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

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Math Expert
Joined: 02 Sep 2009
Posts: 49300
If the average (arithmetic mean) of a and b is j, and the average of  [#permalink]

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24 Aug 2018, 01:26
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Difficulty:

15% (low)

Question Stats:

92% (01:24) correct 8% (01:34) wrong based on 24 sessions

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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

_________________
Manager
Joined: 18 Jul 2018
Posts: 192
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)
If the average (arithmetic mean) of a and b is j, and the average of  [#permalink]

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24 Aug 2018, 01:30
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.
Director
Joined: 31 Oct 2013
Posts: 577
Concentration: Accounting, Finance
GPA: 3.68
WE: Analyst (Accounting)
If the average (arithmetic mean) of a and b is j, and the average of  [#permalink]

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

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