During a sale, a clothing store sold each shirt at a price of $15 and

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.

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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=9 then x>y. Not sufficient.

Answer: A.
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Cost of Shirts = $15
Let shirts be x.
Cost of Sweaters = $25
Let sweaters be y.
Is y>x?

1) \(\frac{15x + 25y}{x+y}\) = 21
=> 15x + 25y = 21x + 21y
=> \(\frac{x}{y}\) = \(\frac{2}{3}\)
This basically means for every 2 shirts sold, 3 sweaters are sold.
y > x. Sufficient.

2) 15x + 25y = 420
3x + 5y = 84
x = 8, y =12 => y > x
x = 18, y = 4 => x > y
Insufficient.

A is the answer.
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WE can do this without even framing eqn.

Stat 1 says that avg of all shirts and sweaters = 21.00$ means its closer to the price of sweaters ($25) vis-a-vis that of shirts ($20) hence more of sweaters must have been sold.

hence statement 1 alone is sufficient to answer the question

stat 2 says that 15 shirts + 25 sweaters = 420 hence cannot be solved.

thus Ans - 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.

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

The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00.

Since the average price of $21 is closer to $25 than it is to $15, there must be more sweaters sold than shirts. Statement one alone is sufficient.

Statement Two Alone:

The total of the prices of all of the shirts and sweaters that the store sold during the sale was $420.00.

It’s possible that 12 sweaters and 8 shirts are sold since 12 x 25 + 8 x 15 = 300 + 120 = $420. It’s also possible that 6 sweaters and 18 shirts are sold since 6 x 25 + 18 x 15 = 150 + 270 = $420. In the former example, more sweaters were sold; however, in the latter example, more shirts were sold. Statement two alone is not sufficient.

Answer: A
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The following solution uses the method of algebraic manipulation. This alternative method is longer, but can provide certainty.

Let the number of shirts sold be H and the number of sweaters sold be W.

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

Average of the prices = (Total price of H shirts and W sweaters)/(Total number of shirts and sweaters).

(15H + 25W)/(H + W) = 21
W/H = 3/2

This ratio tells us that the number of sweaters sold (a multiple of 3) will always be greater than the number of shirts sold (a multiple of 2). SUFFICIENT

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

15H + 25W = 420
3H + 5W = 84 ............. (a)

We have to determine whether W > H.

Start with W > H ? Algebraically manipulate this inequality to make use of equation (a).

W > H ?
5W > 5H ?
5W + 3H > 8H ?
84 > 8H ?

Is H < 21/2 ?

We arrive at the question: is the number of shirts sold less than 21/2? Since we do not have any information about the number of shirts sold, we cannot answer the original question about whether the number of sweaters sold is greater than the number of shirts sold. INSUFFICIENT

ANSWER: (A)

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