Data Sufficiency

A weighted average multiplies each data point by an arbitrary 'weight' and divides by the sum of the weights. Your everyday garden variety average, or arithmetic mean, is actually a special case of a weighted average, except all the weights are equal to 1.

Stuart Kovinsky wrote:

dferm wrote:During a sale, a clothing store sold each shirt at a price of $15 and each sweater at a price of $25. Did the store sell more sweaters than shirts during the sale?

(1) The average (arithmetic mean) of the prices of all the shirts and sweaters that the store sold during the sale was $21.

(2) The total of the prices of all the shirts and sweaters that the store sold during the sale was $420

Please explain..

Thanks.

This question is much simpler if we think of concepts rather than specific arithmetic.

(1) Well, $21 is closer to $25 than it is to $15. Therefore, $25 must have more "weight" in the weighted average equation. The only way that $25 could have more weight is if there were more items sold at $25 than at $15: sufficient.

(2) We still have too many variables (2) and not enough equations (1): insufficient.

(1) is sufficient and (2) isn't: choose (A).