A CAN COMPLETE A PIECE OF WORK IN 10 DAYS, B IN 15 DAYS AND C IN 20 DAYS. A AND C WORKED TOGETHER FOR 2 DAYS AND THEN A WAS REPLACED BY B. IN HOW MANY DAYS, ALTOGETHER WAS THE WORK COMPLETED??

Question

Answer 1

Altogether the work is done in 8 days
first you have to find the LCM of 10,15 and 20 i.e 60
then find out the efficiency i. e 6,4,3 repectively
so firstly A and C work together that means (A+C) in 1 day is 9 work i.e 6+3 so in two days 9*2=18 work
total work is the LCM which is 60
so 18 work is subtracted from 60 so 60-18=42
then A was replace by B so the next we have to find out the (B+C)=4+3=7 which is 1 day work
so finally 42/7= 6 days
total required days is 6+2=8 days

Answer 2

Altogether the work is done in 8 days
first you have to find the LCM of 10,15 and 20 i.e 60
then find out the efficiency i. e 6,4,3 repectively
so firstly A and C work together that means (A+C) in 1 day is 9 work i.e 6+3 so in two days 9*2=18 work
total work is the LCM which is 60
so 18 work is subtracted from 60 so 60-18=42
then A was replace by B so the next we have to find out the (B+C)=4+3=7 which is 1 day work
so finally 42/7= 6 days
total required days is 6+2=8 days

Answer 3

6 days

Answer 4

6 days

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A CAN COMPLETE A PIECE OF WORK IN 10 DAYS, B IN 15 DAYS AND C IN 20 DAYS. A AND C WORKED TOGETHER FOR 2 DAYS AND THEN A WAS REPLACED BY B. IN HOW MANY DAYS, ALTOGETHER WAS THE WORK COMPLETED??