Correct Option: B

Method 1 to solve this question.
Let us assume the numbers of days of tour before extending the tour are D and daily expenses is E.
According to question,
Total expenses on tour = per day expenses x total number of days
Total expenses on tour = E x D = ED
360 = ED
⇒ ED = 360
⇒ E = 360/D .................(1)
Again according to question,
After extending the tour by 4 days, Number of days of tour = D + 4
Expenses will be reduced by 3 rupees, then everyday expenses = E - 3
so total expenses = (D + 4) x (E - 3)
360 = (D + 4) x (E - 3)
(D + 4) x (E - 3) = 360 .............................................(2)
Put the value of E in Equation (2). we will get,
(D + 4) x (360/D - 3) = 360
⇒ (D + 4) x ( (360 - 3D)/D ) = 360
⇒ (D + 4) x ( (360 - 3D) ) = 360 x D
⇒ (D + 4) x (360 - 3D) = 360 x D
After multiplication by algebra law,
⇒ 360 x D - 3D x D + 4 x 360 - 4 x 3D = 360D
⇒ 360 x D - 3D + 1440 - 12D = 360D
⇒ - 3D + 1440 - 12D = 360D - 360 x D
⇒ - 3D + 1440 - 12D = 0
⇒ 3D - 1440 + 12D = 0
⇒ D - 480 + 4D = 0
⇒ D + 4D - 480 = 0
⇒ D + 24D - 20D - 480 = 0
⇒ D(D + 24) - 20(D + 24) = 0
⇒ (D + 24) (D - 20) = 0
Either (D + 24) = 0 or (D - 20) = 0
So D = - 24 or D = 20
But days cannot be negative so D = 20 days.

Method 2 to solve this question.
Let us assume the original day of tour is d days.
Given that his tour is extended for 4 days
Hence daily expenses per days = 360/(d + 4)
Therefore, According to question,
360/d - 360/(d + 4) = 3
⇒ (360(d + 4) - 360d)/d x (d + 4) = 3
⇒ (360(d + 4) - 360d)/d x (d + 4) = 3
⇒ 360(d + 4) - 360d = 3d x (d + 4)
⇒ 360d + 1440 – 360d = 3(d² + 4d)
⇒ 1440 = 3d² + 12d
⇒ 3d² + 12d – 1440 = 0
⇒ d² + 4d – 480 = 0
⇒ d² + 24d – 20d – 480 = 0
⇒ d(d + 24) – 20(d + 24) = 0
⇒ (d + 24)(d – 20) = 0
⇒ (d + 24) = 0 or (d – 20) = 0
? d = –24 or d = 20
Since days cannot be negative, d = 20
Hence his original duration of the tour is 20 days.