Question

A and Bdo a piece of work in 10 days, Band C do it in 20 days and C and A do it in 15 days. In how
many days will A, B and C finish it working together?​

Answer 1

Answer:

Suppose are eating candy

Take LCM of 10,15 and 20 = 60

A + B eat in 10 days means 6 candy/day

B+C eat it in 15 days means 4 candy/day

C+A eat it in 20 days means 3 candy/day

Now, add all with /day work.

2(A+B+C) = 13 candy/day

A+B+C =13/2 per day

Now put the value of A+B per day work we have,

6 + C= 13/2

Now equating we have

C=1/2 means C eats 1/2 candy per day and we have 60 candy to eat.

So,C will it it in 60/1/2 days = 120days

Step-by-step explanation:

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Answer 2

Answer:

If A, B and C are workers, then how quickly a unit of work is completed depends on individual speeds ignoring all other factors.

We are given the reciprocal of the combined speed for each pair. So in terms of speeds (unit work per day):

SA + SB =1/10

SC + SB = 1/15

SC + SA=1/20

Using basic algebra (substitution or subtraction):

SC = 1/120 or 1 Unit Work per 120 days, i.e. it would take C alone 120 days to do one piece of work.

And it follows that:

SB= 7/120 or 1 unit of work in 17.14 days

SA = 5/120 or 1 unit of work in 24 days.