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

[dir1118]

Hi Ron,

I have a small question here, it has made me puzzle for a long time, if the result we calulated is: shirts more than sweaters,
do we still choose A as a answer? Since the question is asked about whether sweaters sell more than shirts.
Thank you so much for your answer.

Paddy

[RonPurewal]

Hi Ron,

I have a small question here, it has made me puzzle for a long time, if the result we calulated is: shirts more than sweaters,
do we still choose A as a answer? Since the question is asked about whether sweaters sell more than shirts.
Thank you so much for your answer.

Paddy

yes, that would still be sufficient. that would be a "no" answer to the yes/no question.

if you get a definite "yes" to a yes/no question, then that's sufficient.
if you get a definite "no" to a yes/no question, then that's also sufficient (although it very rarely happens in practice).

[dir1118]

Hi Ron,

I have a small question here, it has made me puzzle for a long time, if the result we calulated is: shirts more than sweaters,
do we still choose A as a answer? Since the question is asked about whether sweaters sell more than shirts.
Thank you so much for your answer.

Paddy

yes, that would still be sufficient. that would be a "no" answer to the yes/no question.

if you get a definite "yes" to a yes/no question, then that's sufficient.
if you get a definite "no" to a yes/no question, then that's also sufficient (although it very rarely happens in practice).

Thanks for your confirmation, Ron.

[RonPurewal]

sure thing.

[pratik.munjal]

Hello fellas:

After all is said and done, more is said and done. So here are my two cents:

(All mathematical philosophy and smugness aside)

Let the shirts be S
Let the sweaters be W.

When I read the question (during the test that is), the only thing that I scribble on my note pad is 15S + 25W. That's what the question "tells" me.

1-Now I look at statement 1. It says something about averages. More precisely it helps me add something to what I had already written. This is what I write:

15S + 25W/(S+W) = 21

I ponder further. Without knowing the values, can I "solve" it? I try.

It gets to a stage where I have 15S + 25W = 21S + 21W. Further solving it I get 4W = 6S. I divide the whole by 4. I look back at the question-"Did the store sell more sweaters than shirts". I look back at my deduction, and realise that what I've found is sufficient to answer the question. I move on.


2- Now I consider statement 2. I "modify" my original equation to get 15S + 25W = 420. I put my thinking cap on (which I'd never removed by the way).

I ask myself-"Is there anything that I can deduce?". A voice tells me "No". But that's my heart-saying "You've got this right". I use my head-Is it possible to "pick" some values. I think. (The clock is ticking by the way). And I say-no. The heart and head concur. Statement 2 is not sufficient.

I mark the answer as A.

[jnelson0612]

Nice, pratik!

[krishnan.anju1987]

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.

This could be worked out by cross multiplying and then y=(3/2) *x where y is number of sweaters and x the number of shirts. so clearly number of sweaters would be more than the number of shirts.

[tim]

thanks..

[supratim7]

Even I burned hell lot of time while testing the 2nd statement.. Faced severe timing issues on similar problems in the past. These are time guzzlers.. more so if you are not arithmetic savvy

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.

I like the approach taken by "vivekwrites" to prove the 2nd statement (15 x a + 25 x b= 420) insufficient i.e.

a) Find one pair in regular sort of way

20 + 400 = 420 doesn't work
45 + 375 = 420 works i.e. 15 x 3 + 25 x 15

b) Then apply the ratio rule i.e. manipulate the first pair keeping in mind the ratio ( 25/15 = 5/3). In this case, for every decrease of three 25's", there will be an increase of five 15's

1st pair: 15 x 3 + 25 x 15
2nd pair: 15 x 8 + 25 x 12
3nd pair: 15 x 13 + 25 x 9
So, insufficient

I find this approach quicker/better than listing out all the numbers. It saved me from tons of arithmetic in head/paper.

20 + 400 » doesn't work
45+ 375 » works
70 + 350 » doesn't work
95 + 325 » doesn't work
120 + 300 » works
145 + 275 » doesn't work
170 + 250 » doesn't work
195 + 225 » works
etc.
etc.

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.

Testing Ron's example (1)
20 x a + 30 x b= 130

a) Find one pair in regular sort of way
10 + 120 = 130 doesn't work
40 + 90 = 130 works i.e. 20 x 2 + 30 x 3

b) Ratio: 30/20 = 3/2. In this case, for every decrease of two 30's", there will be an increase of three 20's
1st pair: 20 x 2 + 30 x 3
2nd pair: 20 x 5 + 30 x 1
So, insufficient

Testing Ron's example (2)
15 x a + 35 x b= 160

a) Find one pair in regular sort of way
20 + 140 = 160 doesn't work
55 + 105 = 160 doesn't work
90 + 70 = 160 works i.e. 15 x 6 + 35 x 2

b) Ratio: 35/15 = 7/3. In this case, for every decrease of three 35's", there will be an increase of seven 15's
1st pair: 15 x 6 + 35 x 2
2nd pair: 15 x (13) + 35 x (-1) » Not allowed
So, sufficient

May I request instructors to comment on this approach..

Many thanks | Supratim

[RonPurewal]

supratim7, yes, that works.
so, if that method makes immediate sense to you"”with no hesitation whatsoever"”then go ahead and use it.

just make sure you are not trying to avoid arithmetic altogether"”sometimes you're just going to have to sit down and crank out the numbers. that's how it is with these problems.

[supratim7]

Thank you Ron. Appreciate it.
Yes, one cannot avoid arithmetic altogether. Just trying to find an alternative :)

Many thanks | Supratim

[tim]

:)