Number of rooms in the hotel N=100
Average = Occupancy per day/number of days= 80
No. of days the rooms were all occupied=?
1. Let x be the no. of days when the rooms were not all occupied.
80(30)=40x+100(30-x)
suff

2. Sales in the room that was all occupied= 5* sales in the rooms that were not all occupied.
x*p=5y*p
x=5y
==> x:y= 5:1
== insuff (no info on the rooms)


Answer is A

Anyone please explain...how option "D" is correct?

chetan2u, Bunuel, VeritasKarishma, nick1816, why is option D right?

suppose, for x days, all the rooms were occupied

Statement 1-

80*30= 100x + 40(30-x)

We can find x

Sufficient

Statement 2-

suppose an average of N rooms were occupied on days that the rooms weren’t all occupied and charge for each room=1(it will cancel out, so doesn't matter what you wanna assume

100*x= 5*N*(30-x)...(1)

Also,

100x+N*(30-x)= 80*30.....(2)

from(1) and (2)

120x = 2400

x=20

Sufficient

So we have 80*30=2400 rooms occupied in the month.
If charges per room is x, total sales is 2400x
1) An average of 40 rooms were occupied on days that the rooms weren’t all occupied.
We can find answer by weighted average method.
40.....80....100
So all occupied:total days= (80-40):(100-40)=2:3=20:30, as there are total 30 days.
So rooms were completely occupied on 20 days.
You could also make an algebraic equation.

(2) The sales of days that the rooms were all occupied is 5 times greater than the sales of days that the rooms were not all occupied.
If rooms were completely occupied for y days, the sales =100*y*x=100xy
So on the other days it is 2400x-100xy.
Now given that 100xy/(2400x-100xy)=5/1
100y=5(2400-100y)
600y=5*2400
y=20

D