IMO C

statement 1 : total unoccupied seats in front 20 rows = 10*20

statement 2 : total unoccupied seats in last 15 rows = 20*15

common unoccupied

together total unoccupied seats = 300 + 200 - 100 = 400

25*27 - 400 = total occupied seats

pls post the ans as well

I think the answer should be E, here is my reasoning.

We can look at the problem as a Venn diagram, let's call

B15 = number of unoccupied seats in the back 15 rows = 300
F20 = number of unoccupied seats in the front 20 rows = 200

The intersection of this two groups is the number of unoccupied seats in the middle rows, I'll call it M10
So we can split the groups
B15 = B5 + M10 = 300 (1)
F20 = F10 + M10 = 200 (2)

subtracting (2) from (1)

B5 - F10 = 100

Now the total of unoccupied seats in the Theater is B5 + M10 + F10 or
B15 + F20 - M10

Think of these two possible configurations

B5 = 300 and F10 = 200 so the number of total unoccupied seats would be 500
B5 = 200 and F10 = 100 --> M10 = 100 so the number of total unoccupied seats would be 400

why E ? I cudnt get ur point here?

Yeah OA please. I am visualizing E

I'm thinking E too. I went for the Venn diagram too... there is no way to know the number of unoccupied seats in the intersection... or for that matter the number of unoccupied seats in the front 10 or back 5 coz all front 10 rows my be packed and still have an average of 10 unoccupied for the first 20 and the last 5 could be nearly empty and still and average of 15 unoccupied for the last 10

Yep! got the point..

i wud alter my opinion now..

E it is!