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

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Joined: 02 Sep 2009
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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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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 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

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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
1
a+b = 2j

and c+d+e = 3k

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

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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:

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

=$$\frac{2j + 3k + j}{6}$$
= $$\frac{3j + 3k}{6}$$

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

=$$\frac{j + k}{2}$$

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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