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If the elements of set X are a, b, c and d, is the average (

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If the elements of set X are a, b, c and d, is the average (

by Max@Math Revolution » Thu Apr 12, 2018 12:41 am

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[GMAT math practice question]

If the elements of set X are a, b, c and d, is the average (arithmetic mean) of a, b, c, and d contained in set X?

1) The average (arithmetic mean) of every pair of elements of set X is 10.
2) The average (arithmetic mean) of every three elements chosen from set X is 10.
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by GMATGuruNY » Thu Apr 12, 2018 2:57 am

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Max@Math Revolution wrote:[GMAT math practice question]

If the elements of set X are a, b, c and d, is the average (arithmetic mean) of a, b, c, and d contained in set X?

1) The average (arithmetic mean) of every pair of elements of set X is 10.
2) The average (arithmetic mean) of every three elements chosen from set X is 10.
SUM = (COUNT)(AVERAGE).

Statement 1:
Since every pair of elements must have an average of 10, the SUM of every pair of elements = (count)(average) = 2*10 = 20.
For every pair to yield a sum of 20, every value in the set must be equal to 10:
10, 10, 10, 10.
If any pair of values is selected from the set above, the sum will be 20.
No other set will satisfy this constraint.
The average for the set above = (10+10+10+10)/4 = 10.
Since the average is contained within the set, the answer to the question stem is YES.
SUFFICIENT.

Statement 2:
Since every trio of elements must have an average of 10, the SUM of every trio of elements = (count)(average) = 3*10 = 30.
For every trio to yield a sum of 30, every value in the set must be equal to 10:
10, 10, 10, 10.
If any 3 values are selected from the set above, the sum will be 30.
No other set will satisfy this constraint.
The average for the set above = (10+10+10+10)/4 = 10.
Since the average is contained within the set, the answer to the question stem is YES.
SUFFICIENT.

The correct answer is D.

A similar problem in the OG:
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by Max@Math Revolution » Sun Apr 15, 2018 5:29 pm

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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.

Since we have 4 variables (a, b, c and d) and 0 equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) and 2):
Condition 1) gives rise to the following equations:
( a + b ) / 2 = 10 => a + b = 20
( a + c ) / 2 = 10 => a + c = 20
...
( c + d ) / 2 = 10 => c + d = 20

Condition 2) gives rise to the following equations:
( a + b + c ) / 3 = 10 => a + b + c = 30
( a + b + d ) / 3 = 10 => a + b + d = 30
( a + c + d ) / 3 = 10 => a + c + d = 30
( b + c + d ) / 3 = 10 => b + c + d = 30

Combining these equations yields a = b = c = d = 10.
Therefore, the average is 10, which is an element of set X, and conditions 1) and 2) are sufficient, when taken 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)
Condition 1) gives rise to the following equations:
( a + b ) / 2 = 10 => a + b = 20
( a + c ) / 2 = 10 => a + c = 20
...
( c + d ) / 2 = 10 => c + d = 20

Combining these equations yields a = b = c = d = 10.

Condition 2)

Condition 2) gives rise to the following equations:
( a + b + c ) / 3 = 10 => a + b + c = 30
( a + b + d ) / 3 = 10 => a + b + d = 30
( a + c + d ) / 3 = 10 => a + c + d = 30
( b + c + d ) / 3 = 10 => b + c + d = 30

Combining these equations yields a = b = c = d = 10.

Therefore, the answer is D.

Answer: D

Since conditions 1) and 2) are equivalent, D is the answer by Tip 1) of the VA method.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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