dchow23 wrote:
from statement 2,
shirts x
sweaters y
15x +20y = 420
Can we say that since 60 is a common multiple between the 15 and 20, there will be more than one answer that can satisfy the equation?
If there is a common multiple for
Not sure that I understand your question correctly but generally for such kind of question it's always better to check whether an equation whose solution must be integers only has one or more set of variables satisfying it.
As for the question.
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 --> since the average sale price of $21 is closer to $25 than to $15 then store sold more sweaters than shirts (sweaters have more weight in the weighted average and thus pulled it closer to $21). Sufficient.
(2) The total of the prices of all the shirts and sweaters that the store sold during the sale was $420 --> 15x + 25y =420 --> 3x+5y=84. If x=8 and y=12 then x<y but if x=13 and y=5 then x>y. Not sufficient.
Answer: A.
Hi Bunuel there is a typo, it should be x = 13 and y = 9.