Math, asked by raghunath05, 1 year ago

A painter can paint a building in 4 days and his apprentice can do it in 6 days. How long will it take them to paint this building if they work together on it except for 1 day when the painter is ill and the apprentice works alone? ​

Answers

Answered by HashtagNoName
3

Answer:

The area of the building painted by the painter in one day = 1/4

The area of a building painted by the apprentice = 1/6

Assuming that the painter is I'll on the first day,

The apprentice paints 1/6 of the building.

When they work together, they paint

1/6 + 1/4 = 5/12 of the building.

Let the number of days they work together be 'x'.

1/6 + (5/12) x = 1 [ As 1 represents the total area of the building]

(5/12)x = 1 - 1/6 = 5/6

x = (5/6)(12/5) = 2

So they together for two days.

Total number of days taken = 2 + 1 = 3

[ Because the apprentice worked alone for one day]

Answered by bhavya2543
1

Let the amount of total work be 1.

The work done by the painter in one day = 1/4 of the total work.

The work done by the apprentice in one day = 1/6 of the total work.

The work done by the two people together in one day = 1/4 + 1/6 = 5/12 of the total work.

1/6 of the total work is done by the apprentice alone for one day.

The remaining 5/6 of the total work is done by the painter and apprentice together for x days.

Therefore x = amount of work / work done in one day = (5/6)/(5/12) = 2 days.

Therefore, number of days taken to complete the work = 2 + 1 = 3 days [We will add 1 as there was only 1 painter painted the building 1 day ]

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