Two persons A & B together can complete a work in 18 days. If A alone worked for 9 days and then B alone worked for 18 days, they would be left with 1/3rd of the entire work. How many days would A take to complete the whole work working alone?

Correct option is C

Information Given in Question:
A & B together can complete the work in 18 days.
A worked alone for 9 days
B worked alone for 18 days
After that, 1/3rd of the total work is still left.
We need to find in how many days A alone can complete the whole work.

Concept/Formula Used:
Let total work = LCM or any convenient multiple = 1 unit (for fractional calculations)
Work done = Time × Efficiency
Efficiency = 1 / time taken
If A’s efficiency = A, and B’s = B, then:
A + B = 1/18 (combined efficiency)

Detailed Explanation:
Let total work = 1 unit
Let A's 1 day work = a
Let B's 1 day work = b
From the first statement:
a + b = 1/18  (Equation 1)
Work done by A in 9 days = 9a
Work done by B in 18 days = 18b
Total work completed = 9a + 18b
Given: this work = 2/3 of the total work
So,
9a + 18b = 2/3  (Equation 2)
Solve Equation 1 and Equation 2
From Equation 1:
a + b = 1/18 => a = 1/18 – b
Substitute in Equation 2:
9(1/18 – b) + 18b = 2/3
=> (9/18 – 9b + 18b) = 2/3
=> (1/2 + 9b) = 2/3
=> 9b = 2/3 – 1/2
=> 9b = (4 – 3)/6 = 1/6
=> b = 1/54
Now from a + b = 1/18
=> a = 1/18 – 1/54
=> a = (3 – 1)/54 = 2/54 = 1/27
So, A alone can do 1/27 of the work in a day
=> A will complete the whole work in 27 days

Short Trick: 
Let efficiencies (units/day) of A & B be a & b respectively
18(a + b)=(9a+18b)/(2/3)
12a + 12b = 9a +18b
3a = 6b
a/b =2/1=2x/1x
Total work = 18(2x+x)= 54x
Required time = 54x/2x = 27 days