During a sale, a clothing store sold each shirt at a price

[GMAT 5/18]

Hi all,

I am taking the GMAT on 5/18 and would like to know how to derive the answer to this question. I have a feeling it is a prime box type question, but couldn't come up with the correct answer. Thanks!

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 (aritmetic 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.

[JadranLee]

Let x be the number of shirts sold, and let y be the number of sweaters sold. The question can be rephrased as "Is y>x?".
The question also tells us the prices of the shirts and sweaters, so we know that the average price of all the shirts and sweaters that the store sold during the sale was (15x + 25y) / 2.

Statement (1) is sufficient. It tells us that the the average equals 21. Knowing that [(15x + 25y) / 2] is closer to 25 than to 15 tells us that y is greater than x. (If y=x, the average is exactly 20. If y<x, the average is less than 20. If y>x, the average is greater than 20. You can test a few numbers for x and y to persuade yourself of this.)

Statement (2) is not sufficient. It tells us that 15x + 25y =420, which is not enough to tell us whether y>x. That's because y could be greater than x (when y=12 and x=8) or less than x (when y=3 and x=23).

Since statement (1) is sufficient and statement (2) is not sufficient, the answer is (A).

Hi all,

I am taking the GMAT on 5/18 and would like to know how to derive the answer to this question. I have a feeling it is a prime box type question, but couldn't come up with the correct answer. Thanks!

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 (aritmetic 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.

[GMAT 5/18]

Thanks very much, Jad. Perfect explanation!

[ddohnggo]

Sorry to unearth an old thread, but when I saw this problem on the gmat prep exam and tried to apply the first premise, I used the following formula: total/# items = average ----- translated: 15x+25y/(x+y) = 21. However, at this point I couldn't deduce it any further like was shown in the solution above.

My question is that if I used the average formula 15x+25y/(x+y) = 21 what would be the next step to further deduce the answer for the first premise as was done above? Or is it best to go with the formula: 15x+25y/2?
editor: the latter formula is wrong, since there aren't just two quantities; there are (x + y) quantities. you're right the first time.
just multiply both sides of your first effort by the denominator (x + y) and you're in business.

[StaceyKoprince]

You can still use the same process Jad used - the idea is more to understand the concept of weighted averages. You don't even really need to write a formula at all.

The idea is that shirts cost $15 and sweaters cost $25. If an equal number of each is sold, then the average would be right in the middle: $20. If more shirts are sold, then the average will be between $15 and $20. If more sweaters are sold, then the average will be between $20 and $25.

So your formula also allows you to see that, because the average is $21, which is between $20 and $25. But the real key is not in the formula itself - it's in the thinking I outlined in the previous paragraph.

[kramacha1979]

Statement 2 says

15 * SH + 25 * SW = 420

Only after plugging some values I can come up with 2 cases, one in which SW > SH and the other in which SH > SW.

However there could be a equation in which there is just one solution. Can't come up with any now.
But my question is how should I go about solving such cases ? Plugging numbers can burn time..
How do I quickly solve and come to a definite answer w/o picking numbers ?

[RonPurewal]

However there could be a equation in which there is just one solution. Can't come up with any now.

here's one from an official problem.

click here

But my question is how should I go about solving such cases ? Plugging numbers can burn time..
How do I quickly solve and come to a definite answer w/o picking numbers ?

you can't.

no, really, seriously, you can't.

unless, of course, you feel like doing this.
you don't, do you?

[vivekwrites]

Ron, I came up with one solution 12 and 8 and then I realized that for every increase of 3 sweater, there will be decrease of 5 shirts (ratio is 3:5) and vice-versa. So other solutions will be (9,13), (15,3) etc. Clearly B is insufficient.

[randy.hagedorn]

So Ron ... you state in the other thread concerning the pencils that B was the correct answer because it was a a diophantine equation.

You said that "a diophantine equation is just an equation whose solutions have to be whole numbers (0, 1, 2, ..., and yes, 0 is considered a whole number).

it's all based on context: either
(1) the problem statement will directly state that the solutions must be integers, or, more commonly,
(2) the context of the word problems will dictate that the solutions must be integers. i.e., if a problem calls for a school class of x boys and y girls, then obviously x and y must be whole numbers (barring the possibility that you're watching some gory school horror movie)."

The shirt/sweater question really can't be solved by plugging in numbers like the pencil question since the prices for the shirts and sweaters are multiples of 5 thus giving us a myriad of different solutions, right?

[RonPurewal]

The shirt/sweater question really can't be solved by plugging in numbers like the pencil question since the prices for the shirts and sweaters are multiples of 5 thus giving us a myriad of different solutions, right?

you're still basically plugging in numbers.

a few of the posters above, including jadran lee (one of our moderators), have posted some of the mathematical reasons WHY there is more than one solution to this problem, but you still have to go out and FIND more than one solution before you declare the statement insufficient.

there is a fair amount of theory behind the fact that, say, 21x + 23y = 130 has only one whole-number solution while 20x + 30y = 130 has more than one.
if you understand that theory, great.
if you don't, it doesn't matter. just start throwing down numbers and see whether more than one pair works out.

--

in your example, yes, the common factor of 5 makes it EXTREMELY likely that there will be more than one solution.
still doesn't guarantee it, though. for instance, 15x + 35y = 160 only has one whole-number solution (x = 6 and y = 2), despite the common factor of 5 and the relatively large sum.

[subbiah.an]

Sorry to unearth an old thread, but when I saw this problem on the gmat prep exam and tried to apply the first premise, I used the following formula: total/# items = average ----- translated: 15x+25y/(x+y) = 21. However, at this point I couldn't deduce it any further like was shown in the solution above.

My question is that if I used the average formula 15x+25y/(x+y) = 21 what would be the next step to further deduce the answer for the first premise as was done above? Or is it best to go with the formula: 15x+25y/2?

Hi,
If you follow this total/# items = average ----- translated: 15x+25y/(x+y) = 21, then I think,
15X + 25 Y = 21X + 21 Y
4Y = 6X
Y/X = 3/2. Therefore Y is greater i.e More sweaters sold.
[editor: yes. perfect.]

[Ben Ku]

Let me know if there are further questions. Thanks!

[amuppur]

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 (aritmetic 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.

I got a chance to look thru this old post and I have a question. Why can't C be the answer.

Let N1 = number of shirts sold and N2 = number of sweaters sold
Is N2 > N1

(1) 21 = [ (15*N1)+(25*N2)] / N1 + N2
and I arrive at 3N1 = 2N2

(2) (15*N1) + (25*N2) = 420
3N1 +5N2 = 84

Substituting (1) in (2), we get
7N2 = 84 (or) N2 = 12
N1 = 8

Hence, N2 > N1
[editor: you don't bother with "together" if one of the statements is sufficient by itself. see the post below yours.]

[evankapil]

You are misinterpreting the DS format. Option C should only be considered if A or B by themselves are not sufficient.

[RonPurewal]

You are misinterpreting the DS format. Option C should only be considered if A or B by themselves are not sufficient.

yes.