Question

A, b and c can do a piece of work in 30, 45, and 90 days. In how many days can a alone do the work if he is assisted by b and c on every 4th day?

Answer 1

LCM of 30,45 and 90 is 180

So total work is 180 units

A=180/30=6units/day

B=180/45=4units/day

C=180/90=2units/day

A does the work alone for 3 days

So, for 3 days

A=6units×3=18units

A is assisted by B and C on 4th day,

Therefore total work on 4th day = 6+4+2=12units

Total work done by them in a set of 4 days = 12+18=30 units

Therefore to complete 180 units, 180/30=6 such sets of 4 days are needed.

6 sets of 4 days means 6×4=24 days

Therefore A alone can do the work if he is assisted by B and C on every 4th day in 24 days.

Answer 2

Answer: 24 days

Step-by-step explanation:

• Given A,B and C can do a work in 30,45 and 90 days.
• LCM of 30,45 and 90 is 180.
• So, there is a total of 180 units of work.
• Work done by A in a day=$\frac{180}{30}=6\ units$
• Work done by B in a day=$\frac{180}{45}=4\ units$
• Work done by B in a day=$\frac{180}{90}=2\ units$
• So, in a set of 4 days total work done by A,B and C =(6*4)(4+2)

                                                                                              =24+6

                                                                                              =30 units

• So, to complete 180 units of work number of sets of 4 days required=$\frac{180}{30}$

                      =6 units

• Total number of days required=6*4

                                                          =24 days

• Hence, the total number of days required is 24 days.

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