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by tanviet » Mon Jun 08, 2009 5:37 pm
Theater M has 25 rows with 27 seats in each row. How many of the seats were occupied during a certain show?

a, During the show, there was an average (arithmatic mean) of 10 unoccupied seats per row for the front 20 rows

b, Dering the show, there was an average (arithmatic mean) of 20 unoccupied seats per row for the back 15 rows

pls help with this. this is a official question from retired test and need to be studied
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Source: — Data Sufficiency |

by Claret » Mon Jun 08, 2009 6:25 pm
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
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by mikeCoolBoy » Mon Jun 08, 2009 10:44 pm
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
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by Claret » Tue Jun 09, 2009 8:53 am
why E ? I cudnt get ur point here?
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by yogami » Tue Jun 09, 2009 11:12 am
Yeah OA please. I am visualizing E
200 or 800. It don't matter no more.
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by zuleron » Tue Jun 09, 2009 11:30 am
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
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by Claret » Tue Jun 09, 2009 11:39 am
Yep! got the point..

i wud alter my opinion now..

E it is!
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