The average age of the buildings on a certain city block is greater than 40 years old. If four of the buildings were built two years ago and none of the buildings are more than 80 years old, which of the folowing could be the number of buildings on the block?
A. 4
B. 6
C. 8
D. 10
E. 12
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talaangoshtari
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Hi talaangoshtari,
What is the source of this question? And have you transcribed the question correctly? I ask because it technically has 3 correct answers.
GMAT assassins aren't born, they're made,
Rich
What is the source of this question? And have you transcribed the question correctly? I ask because it technically has 3 correct answers.
GMAT assassins aren't born, they're made,
Rich
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talaangoshtari
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Hi Rich,
yes, this question has 3 correct answers, but I didn't understand the process of solving.
1. 4
2. 6
3. 8
4. 10
5. 12
A. 1,2
B. 2,3,4
C. 3,4
D. 3,4,5
E. None
yes, this question has 3 correct answers, but I didn't understand the process of solving.
1. 4
2. 6
3. 8
4. 10
5. 12
A. 1,2
B. 2,3,4
C. 3,4
D. 3,4,5
E. None
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Hi talaangoshtari,
For future posts, you should make sure to include the FULL prompt (including the ACTUAL answer choices to the question).
To answer this question, you can take an Algebraic approach, you can TEST THE ANSWERS or you can just think about the logic involved (and avoid most of the math altogether). Here's the 'logic' approach....
We're given a series of facts to work with:
1) The average age of a group of buildings is > 40.
2) Four of the buildings are 2 years old.
3) NONE of the buildings is over 80 years old.
We're asked for what COULD be the TOTAL number of buildings.
Since we already have 4 "young" buildings, if we made all of the other buildings as old as possible (80 years old), then having 4 really old buildings would bring up the average to above 40 (since the average of 2 and 80 is 41). Thus, 4 or more ADDITIONAL buildings would be needed to bring up the average to the proper level. When combined with the 4 "young" buildings, we could have 4+4, 4+6 or 4+8 buildings.
Final Answer: D
GMAT assassins aren't born, they're made,
Rich
For future posts, you should make sure to include the FULL prompt (including the ACTUAL answer choices to the question).
To answer this question, you can take an Algebraic approach, you can TEST THE ANSWERS or you can just think about the logic involved (and avoid most of the math altogether). Here's the 'logic' approach....
We're given a series of facts to work with:
1) The average age of a group of buildings is > 40.
2) Four of the buildings are 2 years old.
3) NONE of the buildings is over 80 years old.
We're asked for what COULD be the TOTAL number of buildings.
Since we already have 4 "young" buildings, if we made all of the other buildings as old as possible (80 years old), then having 4 really old buildings would bring up the average to above 40 (since the average of 2 and 80 is 41). Thus, 4 or more ADDITIONAL buildings would be needed to bring up the average to the proper level. When combined with the 4 "young" buildings, we could have 4+4, 4+6 or 4+8 buildings.
Final Answer: D
GMAT assassins aren't born, they're made,
Rich
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As Rich has noted, the problem at top has more than one correct answer choice and thus is flawed.
A better problem:
Four of the buildings were built two years ago, implying that the average age of these 4 buildings = 2.
To raise the average age of all of the buildings to a value greater than 40 -- while MINIMIZING THE TOTAL number of buildings -- the average age of the remaining buildings must be the MAXIMUM POSSIBLE VALUE (80).
When the correct answer choice is plugged in, the average age for all of the buildings will be greater than 40.
A: A total of 5 buildings, implying 4 buildings with an average age of 2 and 1 building with an average age of 80
Average for all 5 buildings = (4*2 + 80)/5 = 88/5 ≈ 18.
Too small.
B: A total of 6 buildings, implying 4 buildings with an average age of 2 and 2 buildings with an average age of 80
Average for all 6 buildings = (4*2 + 2*80)/6 = 168/6 = 28.
Too small.
C: A total of 7 buildings, implying 4 buildings with an average age of 2 and 3 buildings with an average age of 80
Average for all 7 buildings = (4*2 + 3*80)/7 = 248/7 ≈ 35.
Too small.
D: A total of 8 buildings, implying 4 buildings with an average age of 2 and 4 buildings with an average age of 80
Average for all 8 buildings = (4*2 + 4*80)/6 = 328/8 = 41.
Success!
The correct answer is D.
Algebraic approach:
SUM = (NUMBER)(AVERAGE).
As noted above:
Four of the buildings were built two years ago, implying that the average age of these 4 buildings = 2.
To raise the average age of all of the buildings to a value greater than 40 -- while MINIMIZING THE TOTAL number of buildings -- the average age of the remaining buildings must be the MAXIMUM POSSIBLE VALUE (80).
Sum of the ages for the four 2-year old buildings = 4*2 = 8.
Let y = the remaining buildings on the block.
If each of these buildings is 80 years old -- the maximum possible age -- then the sum of the ages for these y buildings = 80y.
Since the average age of the 4+y buildings must be greater 40, we get:
(8 + 80y/(4 + y) > 40
8 + 80y > 160 + 40y
40y > 152
y > 3.8.
Since the least possible integer value for y is 4, the least possible value for the total number of buildings = 4+y = 4+4 = 8.
A better problem:
We can PLUG IN THE ANSWERS, which represent the smallest possible value of x.talaangoshtari wrote:The average age of the buildings on a certain city block is greater than 40 years old. Four of the buildings were built two years ago, and none of the buildings is more than 80 years old. If there are a total of x buildings on the block, what is the smallest possible value of x?
A. 5
B. 6
C. 7
D. 8
E. 10
Four of the buildings were built two years ago, implying that the average age of these 4 buildings = 2.
To raise the average age of all of the buildings to a value greater than 40 -- while MINIMIZING THE TOTAL number of buildings -- the average age of the remaining buildings must be the MAXIMUM POSSIBLE VALUE (80).
When the correct answer choice is plugged in, the average age for all of the buildings will be greater than 40.
A: A total of 5 buildings, implying 4 buildings with an average age of 2 and 1 building with an average age of 80
Average for all 5 buildings = (4*2 + 80)/5 = 88/5 ≈ 18.
Too small.
B: A total of 6 buildings, implying 4 buildings with an average age of 2 and 2 buildings with an average age of 80
Average for all 6 buildings = (4*2 + 2*80)/6 = 168/6 = 28.
Too small.
C: A total of 7 buildings, implying 4 buildings with an average age of 2 and 3 buildings with an average age of 80
Average for all 7 buildings = (4*2 + 3*80)/7 = 248/7 ≈ 35.
Too small.
D: A total of 8 buildings, implying 4 buildings with an average age of 2 and 4 buildings with an average age of 80
Average for all 8 buildings = (4*2 + 4*80)/6 = 328/8 = 41.
Success!
The correct answer is D.
Algebraic approach:
SUM = (NUMBER)(AVERAGE).
As noted above:
Four of the buildings were built two years ago, implying that the average age of these 4 buildings = 2.
To raise the average age of all of the buildings to a value greater than 40 -- while MINIMIZING THE TOTAL number of buildings -- the average age of the remaining buildings must be the MAXIMUM POSSIBLE VALUE (80).
Sum of the ages for the four 2-year old buildings = 4*2 = 8.
Let y = the remaining buildings on the block.
If each of these buildings is 80 years old -- the maximum possible age -- then the sum of the ages for these y buildings = 80y.
Since the average age of the 4+y buildings must be greater 40, we get:
(8 + 80y/(4 + y) > 40
8 + 80y > 160 + 40y
40y > 152
y > 3.8.
Since the least possible integer value for y is 4, the least possible value for the total number of buildings = 4+y = 4+4 = 8.
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Solution:talaangoshtari wrote: ↑Thu Oct 15, 2015 6:57 amThe average age of the buildings on a certain city block is greater than 40 years old. If four of the buildings were built two years ago and none of the buildings are more than 80 years old, which of the folowing could be the number of buildings on the block?
A. 4
B. 6
C. 8
D. 10
E. 12
We are given that 4 buildings were built two years ago (that is, each of them has an age of 2) and the average age of all buildings is greater than 40 years old and no buildings are more than 80 years old. Now, let’s check each given Roman numeral choice. That idea is to see if we can show the remaining buildings (i.e., those that were not built two years ago) could have an average age of no more than 80 years old. (Note: In each of the inequalities below, x is the average age of the remaining buildings.)
I. 8 (Since 4 have an age of 2 each, there are 4 remaining buildings.)
[4(2) + 4x]/8 > 40
8 + 4x > 320
4x > 312
x > 78
We see that x is greater than 78; it could still be less than or equal to 80, such as 79 or 80. Therefore, 8 could be the number of buildings.
II. 11 (Since 4 have an age of 2 each, there are 7 remaining buildings.)
[4(2) + 7x]/11 > 40
8 + 7x > 440
7x > 432
x > 61.7
We see that x is greater than 61.7; it could definitely be less than or equal to 80. Therefore, 11 could be the number of buildings.
As we can see, the higher the number of buildings, the lower the average age of the remaining buildings. Therefore, we can safely conclude that 40 could also be the number of buildings.
Answer: E
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