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If the average (arithmetic mean) of a and b is j, and the average of

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If the average (arithmetic mean) of a and b is j, and the average of  [#permalink]

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New post 24 Aug 2018, 01:26
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A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

77% (01:58) correct 23% (01:29) wrong based on 33 sessions

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If the average (arithmetic mean) of a and b is j, and the average of  [#permalink]

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New post 24 Aug 2018, 01:30
1
a+b = 2j

and c+d+e = 3k

\(\frac{a+b+c+d+e+j}{6}\) = \(\frac{3j+3k}{6}\) =\(\frac{k+j}{2}\).

C is the answer.
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If the average (arithmetic mean) of a and b is j, and the average of  [#permalink]

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New post 24 Aug 2018, 01:37
Bunuel wrote:
If the average (arithmetic mean) of a and b is j, and the average of c, d, and e is k, what is the average of a, b, c, d, e and j ?


A. (2k + 3j)/6

B. (k + 2j)/3

C. (k + j)/2

D. (2k + 3j)/5

E. (k + 2j)/4



Given ,

\(\frac{a + b}{2} = j\)
a + b = 2j

and

\(\frac{c + d + e}{3} = k\)
c+ d + e = 3k

Question:

\(\frac{a + b+ c + d + e + j}{6}\)

=\(\frac{2j + 3k + j}{6}\)
= \(\frac{3j + 3k}{6}\)

= \(\frac{3( j + k )}{6}\)

=\(\frac{j + k}{2}\)

The best answer is C.
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If the average (arithmetic mean) of a and b is j, and the average of   [#permalink] 24 Aug 2018, 01:37
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