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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## The average (arithmetic mean) of 5 numbers is 10. If 3 of th tagged by: Max@Math Revolution ##### This topic has 1 expert reply and 0 member replies ### GMAT/MBA Expert ## The average (arithmetic mean) of 5 numbers is 10. If 3 of th [GMAT math practice question] The average (arithmetic mean) of 5 numbers is 10. If 3 of the 5 numbers are removed, is the average (arithmetic mean) of the remaining 2 numbers less than 10? 1) Each of the 3 numbers removed is greater than 10 2) The average of the 3 numbers removed is greater than 10 _________________ Math Revolution Finish GMAT Quant Section with 10 minutes to spare. The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. Only$99 for 3 month Online Course
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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.

Let the five numbers be x1, x2, x3, x4 and x5. Then their average (arithmetic mean) is
( x1 + x2 + x3 + x4 + x5 ) / 5 = 10
" x1 + x2 + x3 + x4 + x5 = 50
Assume the numbers to be removed are x1, x2, and x3.

Since we have 5 variables (x1, x2, x3, x4, x5) and 1 equation, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first.

Condition 1) & 2):
Since x1, x2 and x3 are greater than 10 and x1 + x2 + x3 > 30, we have x4 + x5 < 20 or ( x4 + x5 ) / 2< 10.
Both conditions together are sufficient.

Since this is a statistics problem (one of the key question areas), we should also consider choices A and B by CMT (Common Mistake Type) 4(A).

Condition 1)
Since x1, x2 and x3 are greater than 10, we have x1 + x2 + x3 > 30. So, x4 + x5 < 20 or ( x4 + x5 ) / 2< 10.
Condition 1) is sufficient.

Condition 2)
Since ( x1 + x2 + x3 ) / 3 > 10, we have x1 + x2 + x3 > 30, and x4 + x5 < 20 or ( x4 + x5 ) / 2< 10.
Condition 2) is sufficient, too. In addition, since condition 1) is equivalent to condition 2), D is most likely to be the answer by Tip 1).

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.

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