[GMAT math practice question]

If the average (arithmetic mean) price of apples, bananas and oranges is $3.00 per pound, what is their median price?

1) The price of apples is $3.00 per pound.
2) The price of bananas is $2.97 per pound.
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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 3 variables (a for apples, b for bananas and o for oranges) and 1 equation ( \(\frac{( a + b + o )}{3} = 3\)), C 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.

This question is a new type of GMAT question. The average is provided in the original condition and the question asks for the value of the median. The same average applies for each of the conditions.

Conditions 1) & 2)
If \(a = 3.00\) and \(b = 2.97\), then \(\frac{( a + b + o )}{3} = 3\).
So,
\(3.00 + 2.97 + o = 9\)
and
\(o = 3.03\). Therefore, the median is \(3.00.\)
Thus, both conditions together are sufficient.

Condition 1)
We consider three cases.
Case 1: \(a = b = c = 3.00\)
Since all prices are the same, the median is \(3.00.\)

Case 2: \(b < 3.00\)
If \(b < 3.00\), then we must have \(c > 3.00\).
Therefore, the median is \(3\) since \(b < a < c.\)

Case 3: \(b > 3.00\)
If \(b > 3.00\), then we must have \(c < 3.00.\)
Therefore, the median is \(3\) since \(c < a < b.\)

Thus, condition 1) is sufficient on its own.

Condition 2)
If \(a = 3.00, b = 2.97\) and \(c = 3.03\), then the median is \(3.00.\)
If \(a = 2.00, b = 2.97\) and \(c = 3.00,\) then the median is \(2.97.\)
Since we don’t have a unique solution, condition 2) is not sufficient on its own.

Therefore, A is the answer.

Answer: A

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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