

Answer:

Let Total work is = x

Given :

(A + B) takes 30 days to complete x work

=> work done by (A+B) in 1 day = x/30

(B+C) takes 24 days to complete x work

=> work done by (B+C) in 1 day = x/24

(A+C) takes 40 days to complete x work

=> work done by (A+C) in 1 day = x/40

work done by (A+B)+(B+C)+(A+C) in 1 day = x/30 + x/24 + x/40 = x(4+5+3)/120

work done by 2(A+B+C) in 1 day = 12x/120

= x/10

work done by (A+B+C) in 1 day = x/20

Time taken by (A+B+C) to complete the x work =

= {Total work / work done by (A+B+C) in 1 day}

= {x/x/20} = 20

Time taken by (A+B+C) to complete the total work(x) = 20 Days

Now, Calculating seperately work done by each

=>

Work done by A in 1 day =

= {work done by (A+B+C) in 1 day - work done by (B+C) in 1 day}

= {x/20 - x/24} = x{6–5}/120

= x/120

Time taken by A to complete the work

= (Total work/Work done by A in 1 Day)

= x/x/120 = 120 Days

Time taken by A to complete the work

= 120 Days

Work done by B in 1 day =

= {work done by (A+B+C) in 1 day - work done by (A+C) in 1 day}

= {x/20 - x/40} = x{2–1}/40

= x/40

Time taken by B to complete the work

= (Total work/Work done by B in 1 Day)

= x/x/40 = 40 Days

Time taken by B to complete the work

= 40 Days

Work done by C in 1 day =

= {work done by (A+B+C) in 1 day - work done by (A+B) in 1 day}

= {x/20 - x/30} = x{3–2}/60

= x/60

Time taken by C to complete the work

= (Total work/Work done by C in 1 Day)

= x/x/60 = 60 Days

Time taken by C to complete the work

= 60 Days

Now,

Time taken by (A+B+C) to complete the work = 20 Days

Time taken by A to complete the work = 120 Days

Time taken by B to complete the work = 40 Days

Time taken by C to complete the work = 60 Days

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