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06 Oct 2014, 07:30
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The average (arithmetic mean) of four distinct positive integers is 10. If the average of the smaller two of these four integers is 8, which of the following represents the maximum possible value of the largest integer?
A) 12
B) 14
C) 15
D) 16
E) 17
OE
Query
M13-16
A) 12
B) 14
C) 15
D) 16
E) 17
OE
Given: 0<a<b<c<d. Notice that a,b,c,d are distinct positive integers.
The average of four distinct positive integers is 10: a+b+c+d = 4*10 = 40;
The average of the smaller two of these four integers is 8: a+b = 2*8 = 6. So, 16+c+d = 40, which leads to c+d = 24.
We want to maximize d, so we should minimise c. The minimum value of c is 10, for a=7 and b=9. So, 10+d = 24 and d=14.
The average of four distinct positive integers is 10: a+b+c+d = 4*10 = 40;
The average of the smaller two of these four integers is 8: a+b = 2*8 = 6. So, 16+c+d = 40, which leads to c+d = 24.
We want to maximize d, so we should minimise c. The minimum value of c is 10, for a=7 and b=9. So, 10+d = 24 and d=14.
Query
Hi, can anyone explain why a=7 and b=9, please.
a and b are distinct positive integers.
a and b are distinct positive integers.
M13-16
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06 Oct 2014, 07:38
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goodyear2013
The average (arithmetic mean) of four distinct positive integers is 10. If the average of the smaller two of these four integers is 8, which of the following represents the maximum possible value of the largest integer?
A) 12
B) 14
C) 15
D) 16
E) 17
OE
Query
M13-16
A) 12
B) 14
C) 15
D) 16
E) 17
OE
Given: 0<a<b<c<d. Notice that a,b,c,d are distinct positive integers.
The average of four distinct positive integers is 10: a+b+c+d = 4*10 = 40;
The average of the smaller two of these four integers is 8: a+b = 2*8 = 6. So, 16+c+d = 40, which leads to c+d = 24.
We want to maximize d, so we should minimise c. The minimum value of c is 10, for a=7 and b=9. So, 10+d = 24 and d=14.
The average of four distinct positive integers is 10: a+b+c+d = 4*10 = 40;
The average of the smaller two of these four integers is 8: a+b = 2*8 = 6. So, 16+c+d = 40, which leads to c+d = 24.
We want to maximize d, so we should minimise c. The minimum value of c is 10, for a=7 and b=9. So, 10+d = 24 and d=14.
Query
Hi, can anyone explain why a=7 and b=9, please.
a and b are distinct positive integers.
a and b are distinct positive integers.
M13-16
Given: \(0 \lt a \lt b \lt c \lt d\). Notice that \(a, b, c\) and \(d\) are distinct positive integers.
The average of four distinct positive integers is 10: \(a+b+c+d=4*10=40\);
The average of the smaller two of these four integers is 8: \(a+b=2*8=16\). So, \(16+c+d=40\), which leads to \(c+d=24\).
We want to maximize \(d\), so we should minimize \(c\). The minimum value of \(c\) is 10, for \(a=7\) and \(b=9\). So, \(10+d=24\) and \(d=14\).
Answer: B.
As for your question: we have that a + b = 16. Since a and b are distinct integers, then a = b = 8 is not possible, so the least value of b is 9 (7 + 9 = 16). Hence the least value of c is 10.
Hope it's clear.
General Discussion
15 Jan 2016, 17:43
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Hi All,
This question is an example of a 'limit' question. To maximize the value of one variable, we will likely have to minimize the values of all of the others (while making sure to follow all of the 'restrictions' described in the prompt).
We're told that the AVERAGE of four DISTINCT POSITIVE INTEGERS is 10...
Since the average is 10, the SUM = 4(10) = 40. So to start, we know that we have 4 DIFFERENT POSITIVE integers that add up to 40.
Next, we're told that the average of the SMALLER two of these four integers is 8...
This means that the sum of those 2 integers is 2(8) = 16. Since the total sum is 40, the sum of the LARGER two integers is 40 - 16 = 24. Thus, these two integers are BOTH bigger than the smaller two integers AND they sum to 24.
We're asked which of the following represents the MAXIMUM possible value of the LARGEST integer.
To start, we have to take the smallest two integers and make them as small as possible AND make them DIFFERENT. Since the integers CANNOT BOTH be 8, one has to be 7 and the other has to be 9. This gives us....
7 9 _ _
The remaining two integers are BOTH greater than 9 and sum to 24. To maximize the largest value, we have to minimize the other value. That happens when the third integer is 10...
7 9 10 _
7+9+10 = 26, so the remaining (largest) integer is 14.
Final Answer:
GMAT assassins aren't born, they're made,
Rich
This question is an example of a 'limit' question. To maximize the value of one variable, we will likely have to minimize the values of all of the others (while making sure to follow all of the 'restrictions' described in the prompt).
We're told that the AVERAGE of four DISTINCT POSITIVE INTEGERS is 10...
Since the average is 10, the SUM = 4(10) = 40. So to start, we know that we have 4 DIFFERENT POSITIVE integers that add up to 40.
Next, we're told that the average of the SMALLER two of these four integers is 8...
This means that the sum of those 2 integers is 2(8) = 16. Since the total sum is 40, the sum of the LARGER two integers is 40 - 16 = 24. Thus, these two integers are BOTH bigger than the smaller two integers AND they sum to 24.
We're asked which of the following represents the MAXIMUM possible value of the LARGEST integer.
To start, we have to take the smallest two integers and make them as small as possible AND make them DIFFERENT. Since the integers CANNOT BOTH be 8, one has to be 7 and the other has to be 9. This gives us....
7 9 _ _
The remaining two integers are BOTH greater than 9 and sum to 24. To maximize the largest value, we have to minimize the other value. That happens when the third integer is 10...
7 9 10 _
7+9+10 = 26, so the remaining (largest) integer is 14.
Final Answer:
GMAT assassins aren't born, they're made,
Rich
