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# If 20 Swiss Francs is enough to buy 9 notebooks and 3

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If 20 Swiss Francs is enough to buy 9 notebooks and 3 [#permalink]

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15 Jan 2010, 10:01
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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.
[Reveal] Spoiler: OA

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15 Jan 2010, 10:27
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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.
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15 Jan 2010, 10:45
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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
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15 Jan 2010, 12:06
chetan2u wrote:
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??
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15 Jan 2010, 14:16
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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. Math Expert Joined: 02 Sep 2009 Posts: 32625 Followers: 5655 Kudos [?]: 68662 [42] , given: 9817 Re: Tricky Inequality [#permalink] ### Show Tags 15 Jan 2010, 15:06 42 This post received KUDOS Expert's post 22 This post was BOOKMARKED Hussain15 wrote: 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. _________________ Current Student Joined: 15 Dec 2009 Posts: 289 Schools: Duke Followers: 5 Kudos [?]: 49 [0], given: 12 Re: Tricky Inequality [#permalink] ### Show Tags 15 Jan 2010, 20:30 So my answer would be D. 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. What's the OA? _________________ Retired Moderator Status: The last round Joined: 18 Jun 2009 Posts: 1310 Concentration: Strategy, General Management GMAT 1: 680 Q48 V34 Followers: 74 Kudos [?]: 805 [2] , given: 157 Re: Tricky Inequality [#permalink] ### Show Tags 15 Jan 2010, 21:51 2 This post received KUDOS Bunuel wrote: Hussain15 wrote: 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. _________________ Intern Joined: 05 May 2008 Posts: 4 Followers: 0 Kudos [?]: 1 [0], given: 1 Re: Tricky Inequality [#permalink] ### Show Tags 15 Jan 2010, 23:54 Please site the source of this Question as B does not seem to be the right Sorry Brusel i Know you are always right but I am not sure if B can be the right answer. So please share the source hence to double check Math Expert Joined: 02 Sep 2009 Posts: 32625 Followers: 5655 Kudos [?]: 68662 [1] , given: 9817 Re: Tricky Inequality [#permalink] ### Show Tags 17 Jan 2010, 02:09 1 This post received KUDOS Expert's post Hussain15 wrote: 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. 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. _________________ CEO Status: Nothing comes easy: neither do I want. Joined: 12 Oct 2009 Posts: 2797 Location: Malaysia Concentration: Technology, Entrepreneurship Schools: ISB '15 (M) GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35 Followers: 217 Kudos [?]: 1401 [0], given: 235 Re: Tricky Inequality [#permalink] ### Show Tags 26 Jan 2010, 10:56 Bunuel wrote: Hussain15 wrote: 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? _________________ Fight for your dreams :For all those who fear from Verbal- lets give it a fight Money Saved is the Money Earned Jo Bole So Nihaal , Sat Shri Akaal Support GMAT Club by putting a GMAT Club badge on your blog/Facebook GMAT Club Premium Membership - big benefits and savings Gmat test review : 670-to-710-a-long-journey-without-destination-still-happy-141642.html GMAT Tutor Joined: 24 Jun 2008 Posts: 1181 Followers: 368 Kudos [?]: 1206 [8] , given: 4 Re: Tricky Inequality [#permalink] ### Show Tags 30 Jan 2010, 15:27 8 This post received KUDOS Expert's post Hussain15 wrote: 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.

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15 Feb 2010, 05:14
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

so choice 'B' is sufficient
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11 Dec 2010, 12:09
this is one of the more challenging questions I have ever faced.
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11 Dec 2010, 19:31
Wow. This is tough.

Thanks for all the explanations
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24 Jan 2011, 07:23
Hi Bunuel

Could you please explain this bit to me :

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).

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24 Jan 2011, 10:41
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).
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03 Mar 2011, 11:07
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
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03 Mar 2011, 22:31
Bunuel wrote:
Hussain15 wrote:
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.

Thanks Bunuel! That was a fantastically clear explaination!
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20 May 2011, 00:21
B it is.
Bunuel u rock.
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Re: Tricky Inequality   [#permalink] 20 May 2011, 00:21

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