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If the average (arithmetic mean) of 24 consecutive odd integers is 48,

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Bunuel
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If the average (arithmetic mean) of 24 consecutive odd integers is 48, [#permalink] New post   21 Aug 2018, 06:01
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If the average (arithmetic mean) of 24 consecutive odd integers is 48, what is the median of the 24 numbers?

A. 36
B. 47
C. 48
D. 49
E. 72
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C
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generis
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If the average (arithmetic mean) of 24 consecutive odd integers is 48, [#permalink] New post   21 Aug 2018, 08:47
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Bunuel wrote:
If the average (arithmetic mean) of 24 consecutive odd integers is 48, what is the median of the 24 numbers?

A. 36
B. 47
C. 48
D. 49
E. 72

I. Rule: In ANY evenly spaced set of integers, the median equals the mean

Median = mean = 48

ANSWER C

II. Test a smaller but similar sample

Use an even number of terms. Four consecutive odd integers: {1, 3, 5, 7}

Median = "middle value." If the number of terms is even, median =
average of the two middle terms: \(\frac{(3+5)}{2}=4\)

Mean = \(\frac{(1+3+5+7)}{4}=4\)

So median = mean. In prompt, mean = 48
Median also = 48

ANSWER C

Further, in any evenly spaced set --
median AND mean = \(\frac{FirstTerm+ LastTerm}{2}\)
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If the average (arithmetic mean) of 24 consecutive odd integers is 48, [#permalink] New post   21 Aug 2018, 06:17
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Bunuel wrote:
If the average (arithmetic mean) of 24 consecutive odd integers is 48, what is the median of the 24 numbers?

A. 36
B. 47
C. 48
D. 49
E. 72


As the number of elements are consecutive and is an Even number 24, hence the median should be equal to mean because the distribution is symmetric

Median will be the average of 12th & 13th term, which are odd numbers, hence their average will be an Even number.

Among the options only \(48\) is close to mean. Hence our answer

Option C

---------------------

Method 2: Using progression

\(Sum = \frac{n}{2}*[2a+(n-1)*d]\), here

\(Sum = 48*24\), \(n=24\), \(d=2\) we need \(a\)

\(=>48*24=\frac{24}{2}*(2a+23*2)\)

\(=>a=25\)

Now, \(T_{12}=a+(n-1)d=>a+11*2=25+22=47\)

therefore, \(T_{13}=47+2=49\)

Hence median \(= \frac{(47+49)}{2}=48\)

Option C
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Re: If the average (arithmetic mean) of 24 consecutive odd integers is 48, [#permalink] New post   21 Aug 2018, 11:35
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generis wrote:
Bunuel wrote:
If the average (arithmetic mean) of 24 consecutive odd integers is 48, what is the median of the 24 numbers?

A. 36
B. 47
C. 48
D. 49
E. 72

I. Rule: In ANY evenly spaced set of integers, the median equals the mean

Median = mean = 48

ANSWER C

II. Test a smaller but similar sample

Use an even number of terms. Four consecutive odd integers: {1, 3, 5, 7}

Median = "middle value." If the number of terms is even, median =
average of the two middle terms: \(\frac{(3+5)}{2}=4\)

Mean = \(\frac{(1+3+5+7)}{4}=4\)

So median = mean. In prompt, mean = 48
Median also = 48

ANSWER C

Further, in any evenly spaced set --
median AND mean = \(\frac{FirstTerm+ LastTerm}{2}\)



i join the above solution. :-) everything simple is unique :)
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Re: If the average (arithmetic mean) of 24 consecutive odd integers is 48, [#permalink] New post   23 Aug 2018, 09:52
Bunuel wrote:
If the average (arithmetic mean) of 24 consecutive odd integers is 48, what is the median of the 24 numbers?

A. 36
B. 47
C. 48
D. 49
E. 72


In an Arithmetic Progression, that is, in an evenly spaced sequence of terms,
Mean=Median
=> Median = 48.
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Re: If the average (arithmetic mean) of 24 consecutive odd integers is 48, [#permalink] New post   23 Aug 2018, 17:06
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Bunuel wrote:
If the average (arithmetic mean) of 24 consecutive odd integers is 48, what is the median of the 24 numbers?

A. 36
B. 47
C. 48
D. 49
E. 72


In any set of any number of consecutive odd integers, the average is always equal to the median. Since we are given that the average is 48, the median is also 48.

Answer: C
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Re: If the average (arithmetic mean) of 24 consecutive odd integers is 48, [#permalink] New post   29 Oct 2019, 06:19
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Re: If the average (arithmetic mean) of 24 consecutive odd integers is 48, [#permalink]
29 Oct 2019, 06:19
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