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Re: Theater M has 25 rows with 27 seats in each row. How many of [#permalink]
20 Jun 2008, 01:38

1

This post received KUDOS

quantum wrote:

Theater M has 25 rows with 27 seats in each row. How many of the seats were occupied during a certain show? (1) During the show, there was an average (arithmetic mean) of 10 unoccupied seats per row for the front 20 rows. (2) During the show, there was an average (arithmetic mean) of 20 unoccupied seats per row for the back 15 rows.

Hope you like that one!

1 not suff, because you have info. only about the front 20 rows, the back 5 rows, you dont know whether it is occupied?

Re: Theater M has 25 rows with 27 seats in each row. How many of [#permalink]
20 Jun 2008, 01:49

Yes C cannot be the case - mathematically: Let 0-10 be o1 and v1 for occupied & Vacant respectively similary for 10-20 o2 & v2 and the 20-25 as o3 & v3

From Q stem: o1+v1=270, o2+v2=270 & o3+v3=135 Target: To find o1+o2+o3

Now frm st 1: v1+v2=200 -------insuffi

St2: v2+v3=300 ---Insuffi

St1+2: We end up with eqs: o1+o2=270 & o2+o3=105 ---- no way can we find the target hence E.

Re: Theater M has 25 rows with 27 seats in each row. How many of [#permalink]
20 Jun 2008, 02:03

quantum wrote:

But can someone explain in details why C is wrong?

Oh, just think that you can figure out!

C is combine 1 and 2. You see the intersection between group1(statement1) and group2(statement2) has NOT-fixed number of seats unoccupied. You can be sure only that the number of rows in the intersection is 5 rows, but statement1 say average per row is 10 unoccupied and statement2 says everage per row is 20 unoccupied. So the total number of seats unoccupied in 5 rows in the intersection can be 50 or 100 seats.

Re: Theater M has 25 rows with 27 seats in each row. How many of [#permalink]
20 Jun 2008, 13:02

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This post received KUDOS

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This post was BOOKMARKED

quantum wrote:

Theater M has 25 rows with 27 seats in each row. How many of the seats were occupied during a certain show? (1) During the show, there was an average (arithmetic mean) of 10 unoccupied seats per row for the front 20 rows. (2) During the show, there was an average (arithmetic mean) of 20 unoccupied seats per row for the back 15 rows.

Hope you like that one!

E

Attachments

theater.gif [ 9.82 KiB | Viewed 6920 times ]

_________________

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Re: Theater M has 25 rows with 27 seats in each row. How many of [#permalink]
21 Jun 2008, 01:17

quantum wrote:

Theater M has 25 rows with 27 seats in each row. How many of the seats were occupied during a certain show? (1) During the show, there was an average (arithmetic mean) of 10 unoccupied seats per row for the front 20 rows. (2) During the show, there was an average (arithmetic mean) of 20 unoccupied seats per row for the back 15 rows.

Hope you like that one!

Stem :- Total number is seats = 25 * 27 = 675 To find :- No of seats occupied 'x'.

Option1 ) Avg of 10 unoccupied seats in the front row Number of seats unocupied in the front 20 rows = 200 seats. Total number of occupied seats = 675 -200 = 475 . SUFF

Option 2 ) Avg of 10 unoccupied seats in the back 15 rows:- 20 Total number of seats unoccupied = 20 *15 = 300 Total number of occupied seats = 675 -300 = 375 . SUFF

Theater M has 25 rows with 27 seats in each row. How many of [#permalink]
08 Jan 2010, 13:23

1

This post was BOOKMARKED

Theater M has 25 rows with 27 seats in each row. How many of the seats were occupied during a certain show?

(1) During the show, there was an average (arithmetic mean) of 10 unoccupied seats per row for the front 20 rows. (2) During the show, there was an average (arithmetic mean) of 20 unoccupied seats per row for the back 15 rows.

Re: Theater M has 25 rows with 27 seats in each row. [#permalink]
26 Apr 2013, 21:53

Theater M has 25 rows with 27 seats in each row. How many of the seats were occupied during a certain show?

(1) During the show, there was an average (arithmetic mean) of 10 unoccupied seats per row for the front 20 rows. Clearly not sufficient, no info about other rows

(2) During the show, there was an average (arithmetic mean) of 20 unoccupied seats per row for the back 15 rows. Clearly not sufficient, no info about other rows

1+2)With 1 we know that there are 200 unoccupied in the first 20 rows, with 2 we know that there are 300 unoccupied in the last 15 rows. Because we don't know how many seats there are in the common part (rows 10-20) both statements are still not sufficient. E

Hope it's clear, let me know _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: Theater M has 25 rows with 27 seats in each row. How many of [#permalink]
11 Oct 2013, 05:32

sagarsabnis wrote:

Theater M has 25 rows with 27 seats in each row. How many of the seats were occupied during a certain show?

(1) During the show, there was an average (arithmetic mean) of 10 unoccupied seats per row for the front 20 rows. (2) During the show, there was an average (arithmetic mean) of 20 unoccupied seats per row for the back 15 rows.

Statement 1 - Clearly insfficient, we only have information about 20 rows Statement 2- Same here, we are missing information on other rows

Now both (1) and (2) together - So we have from the 1st statement that the total number of seats for the first 20 rows that were unnocupied totaled 200. And also the unnocpied for the back 15 rows were 300. But we have an overlap here between rows. If we don't have more information it will be impossible to know. Imagine a venn diagram and visualize the intersection between both. As long as you don't have the exact value for the intersection it will be impossible to know the total number of unnocupied seats and hence know what the occupied seats are.

Re: Theater M has 25 rows with 27 seats in each row. How many of [#permalink]
22 Nov 2014, 21:50

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Re: Theater M has 25 rows with 27 seats in each row. How many of [#permalink]
17 Jan 2015, 12:09

1

This post received KUDOS

E

Try seeing extremes.... statements didnt concretize the seating pattern so

divide rows into 3 parts first 10 + 10 + 5 rows

extreme 1: Middle set takes average of say 30 (higher side)

first 10 has average of <10 unoccupied (say 5..u can calculate actual possibility) + next 10 say abt 30 + last 5 has average of <20 unoccupied => would have a overall count on lower side

extreme 2: Middle set takes average of say 5

first 10 has average of >10 unoccupied (say 20..u can calculate actual possibility) + next 10 say abt 5 + last 5 has average of >20 unoccupied => would have a overall count on higher side

The logic is the middle set can act as cushion so we have a range of possibilities

Had their been no overlap....then we can find the total

Re: Theater M has 25 rows with 27 seats in each row. How many of [#permalink]
17 Jan 2015, 12:09

2

This post received KUDOS

E

Try seeing extremes.... statements didnt concretize the seating pattern so

divide rows into 3 parts first 10 + 10 + 5 rows

extreme 1: Middle set takes average of say 30 (higher side)

first 10 has average of <10 unoccupied (say 5..u can calculate actual possibility) + next 10 say abt 30 + last 5 has average of <20 unoccupied => would have a overall count on lower side

extreme 2: Middle set takes average of say 5

first 10 has average of >10 unoccupied (say 20..u can calculate actual possibility) + next 10 say abt 5 + last 5 has average of >20 unoccupied => would have a overall count on higher side

The logic is the middle set can act as cushion so we have a range of possibilities

Had their been no overlap....then we can find the total

Re: Theater M has 25 rows with 27 seats in each row. How many of [#permalink]
10 Mar 2015, 20:30

Expert's post

Hi solitaryreaper,

The type of situation that this DS question is based on is relatively rare, but the most efficient way to deal with it is to think about the "extreme" possibilities in the numbers.

I assume that you recognized that each Fact was insufficient on its own (since neither Fact gives you enough information about ALL 25 rows. When combining Facts, you have to recognize that there's an "overlap" - the "front 20 rows" mentioned in Fact 1 and the "back 15 rows" mentioned in Fact 2 include a series of 10 rows that are a part of BOTH calculations (remember that there are only 25 rows in TOTAL).

It's in those 10 overlapping rows that you can greatly impact the calculations. What if ALL of those seats were full? What if ALL of those seats were empty? In those two examples, you will find 2 different answers to the question. If you've comfortable drawing pictures and 'visualizing' the calculations, you can actually avoid most of the 'math' altogether.

Re: Theater M has 25 rows with 27 seats in each row. How many of [#permalink]
16 Mar 2015, 23:38

EMPOWERgmatRichC wrote:

Hi solitaryreaper,

The type of situation that this DS question is based on is relatively rare, but the most efficient way to deal with it is to think about the "extreme" possibilities in the numbers.

I assume that you recognized that each Fact was insufficient on its own (since neither Fact gives you enough information about ALL 25 rows. When combining Facts, you have to recognize that there's an "overlap" - the "front 20 rows" mentioned in Fact 1 and the "back 15 rows" mentioned in Fact 2 include a series of 10 rows that are a part of BOTH calculations (remember that there are only 25 rows in TOTAL).

It's in those 10 overlapping rows that you can greatly impact the calculations. What if ALL of those seats were full? What if ALL of those seats were empty? In those two examples, you will find 2 different answers to the question. If you've comfortable drawing pictures and 'visualizing' the calculations, you can actually avoid most of the 'math' altogether.

GMAT assassins aren't born, they're made, Rich

Thanks Rich, that helps!

I was looking for a logical approach to tackle such problems under 2 minutes. I think it's not always fruitful to approach such problems in pure mathematical way(because one might end up loosing precious time and energy) . That's where GMAT surprise you by throwing such problems that demand for a more logical approach. IMO such problems and methodologies should be discussed in a greater number !

Re: Theater M has 25 rows with 27 seats in each row. How many of [#permalink]
18 Mar 2015, 10:40

Expert's post

Hi Arun,

You've hit on an important point - knowing MORE than one way to approach questions can be quite helpful on Test Day. For most of your life, you were taught just 1 way to approach a question. When you took a test in school, you were supposed to use that 1 way and so you studied just that 1 way. On the GMAT though, many questions are designed to 'reward' a Test Taker for being is a critical thinker or pattern-matcher more than for being a mathematician. During your studies, it's important to practice more than one approach so that you have the flexibility on Test Day to decide which method is easiest/fastest.

Another thing worth noting is that you should NOT be trying to answer every Quant question within 2 minutes. The AVERAGE amount of time that you'll spend per question over the course of the entire Quant section is about 2 minutes, but your GOAL is actually to work in an efficient fashion. A question might require 1 minute or 3 minutes - and that's fine as long as you're not wasting time. If this question takes you 2.5 minutes, then that's not a problem.

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