The two columns can be written as \(\frac{67+78+x+101}{4}\) and \(\frac{66+79+x+102}{4}\). Using matching operations method, and after rearranging, we get column A equal to \(1+x\) and column B equal to \(2+x\), thus whatever value of X, column B is greater!

We can solve this question using matching operations (my favorite Quantitative Comparison strategy)

Given:
QUANTITY A: The average (arithmetic mean) of 67, 78, x, and 101
QUANTITY B: The average (arithmetic mean) of 66, 79, x, and 102

Rewrite as follows:
QUANTITY A: \(\frac{67+78+x+101}{4}\)

QUANTITY B: \(\frac{66+79+x+102}{4}\)

Multiply both quantities by \(4\) to get:
QUANTITY A: \(67+78+x+101\)
QUANTITY B: \(66+79+x+102\)

Subtract \(66\) from both quantities to get:
QUANTITY A: \(1+78+x+101\)
QUANTITY B: \(79+x+102\)

Subtract \(78\) from both quantities to get:
QUANTITY A: \(1+x+101\)
QUANTITY B: \(1+x+102\)

Subtract \(101\) from both quantities to get:
QUANTITY A: \(1+x\)
QUANTITY B: \(1+x+1\)

Subtract \(x\) from both quantities to get:
QUANTITY A: \(1\)
QUANTITY B: \(1+1\)

Answer: B

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Quantity A is
(67+78+x+101)/4

Quantity B is
(66+79+x+102)/4

But we can rewrite quantity B as (66+(1+78)+x+(101+1))/4 = (67+78+x+101+1)/4

Notice now that Quantity B is the same as Quantity A but with the numerator one more than the numerator of Quantity A.

Final Answer B

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