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# Suppose that the average (arithmetic mean) of ?, ?, and ? is ℎ, the

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Re: Suppose that the average (arithmetic mean) of ?, ?, and ? is ℎ, the [#permalink]
Average * Total = sum

Average (arithmetic mean) of 𝑎, 𝑏, and 𝑐 is ℎ: a + b + c = 3h ------ (1)

Average (arithmetic mean) of 𝑏, 𝑐, and 𝑑 is 𝑗: b + c + d = 3j : b + c = 3j - d ------ (2)

Average of 𝑑 and 𝑒 is 𝑘: d + e = 2k: d = 2k - e ----------(3)

From (1) and (2), we get a + 3j - d = 3h ------(4)

Plugging d = 2k - e in (4)

=> a + 3j - (2k - e) = 3h

=> a + e + 3j - 2k = 3h

=> a + e = 3h - 3j + 2k

Average of a and e : $$\frac{a + e}{ 2} = \frac{3h - 3j + 2k }{ 2}$$

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Re: Suppose that the average (arithmetic mean) of ?, ?, and ? is ℎ, the [#permalink]
Given: i) (a+b+c)/3 = h, ii) (b+c+d)/3 = j, iii) (d+e)/2 = k
From iii) d + e = 2k
d = 2k - e -> iv)
From ii) b + c + d = 3j
Put iv) here
b + c + 2k -e = 3j
b + c = 3j + e -2k -> v)
From i) a + b + c = 3h
Put v) here
a + 3j +e - 2k = 3h
a+e = 3h -3j + 2k
(a + e)/2 = (3h -3j + 2k)/2
Therefore ans is option c
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Re: Suppose that the average (arithmetic mean) of ?, ?, and ? is , the [#permalink]
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Re: Suppose that the average (arithmetic mean) of ?, ?, and ? is , the [#permalink]
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