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The sum of 49 consecutive integers is 7^5. What is their median?

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The sum of 49 consecutive integers is 7^5. What is their median?  [#permalink]

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New post 21 Mar 2019, 04:44
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A
B
C
D
E

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The sum of 49 consecutive integers is 7^5. What is their median?

(A) 7
(B) 7^2
(C) 7^3
(D) 7^4
(E) 7^5
Spoiler: OA

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The sum of 49 consecutive integers is 7^5. What is their median?  [#permalink]

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New post Updated on: 21 Mar 2019, 08:53
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Bunuel wrote:
The sum of 49 consecutive integers is 7^5. What is their median?

(A) 7
(B) 7^2
(C) 7^3
(D) 7^4
(E) 7^5



49 is an odd number. thus median = mean.

Sum is given.

Average/median: \(\frac{7^5}{49}\)

\(\frac{7^5}{7^2}\)

\(7^3\)

C is the correct answer.

Originally posted by KSBGC on 21 Mar 2019, 05:43.
Last edited by KSBGC on 21 Mar 2019, 08:53, edited 2 times in total.
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Re: The sum of 49 consecutive integers is 7^5. What is their median?  [#permalink]

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New post 21 Mar 2019, 08:45
Bunuel wrote:
The sum of 49 consecutive integers is 7^5. What is their median?

(A) 7
(B) 7^2
(C) 7^3
(D) 7^4
(E) 7^5


median of consecutive sequence=sum of sequence/number of terms
7^5/7^2=7^3
C
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Re: The sum of 49 consecutive integers is 7^5. What is their median?  [#permalink]

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New post 21 Mar 2019, 09:24
Bunuel wrote:
The sum of 49 consecutive integers is 7^5. What is their median?

(A) 7
(B) 7^2
(C) 7^3
(D) 7^4
(E) 7^5


solve the given information
7^5/7^2
IMO C
7^3
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Re: The sum of 49 consecutive integers is 7^5. What is their median?  [#permalink]

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New post 21 Mar 2019, 09:30
For consecutive numbers, median = mean

Mean = Sum of all integers / number of integers

Mean = \(7^5\)/49 = \(7^5\)/\(7^2\) = \(7^3\)

Answer C.
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Re: The sum of 49 consecutive integers is 7^5. What is their median?  [#permalink]

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New post 21 Mar 2019, 09:59
In case of consecutive numbers(and evenly spaced sets), mean=median.

So here, Median=Mean =(sum/number of terms)=(7^5/49)=(7^5/7^2)
=7^(5-3)=7^2.

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Re: The sum of 49 consecutive integers is 7^5. What is their median?  [#permalink]

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New post 24 Mar 2019, 18:18
Bunuel wrote:
The sum of 49 consecutive integers is 7^5. What is their median?

(A) 7
(B) 7^2
(C) 7^3
(D) 7^4
(E) 7^5


In a set of evenly spaced integers, the median is equal to the average.

Since average x quantity = sum, we see that average = median = sum/quantity, thus:

Median = 7^5/49 = 7^5/7^2 = 7^3

Answer: C
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Re: The sum of 49 consecutive integers is 7^5. What is their median?   [#permalink] 24 Mar 2019, 18:18
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