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The average (arithmetic mean) of the values in d : Quantitative Comparison Questions

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Re: The average (arithmetic mean) of the values in d [#permalink]
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Re: The average (arithmetic mean) of the values in d [#permalink]
can anyone explain the answer please?
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Re: The average (arithmetic mean) of the values in d [#permalink]
can anyone tell if the given method is correct and how to proceed further? I am stuck at the last step with simplifying the fraction
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Re: The average (arithmetic mean) of the values in d [#permalink]
Given that The average (arithmetic mean) of the values in data set R is m. Data set S consists of the values in data set R and the value k. and the average of the values in data set S is m + a. where a > 0. Data set T consists of the values in data set S and the value k.

Let data set R has x values
Given that average of the values in data set R is m
=> Sum of values in data set R = Average * x = m*x = mx

=> Data Set S = {R ,k} with mean m + a
Sum of values in data set S = Mean * (x+1) [ as data set R had x values ]
= (m+a) * (x+1) = Sum of values in data set R + k = mx + k
=> mx + ax + m + a = mx + k
=> k = ax + m + a

=> Data Set T = {S ,k}
Sum of values in data set T = Sum of values in data set S + k = mx + k + k = mx + 2*(ax + m + a) = mx + 2ax + 2m + 2a

The average of the values in data set T = \(\frac{Sum }{ x + 2}\) = \(\frac{mx + 2ax + 2m + 2a }{ x + 2}\) = \(\frac{mx + 2ax + 2m + 2a + 2a - 2a }{ x + 2}\)
= \(\frac{mx + 2ax + 2m + 4a - 2a }{ x + 2}\) = \(\frac{x*(m + 2a) + 2*(m + 2a) - 2a }{ x + 2}\) = \(\frac{(x + 2)*(m + 2a)}{ x + 2}\) - \(\frac{2a }{ x+2}\) = m + 2a - \(\frac{2a }{ x+2}\)
Now a and x are positive
=> - \(\frac{2a }{ x+2}\) will be negative
=> m + 2a - \(\frac{2a }{ x+2}\) < m + 2a

Clearly, Quantity A(m + 2a - \(\frac{2a }{ x+2}\)) < Quantity B(m + 2a)

So, Answer will be B
Hope it helps!

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Re: The average (arithmetic mean) of the values in d [#permalink]
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