Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Question 40>12n+12p? From the stem: 20>9n+3p I) 20>7n+5p adding stem and I: 40>16n+8p NS since we can find combinations of n and p that satisfy or violate the inequality

II) 20>4n+8p adding stem and II: 40>13n+11p NS since we can find combinations of n and p that satisfy or violate the inequality

I+II) adding the stem, I, and II: 60>20n+16p. The question can be rewritten as 60>18n+18p. Again, we can find combinations of n and p that satisfy or violate the inequality, hence E.

B.... stem tells us _20>9n+3p... or 40>18n+6p.... we are asked is 40>12n+12p?.. it will be true if 6n<=6p.. si.. 20>7n+5p or 40>14n+10p.. from stem and this eq , we can say that 4n<=4p... we cant say abt 6n<=6p?..insuff sii.. 20>4n+8p or 40>8n+16p..from stem and this eq , we can say that 10n<=10p...so, we can say 6n<=6p.. suff
_________________

B.... stem tells us _20>9n+3p... or 40>18n+6p.... we are asked is 40>12n+12p?.. it will be true if 6n<=6p.. si.. 20>7n+5p or 40>14n+10p.. from stem and this eq , we can say that 4n<=4p... we cant say abt 6n<=6p?..insuff sii.. 20>4n+8p or 40>8n+16p..from stem and this eq , we can say that 10n<=10p...so, we can say 6n<=6p.. suff

Two queries: How did you decide from 40>12n + 12p that it will be true if 6n<=6p??

You wrote : 4n<=4p .... cant say 6n<=6p but then you wrote 10n<=10p... we can say 6n<=6p Can you kindly explain in detail??
_________________

The answer is D. The price for both notebooks and pencils is the same: $1.67 One can inject this figure into any of the statements to find these are all true statements. In fact, we not need neither s1 nor s2 to solve this question, but since there is no F, D is the answer.

If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40 Swiss Francs enough to buy 12 notebooks and 12 pencils?

(1) 20 Swiss Francs is enough to buy 7 notebooks and 5 pencils. (2) 20 Swiss Francs is enough to buy 4 notebooks and 8 pencils.

Given \(9n+3p\leq20\), question is \(12n+12p\leq40\) true? Or is \(6n+6p\leq20\) true? So basically we are asked whether we can substitute 3 notebooks with 3 pencils. Now if \(p<n\) we can easily substitute notebooks with pencils (equal number of notebooks with pencils ) and the sum will be lees than 20. But if \(p>n\) we won't know this for sure.

But imagine the situation when we are told that we can substitute 2 notebooks with 2 pencils. In both cases (\(p<n\) or \(p>n\)) it would mean that we can substitute 1 (less than 2) notebook with 1 pencil, but we won't be sure for 3 (more than 2).

(1) \(7n+5p\leq20\). We can substitute 2 notebooks with 2 pencils, but this not enough. Not sufficient.

(2) \(4n+8p\leq20\). We can substitute 5 notebooks with 5 pencils, so in any case (\(p<n\) or \(p>n\)) we can substitute 3 notebooks with 3 pencils. Sufficient.

Without any math, my reasoning is 2 variables need separate 2 equations:

1) Stem + 1) sufficient 2 different equations defining n and p. Sufficient

2) Stem + 2) sufficient 2 different equations defining n and p. Sufficient

Now because these are >= and not just = in most other cases you would have to consider that n and p could be negative. But here you would not buy negative quantities of n and p, thus either statement should be sufficient and the answer in my opinion is B.

If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40 Swiss Francs enough to buy 12 notebooks and 12 pencils?

(1) 20 Swiss Francs is enough to buy 7 notebooks and 5 pencils. (2) 20 Swiss Francs is enough to buy 4 notebooks and 8 pencils.

Given \(9n+3p\leq20\), question is \(12n+12p\leq40\) true? Or is \(6n+6p\leq20\) true? So basically we are asked whether we can substitute 3 notebooks with 3 pencils. Now if \(p<n\) we can easily substitute notebooks with pencils (equal number of notebooks with pencils ) and the sum will be lees than 20. But if \(p>n\) we won't know this for sure.

But imagine the situation when we are told that we can substitute 2 notebooks with 2 pencils. In both cases (\(p<n\) or \(p>n\)) it would mean that we can substitute 1 (less than 2) notebook with 1 pencil, but we won't be sure for 3 (more than 2).

(1) \(7n+5p\leq20\). We can substitute 2 notebooks with 2 pencils, but this not enough. Not sufficient.

(2) \(4n+8p\leq20\). We can substitute 5 notebooks with 5 pencils, so in any case (\(p<n\) or \(p>n\)) we can substitute 3 notebooks with 3 pencils. Sufficient.

Answer: B.

Thanks Bunuel!! OA is "B"

Kudos!!

Just wanna know that you solved this problem conceptually and no algebric approach is used. How did you know that this will be done without using algebra.
_________________

Just wanna know that you solved this problem conceptually and no algebric approach is used. How did you know that this will be done without using algebra.

This can be done with algebra or graphic approach as well. Choosing the way you solve depends which approach suits you personally the most, don't think that there is some ground rule for which way to choose.
_________________

If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40 Swiss Francs enough to buy 12 notebooks and 12 pencils?

(1) 20 Swiss Francs is enough to buy 7 notebooks and 5 pencils. (2) 20 Swiss Francs is enough to buy 4 notebooks and 8 pencils.

Given \(9n+3p\leq20\), question is \(12n+12p\leq40\) true? Or is \(6n+6p\leq20\) true? So basically we are asked whether we can substitute 3 notebooks with 3 pencils. Now if \(p<n\) we can easily substitute notebooks with pencils (equal number of notebooks with pencils ) and the sum will be lees than 20. But if \(p>n\) we won't know this for sure.

But imagine the situation when we are told that we can substitute 2 notebooks with 2 pencils. In both cases (\(p<n\) or \(p>n\)) it would mean that we can substitute 1 (less than 2) notebook with 1 pencil, but we won't be sure for 3 (more than 2).

(1) \(7n+5p\leq20\). We can substitute 2 notebooks with 2 pencils, but this not enough. Not sufficient.

(2) \(4n+8p\leq20\). We can substitute 5 notebooks with 5 pencils, so in any case (\(p<n\) or \(p>n\)) we can substitute 3 notebooks with 3 pencils. Sufficient.

Answer: B.

This is awesome approach, but this will hold true when 7+5 =4+8 =9+3

what if these were not summing to 12? then how substitution method could have followed? In that case is there any alternate general logic?
_________________

If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40 Swiss Francs enough to buy 12 notebooks and 12 pencils?

(1) 20 Swiss Francs is enough to buy 7 notebooks and 5 pencils. (2) 20 Swiss Francs is enough to buy 4 notebooks and 8 pencils.

I like Bunuel's approach above. One can also do:

S1: Say notebooks are free, and pencils cost $4 each. Then the answer is 'no'. On the other hand, if pencils and notebooks are both free, the answer is 'yes'. Not sufficient.

S2: * From the stem, we can buy 3 notebooks and 1 pencil for < 20/3 francs. * From S2 we can buy 2 notebooks and 4 pencils for < 10 francs. * Adding, we can buy 5 notebooks and 5 pencils for < 50/3 francs. * Thus, we can buy 12 notebooks and 12 pencils for < (12/5)(50/3) = 40 francs.

The answer is B.
_________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

stmt 1: notebooks free and pencil $4 each then ans is no both notebooks and pencils free then ans is yes

stm2 ; note books free then pencils can have maximum of 2.5 francs each then 12 pencils costs around 30 francs so the ans is free if both notebooks and pencil are free then the ans is yes

But imagine the situation when we are told that we can substitute 2 notebooks with 2 pencils. In both cases ( or ) it would mean that we can substitute 1 (less than 2) notebook with 1 pencil, but we won't be sure for 3 (more than 2).

Regards, Subhash
_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

I got this wrong. Thanks Bunuel for the crisp explanation.

Subhash, Try these numbers. n=1 and p=2.5

For the (1) 7n+5p, this works out to 7 + 12.5 = 19.5 so it holds true (less than or equal to 20) If you substitute one more notebook with a pencil however (6n+6p), it does not hold true.

n=1 and p=2 will hold in both scenarios. So (1) is NS (not sufficient).

Bunnel :Can you please elaborate the P>N & P<N concept below

So basically we are asked whether we can substitute 3 notebooks with 3 pencils. Now if P<N we can easily substitute notebooks with pencils (equal number of notebooks with pencils ) and the sum will be lees than 20. But if P>Nwe won't know this for sure.

But imagine the situation when we are told that we can substitute 2 notebooks with 2 pencils. In both cases ( P<N orP>N ) it would mean that we can substitute 1 (less than 2) notebook with 1 pencil, but we won't be sure for 3 (more than

If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40 Swiss Francs enough to buy 12 notebooks and 12 pencils?

(1) 20 Swiss Francs is enough to buy 7 notebooks and 5 pencils. (2) 20 Swiss Francs is enough to buy 4 notebooks and 8 pencils.

Given \(9n+3p\leq20\), question is \(12n+12p\leq40\) true? Or is \(6n+6p\leq20\) true? So basically we are asked whether we can substitute 3 notebooks with 3 pencils. Now if \(p<n\) we can easily substitute notebooks with pencils (equal number of notebooks with pencils ) and the sum will be lees than 20. But if \(p>n\) we won't know this for sure.

But imagine the situation when we are told that we can substitute 2 notebooks with 2 pencils. In both cases (\(p<n\) or \(p>n\)) it would mean that we can substitute 1 (less than 2) notebook with 1 pencil, but we won't be sure for 3 (more than 2).

(1) \(7n+5p\leq20\). We can substitute 2 notebooks with 2 pencils, but this not enough. Not sufficient.

(2) \(4n+8p\leq20\). We can substitute 5 notebooks with 5 pencils, so in any case (\(p<n\) or \(p>n\)) we can substitute 3 notebooks with 3 pencils. Sufficient.

Answer: B.

Thanks Bunuel! That was a fantastically clear explaination!
_________________

"All that is really worthwhile is action. Personal success or personal satisfaction are not worth another thought." -Pierre Teilhard de Chardin

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...

[rss2posts title=The MBA Manual title_url=https://mbamanual.com/2016/11/22/mba-vs-mim-guest-post/ sub_title=MBA vs. MiM :3qa61fk6]Hey, guys! We have a great guest post by Abhyank Srinet of MiM-Essay . In a quick post and an...

Marketing is one of those functions, that if done successfully, requires a little bit of everything. In other words, it is highly cross-functional and requires a lot of different...