Data Sufficiency

During a sale, a clothing store sold each shirt at a price of $15 and each sweater at a price of $25.00. Did the store sell more sweaters than shirts during the sale?

1) The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00.
2) The total of the prices of all of the shirts and sweaters that the store sold during the sale was $420.00.

The OA is A.

15 sh + 25 sw = 21(sh+sw)
4sw = 6sh
2sw = 3sh. Sufficient.

15 sh + 25 sw = 420
3sh + 5sw = 84

5*16 = 80 (max sw = 16)
Hence by trial method sw = 15, sh = 3.

3*28 = 84 (max sh = 28)
hence by trial method 3*sh = 54 gives sw = 6
meaning sh = 18, sw = 6. Hence not sufficient.

Thus A.

AAPL wrote:During a sale, a clothing store sold each shirt at a price of $15 and each sweater at a price of $25.00. Did the store sell more sweaters than shirts during the sale?

1) The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00.
2) The total of the prices of all of the shirts and sweaters that the store sold during the sale was $420.00

Statement 1:
Since the average price for the MIXTURE of shirts and sweaters (21) is closer to the sweater price (25) than to the shirt price (15), the store sold more sweaters than shirts.
Thus, the answer to the question stem is YES.
SUFFICIENT.

Statement 2:
Let x = shirts and y = sweaters.
Since the $15 shirts and $25 sweaters earn a total of $420 in revenue, we get:
15x + 25y = 420
3x + 5y = 84.
Since the two values in blue are each a multiple of 3, 5y must also be a multiple of 3.
Implication:
y can be any multiple of 3 such that 5y ≤ 84.
Case 1: y=0, with the result that x=28
In this case, the store sells more shirts than sweaters, so the answer to the question stem is NO.
Case 2: y=15, with the result that x=3
In this case, the store sells more sweaters than shirts, so the answer to the question stem is YES.
Since the answer is NO in Case 1 but YES in Case 2, INSUFFICIENT.

The correct answer is A.