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# Four integers have a mean (arithmetic average) of 36. Three of the int

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Math Expert
Joined: 02 Sep 2009
Posts: 50049
Four integers have a mean (arithmetic average) of 36. Three of the int  [#permalink]

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18 Apr 2018, 18:17
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15% (low)

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95% (01:04) correct 5% (00:34) wrong based on 57 sessions

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Four integers have a mean (arithmetic average) of 36. Three of the integers are 6, 30, and 64. If the fourth integer is 2K, then K =

(A) 20
(B) 22
(C) 24
(D) 40
(E) 44

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Re: Four integers have a mean (arithmetic average) of 36. Three of the int  [#permalink]

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18 Apr 2018, 20:39
Bunuel wrote:
Four integers have a mean (arithmetic average) of 36. Three of the integers are 6, 30, and 64. If the fourth integer is 2K, then K =

(A) 20
(B) 22
(C) 24
(D) 40
(E) 44

(6+ 30 + 64 + 2K)/4 = 36 => K= 22. Answer (B)
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Re: Four integers have a mean (arithmetic average) of 36. Three of the int  [#permalink]

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19 Apr 2018, 01:28
Four integers have a mean (arithmetic average) of 36. Three of the integers are 6, 30, and 64. If the fourth integer is 2K, then K =

(A) 20
(B) 22
(C) 24
(D) 40
(E) 44

Arithmetic Mean(AM) = Sum/No. of terms
Sum= 6+30+64+2K = 100 +2K
No. of terms = 4
AM= 36 = (100 +2k)/4
144 = 100+2k
44=2k
22=k
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Four integers have a mean (arithmetic average) of 36. Three of the int  [#permalink]

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19 Apr 2018, 11:41
Bunuel wrote:
Four integers have a mean (arithmetic average) of 36. Three of the integers are 6, 30, and 64. If the fourth integer is 2K, then K =

(A) 20
(B) 22
(C) 24
(D) 40
(E) 44

$$A*n = S$$ and $$\frac{Sum}{n}=A$$

Four integers have a mean (arithmetic average) of 36.

Three of the integers are 6, 30, and 64. Fourth integer = 2K

$$A = 36$$, $$n = 4$$

$$\frac{6 + 30 + 64 + 2K}{4} = 36$$
$$100 + 2K = 144$$
$$2K = 44$$
$$K = 22$$

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Re: Four integers have a mean (arithmetic average) of 36. Three of the int  [#permalink]

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20 Apr 2018, 10:24
30+6+64=100
100+2k / 4 =36
100+2k=144
2k=44
k=22
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Re: Four integers have a mean (arithmetic average) of 36. Three of the int  [#permalink]

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20 Apr 2018, 11:17
Bunuel wrote:
Four integers have a mean (arithmetic average) of 36. Three of the integers are 6, 30, and 64. If the fourth integer is 2K, then K =

(A) 20
(B) 22
(C) 24
(D) 40
(E) 44

6 + 30 + 64 + 2k = 36*4

Or, 100 + 2k = 144

So, 2k = 44

Or, k = 22, answer must be (B)
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Re: Four integers have a mean (arithmetic average) of 36. Three of the int  [#permalink]

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23 Apr 2018, 16:53
Bunuel wrote:
Four integers have a mean (arithmetic average) of 36. Three of the integers are 6, 30, and 64. If the fourth integer is 2K, then K =

(A) 20
(B) 22
(C) 24
(D) 40
(E) 44

Since the average of 4 integers is 36, the sum of the integers is 4 x 36 = 144. We can create the equation:

6 + 30 + 64 + 2K = 144

2K = 44

K = 22

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Re: Four integers have a mean (arithmetic average) of 36. Three of the int &nbs [#permalink] 23 Apr 2018, 16:53
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