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The average (arithmetic mean) of a list of 5 numbers (arithmetic mean)

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The average (arithmetic mean) of a list of 5 numbers (arithmetic mean) [#permalink]

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New post 23 Nov 2017, 01:21
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[GMAT math practice question]

The average (arithmetic mean) of a list of 5 numbers (arithmetic mean) is 15. If 2 numbers are removed from list, is the average (arithmetic mean) of the 3 remaining numbers less than 15?

1) Each of the 2 numbers removed is greater than 15.
2) The average of the 2 numbers removed is greater than 15.
[Reveal] Spoiler: OA

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Re: The average (arithmetic mean) of a list of 5 numbers (arithmetic mean) [#permalink]

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New post 23 Nov 2017, 10:06
Average of 5 numbers is 15. This means their sum is = 15*5 = 75

(1) Each of the two numbers removed is greater than 15. So sum is reduced by a number greater than 30. Sum of remaining three numbers is thus less than (75-30) = 45. Sum of remaining three numbers is < 45. Thus average < 45/3 or average < 15. Sufficient.

(2) Average of two numbers removed > 15. Thus their sum > 30. This becomes same as previous statement.
Sufficient.

Hence D answer

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Re: The average (arithmetic mean) of a list of 5 numbers (arithmetic mean) [#permalink]

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New post 26 Nov 2017, 17:23
=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Assume that the five numbers are, \(a, b, c, d\), and \(e\). Then \(\frac{a + b + c + d + e}{5} = 15\), or \(a + b + c + d + e = 75\).

Since we have 5 variables (\(a,b,c,d\), and \(e\)) and 1 equation, E is most likely to be the answer. Thus, we should consider
both conditions together first.

Conditions 1) and 2):
Suppose that a and b are the numbers removed, and that a > 15 and b > 15. Then
\(\frac{( a + b )}{2} > 15\)
⟺\(a + b > 30\)
So,
\(c + d + e = 75 – ( a + b ) < 45\), since \(a + b > 30\). Thus,
\(\frac{c + d + e}{3}\) <\(\frac{45}{3}\) = \(15\),
and the answer is ‘yes’.
Therefore, both conditions are sufficient when applied together.

Since this question is a statistics question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
Assume that \(a\) and \(b\) are the numbers removed, and that \(a > 15\) and \(b > 15\). Then
\(a + b > 30\), and so \(c + d + e = 75 – ( a + b ) < 45\), since \(a + b > 30\).
Therefore,
\(\frac{( c + d + e )}{3}\) <\(\frac{45}{3}\) = \(15\),
and the answer is ‘yes’.
This is sufficient.

Condition 2)

Assume that \(a\) and \(b\) are the numbers removed, and that their average is greater than \(15\). That is,
\(\frac{a + b}{2} > 15\). Then \(a + b > 30\), and so
\(c + d + e = 75 – ( a + b ) < 45\).
Therefore,
\(\frac{c + d + e}{3}\) <\(\frac{45}{3}\) = \(15\),
and the answer is ‘yes’.
This is sufficient.

Therefore, the answer is D.

Note: By Tip 1) of the VA method, if conditions 1) and 2) provide the same information, we can just select answer D.

Answer: D
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Re: The average (arithmetic mean) of a list of 5 numbers (arithmetic mean)   [#permalink] 26 Nov 2017, 17:23
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