Answer: Option B
Let the total area of the tank be 'A' square metres, and the area of the plastered portion of the tank be 'x' square metres. Therefore, the remaining area to be plastered is (A - x) square metres.
Given that the workers completed 1/2 of the work in one day, therefore the plastered area on the first day is (1/2)*A.
The remaining area to be plastered after the first day is (A - (1/2)*A) = (1/2)*A.
Given that the workers completed 1/3rd of the remaining work on the next day, therefore the area plastered on the second day is (1/3)*((1/2)*A) = (1/6)*A.
The total area plastered is the sum of the areas plastered on both days, i.e.,
x = (1/2)*A + (1/6)*A = (2/3)*A.
The area remaining to be plastered is given as 60 square metres, i.e.,
A - x = 60.
Substituting the value of x from above, we get:
A - (2/3)*A = 60,
which gives us (1/3)*A = 60.
Therefore, the total area of the tank is:
A = 180 square metres.
Hence, the area of the tank that is to be plastered is:
(A - x) = A - (2/3)*A = (1/3)A = 60(3) = 180 square metres.
Therefore, option B, i.e., 180 sq. m is the correct answer.If you think the solution is wrong then please provide your own solution below in the comments section .