The attached chart graphs the average attendance per game for a certai

Question

https://gmatclub.com/forum/download/file.php?mode=view&id=68498 The attached chart graphs the average attendance per game for a certain sports franchise for 5 seasons, represented as percentage of stadium capacity. The stadium’s capacity was increased from 60,000 seats to 90,000 seats before the 2004 season. 1. The franchise’s highest average attendance occurred in GRAPH_1_PLACEHOLDER . 2. The highest total attendance for a season (the sum of individual game attendances) occurred in GRAPH_2_PLACEHOLDER . The Last question based on this stem can be found here. 7.jpg

Shankey94 · Answer

The attached chart graphs the average attendance per game for a certain sports franchise for 5 seasons, represented as percentage of stadium capacity. The stadium’s capacity was increased from 60,000 seats to 90,000 seats before the 2004 season.

1. The franchise’s highest average attendance occurred in ____________.

A. 2002
B. 2003
C. 2004
D. 2005

We got to compare highest average attendance (in absolute numbers)
Now for years before 2004 and after 2003 we can compare directly via avg percent attendance as the base (total capacity) is same. Hence, we need to compare between 2003 and 2005
for 2003: Avg attendance = ~82%
Avg attendance in absolute numbers: 0.82*60000 = ~49000

for 2005: Avg attendance = ~78%
Avg attendance in absolute numbers: 0.78*90000 = ~72000
D

2. The highest total attendance for a season (the sum of individual game attendances) occurred in ____________.

A. 2002
B. 2003
C. 2004
D. It cannot be determined

To calculate the total absolute attendance we will need the total number of games played in the season. Without this information we cannot compute the total attendance for a season.
Although we can compare the total attendance among different years based on avg attendance:
from 2002
D

3. From 2003 to 2004, the average attendance per game ____________.

A. Increased by more than 5%
B. Decreased by more than 5%
C. Stayed approximately the same

%Increase in attendance = (Attendance in 2004-Attendance in 2003)/Attendance in 2003
Attendance in 2003 = %Attendance*capacity
 = (0.58*90000-0.82*60000)/0.82*60000
~6%
hence A

UtkarshAnand · Answer

1) Franchise's highest average attendance occurred in: ANSWER D.

Approach:  As we are given, that stadium capacity was increased from 60,000 to 90,000 prior to 2004 game. We can segregate the years into 2 blocks, one block of years 2001, 2002, 2003 and the other block of years 2004 and 2005. Now amongst both the blocks we will select the year with highest average attendance giving us 2 final contenders. Until this step visual analysis can be performed. Now we will calculate the values for the final 2 contenders and the year with the highest value will be our answer.

Approach in action: We can see from the graph that out of 1st block of years (2001, 2002 and 2003) having stadium capacity 60,000, the year with highest average attendance is 2003. Now repeating the same exercise for 2nd block of years (2004, 2005) having stadium capacity 90,000, the year with highest average attendance is 2005. So, we have years 2003 and 2005 as final contenders.
 
 2003: 82.5% of 60,000 = 49,500.
2005: 77.5% of 90,000 = 69,750.
 
Thus,  franchise's highest average attendance occurred in the year 2005.

2) The highest total attendance for a season (the sum of individual game attendances) occurred in: ANSWER D
 
As we don't know the number of games played in a season, we cannot find the sum of individual game attendances. Hence, the year for the highest total attendance for a season  cannot be determined.

3) From 2003 to 2004, the average attendance per game stayed approximately the same. ANSWER D. 
 2003: 82.5% of 60,000 = 49,500.
2004: 55% of 90,000 = 49,500. 
Hence, the figures of average attendance per game for 2003 and 2004 stayed the same.

Kushchokhani · Answer

Stadium's capacity from 2001 to 2003=60,000 ---(i)
Stadium's capacity from 2004 to 2005=90,000 ---(ii)

1. The franchise’s highest average attendance occurred in 2005.
Since 2003 has the highest average attendance % among equation (i) and 2005 has the highest average attendance % among equation (ii), we just need to compare these 2 years to arrive at the correct answer.
2003- ~82% of 60,000=49,200
2005- ~78% of 90,000=70,200
Hence, answer is D

2. The highest total attendance for a season (the sum of individual game attendances) occurred in
Neither the bar chart nor the info below it provide anything to determine answer to this ques. Also, the highest ans to ques 1 i.e. 2005 is not an option for this ques. Further, highest average may not lead us to conclude about highest total.
Hence, answer is D

3. From 2003 to 2004, the average attendance per game stayed approximately the same
2003- ~82% of 60,000=49,200
2004- ~57% of 90,000=51,300
Increase in 2004 from 2003=4.27% which is less than 5%
Hence, answer is C

MiloMia · Answer

1. The franchise’s highest average attendance occurred in ____________.

D. 2005

Capacity: 60,000

2001 65% 39k

2002 75% 45k

2003 81% 48.6k

Capacity: 90,000

2004 55% 49.5k

2005 79% 71k

2. The highest total attendance for a season (the sum of individual game attendances) occurred in ____________.

D. It cannot be determined

- Do not know how many games are in each season

3. From 2003 to 2004, the average attendance per game ____________.

B. Decreased by more than 5%

49.5k-48.6k/48.6k = 1.82%

Catman · Answer

Let average attendance (% of capacity) in the year 2001,2002,2003,2004,2005 be (65%,75%,82.5%,55%,78%) as per the approximate values from graph.
Now the total stadium capacity are 60,000 in the years 2001,2002,2003 and 90,000 in the years 2004,2005.

Calculating the average attendance by multiplying the capacity with the average attendance we get average attendance in respective years.
2001= 65% * 60,000 = 39,000
2002=75% * 60,000 = 45,000
2003=82.5% * 60,000 = 49,500
2004=55% * 90000 = 49,500
2005=78% * 90,000 = 70,200

Thus, Highest average attendance is in the year 2005.
The highest total attendance can't be determined as we don't know how many individual games were played in 1 season.
In 2003 and 2004 the average attendance per game is almost same.

Thus, 1D, 2D and 3C IMO.

Arm007 · Answer

1Q: 
Lets split the years into two half
pre-expansion: 2001-2003
The highest avg attendance % during this time is around 80-85% in 2003

Post-expansion: 2004-2005
The highest avg attendance % during this time is around 75-80% in 2005

Note that the question asks about average attendance and not for average attendance %.
Now comparing between 2003 and 2005: 
The avg number of people attending a game in 2003 = 60000*0.8 to 60000*0.85 = 48000 to 51000 
The avg number of people attending a game in 2005 = 90000*0.75 to 90000*0.80 = 67500 to 72000

Hence the highest average attendance occurred in 2005.  Option D

2Q:
Since the number of games per season need not be constant, we won't be able to determine a solution for this.

For ex: 
2003 season could have had 10 matches, while 2005 season had only 1 match. So net attendance will be higher in 2003 than 2005. But if both the season had equal matches, the net attendance in 2005 would be higher.

Hence  Option D .

3Q:
Now comparing between 2003 and 2004: 
The average attendance % in 2003 = 80-85%
The average attendance % in 2004 = 55-60%
The avg number of people attending a game in 2003 = 60000*0.8 to 60000*0.85 = 48000 to 51000 
The avg number of people attending a game in 2004 = 90000*0.55 to 90000*0.60 = 49500 to 54000

If we take average of range in 2003 = (48000+51000)/2 = 49500
If we take average of range in 2004 = (49500+54000)/2 = 51725

% increase of attendance in 2004 = (51725-49500)*100/49500 = ~4.5%

Since 4.5% is not greater than 5%, we choose  Option C .

JDLaw · Answer

Lets make an average occupancy table
2001=70% of 60000= 42000
2002=75% of 60000= 45000
2003=85% of 60000= 51000
2004=55% of 90000= 49500
2005=80% of 90000= 72000

1) From created table, Answer-D
2) No value to indicate number of individual games and their attendance values, so no way to calculate actual total value. Therefore Ans-D
3) The values are approximately same as seen from table, if we were to take option value of 5% change, The 2004 value would be approx 53500 in case of increase and 48500 in case of dec. So it is safe to say the values remain constant. The variation in our made table is due to not having the exact average % value of attendance. Answer-C

Archit3110 · Answer

given info before 2004 games were 60,000 and after 90,000
  1. The franchise’s highest average attendance occurred in ____________.

A. 2002 : 42,000
B. 2003 : 49,200
C. 2004: 52,200
 D. 2005: 71,100

2. The highest total attendance for a season (the sum of individual game attendances) occurred in ____________.

A. 2002
B. 2003
C. 2004
  D. It cannot be determined   
given is the info about the average attendance per game for a certain sports franchise 
 so asked info cannot be determined

3. From 2003 to 2004, the average attendance per game ____________.

A. Increased by more than 5%  
B. Decreased by more than 5%
C. Stayed approximately the same

2003 games attendance 49,200
2004 attendance ; 52,200
% change ; 3000/49200 ; ~6%
  A. Increased by more than 5%

rajtare · Answer

Q1: we have two big bar graph one in 2003, and another in 2005. Let's evaluate each of them

2003: 82% of 60,000 = 49,200
2005: 78% of 90,000 = 70,200

hence answer is option D

Q2: Average = Total individual attendance in each game/ Total number of games

We have Average, but since we don't have total number of games played in each season hence we cannot determine the answer for question 2

Q3:
2003: 82% of 60,000 = 49,200
2004: 47% of 90,000 = 42,300

Which is approximately the same hence Option C

av1901 · Answer

Stadium's capacity till 2003 season = 60,000
Stadium's capacity in 2004 and 2005 season = 90,000

1. The franchise’s highest average attendance occurred in ____________.

A. 2002
B. 2003
C. 2004
D. 2005

2002: 75% of 60,000 = 45,000 
 2003: 82% of 60,000 = 50,000 (approximately) 
 2004: 55% of 90,000 = 49,500 
  2005: 78% of 90,000 = 70,000 (approximately)

2. The highest total attendance for a season (the sum of individual game attendances) occurred in ____________.

A. 2002
B. 2003
C. 2004
D. It cannot be determined

The chart represents the average attendance per game for each season from 2001 to 2005 (both inclusive). Now, in order to calculate total attendance for any season, we need to know the number of games played in that particular season (because Average = Sum/Number) but since we do not know the number of games played in any season, we cannot determine the highest TOTAL attendance for any season, as number of games played can be different in different seasons.

Answer: D (It cannot be determined)

3. From 2003 to 2004, the average attendance per game ____________.

A. Increased by more than 5%
B. Decreased by more than 5%
C. Stayed approximately the same

Average attendance in 2003 = 82% of 60,000 = 50,000 (approximately)
Average attendance in 2004 = 55% of 90,000 = 49,500 (approximately)

As we can see, that both have stayed approximately the same

Answer: C (Stayed approximately the same)

pintukr · Answer

For the years,

Attendance (%) Attendance Nos) 
2001 (60000) 67% 40200
2002 (60000) 75% 45000
2003 (60000) 82% 49200
2004 (90000) 58% 52200
2005 (90000) 78% 70200

1. The franchise’s highest average attendance occurred in ____________.

A. 2002
B. 2003
 C. 2004

(C) is the CORRECT answer

D. 2005

2. The highest total attendance for a season (the sum of individual game attendances) occurred in ____________.

A. 2002
B. 2003
C. 2004
 D. It cannot be determined

Only average attendance (in %) pr0vided per game; no information about games played every year
Hence, CANNOT BE DETERMINED..(D) is the CORRECT answer

3. From 2003 to 2004, the average attendance per game ____________.

A. Increased by more than 5% 
B. Decreased by more than 5%
C. Stayed approximately the same

As per the data above , increase is by more than 5% (for yr 2003 to 2004)
(A) is the CORRECT answer

Sajjad1994 · Answer

Hello Everyone!

Here is the OA to this question:

1. D
2. D
3. C

The same is posted in the question's main thread.

Sajjad1994 · Answer

Official Explanation

While it’s tempting to use the height of the bars to choose 2003, it is important to note that stadium capacity went from 60,000 to 90,000 before 2004. 2005 has a lower percentage of capacity, but ~75% of 90,000 is more than ~80% of 60,000.

Answer: D

• It cannot be determined.  We know the average attendance per game, but we don’t know how many games took place in a given season. 2005 has the highest per game attendance, but if only 3 games were played, it would have a lower total attendance than 2003 if 10 games were played that season.

Answer: D

• Stayed approximately the same.  Again, we must avoid visual temptation. 2003 saw ~80% of 60,000, and 2004 saw ~55% of 90,000, so average attendance per game was actually constant.

Answer: C

trishala98 · Answer

But if the question stem is defining the average attendance as % of stadium capacity, and the question is asking us about average attendance, shouldn't the answer be 2003?

cheshire · Answer

If you read the question it says the stadium capacity was increased in 2004

The attached chart graphs the average attendance per game for a certai

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