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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 [#permalink]
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15 Jul 2012, 05:50
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sayak636 wrote: Bunuel... I have a doubt in regard to this question. Is it possible to add or subtract two equations, each with an inequality? Say, for example, if X + Y <= A and Z+ W<=B, then can we say that XZ+YW<=AB and X+Y+Z+W<=A+B?
If so, then S1 tells us that 7n+5p<=20 and we know that 3n+p<=20/3; if we subtract one from the other, we get 4n+4p<=40/3, or n +p<=10/3. Does this mean that S1 is sufficient?
Thanks. OA for this question is B and it's correct. As for your solution, notice that: You can only add inequalities when their signs are in the same direction:If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) > \(a+c>b+d\). Example: \(3<4\) and \(2<5\) > \(3+2<4+5\). You can only apply subtraction when their signs are in the opposite directions:If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) > \(ac>bd\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) > \(35<41\). So, we can not subtract 3n+p<=20/3 from 7n+5p<=20 since the signs are in the same direction. Hope it helps.
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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 [#permalink]
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21 Aug 2012, 05:41
Guys...sorry for bothering but can anyone help me in understanding the bunuel's explanation. I just don't understand that if we can replace 2 notebooks with 2 pencils, then why can't be substitute 3 notebooks with 3 pencils.
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Re: Tricky Inequality [#permalink]
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13 Sep 2013, 05:11
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. Responding to a pm: First of all, it is a tricky question. What makes it different from the run of the mill similar questions is the use of "is enough". Had the question said 9N and 3P cost 20 Swiss francs, life would have been much easier. Then your complain "I just don't understand that if we can replace 2 notebooks with 2 pencils, then why can't be substitute 3 notebooks with 3 pencils" would be totally justified. The problem here is that we know that 9N and 3P cost less than or equal to 20 Swiss Francs. So why does Bunuel say "We can substitute 2 notebooks with 2 pencils, but this not enough."? I can try to answer this question of yours using an example. Say 1 N costs 1.5 SF and 1 P costs 1.85 SF Then 9N and 3P cost 19.05 SF (which is less than 20 SF) 7N and 5P cost 19.75 SF. (So even though 1P costs more than 1N, 2P can substitute 2N because total cost is less than 20 SF. Obviously 1P can substitute 1N since there was enough leeway for even 2P in place of 2N) But 6N and 6P cost 20.1 SF. (Now we see that 3P cannot substitute 3N. This time it crossed 20 SF) I hope this answers why we can substitute 2P in place of 2N but not 3P in place of 3N.
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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 [#permalink]
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03 Nov 2013, 13:02
Hi, I am still having some problems with this question.
If 20 SF can buy 9N and 3P, with statement 1 20 SF can buy 7N and 5P, it shows that we can switch 2N for 2P?
So if we double it to 40SF then with 40SF we can buy 14N and 10P with statement 1, so why cant we still change 2N for 2P and get 40SF to buy 12N and 12P



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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 [#permalink]
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03 Nov 2013, 13:04



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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 [#permalink]
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06 Nov 2013, 21:08
Allen122 wrote: Hi, I am still having some problems with this question.
If 20 SF can buy 9N and 3P, with statement 1 20 SF can buy 7N and 5P, it shows that we can switch 2N for 2P?
So if we double it to 40SF then with 40SF we can buy 14N and 10P with statement 1, so why cant we still change 2N for 2P and get 40SF to buy 12N and 12P Yes, the word 'enough' makes it a challenging question. Check out the detailed discussion of this concept here: http://www.veritasprep.com/blog/2013/09 ... ofwords/
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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 [#permalink]
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26 Nov 2013, 06:26
VeritasPrepKarishma wrote: Allen122 wrote: Hi, I am still having some problems with this question.
If 20 SF can buy 9N and 3P, with statement 1 20 SF can buy 7N and 5P, it shows that we can switch 2N for 2P?
So if we double it to 40SF then with 40SF we can buy 14N and 10P with statement 1, so why cant we still change 2N for 2P and get 40SF to buy 12N and 12P Yes, the word 'enough' makes it a challenging question. Check out the detailed discussion of this concept here: http://www.veritasprep.com/blog/2013/09 ... ofwords/THANK YOU! This is the best explanation I have seen so far! Thks!
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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 [#permalink]
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24 Jan 2014, 03:11
Thanks Bunuel, your insight is truly invaluable.
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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 [#permalink]
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16 Feb 2014, 16:57
He is my solution which I find more formal.
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.
What we have: (A) 20 ≥ 9n+3p → 20 ≥ 3(n+p) + 6n
What we need to prove is: (B) 40 ≥ 12n+12p → 40/12 ≥ n+p → n+p ≤ 10/3
Let's see the first statements: (1) 20 ≥ 7n+5p → 20 ≥ 5(n+p)+2n we can not combine (1) with (A) to prove or disprove (B) Unsufficient
The second statement is: (2) 20 ≥ 4n+8p → 20 ≥ 4(n+p)+4p Let's combine it with (A) so we can find the solution: 20 ≥ 3(n+p)+6n x 2 20 ≥ 4(n+p)+4p x 3 → 40 ≥ 6(n+p)+12n + 60 ≥ 12(n+p)+12p = 100 ≥ 18(n+p)+12p+12n → 100 ≥ 30(n+p) → n+p ≤ 10/3 Which is exactly what we need to prove (B). (2) is Sufficient.
So the answer is B.



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Re: Tricky Inequality [#permalink]
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13 Mar 2014, 05:42
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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.  From given equation 9n + 3p < 20 i.e 3n + 1p < 20/3  eq 1 (1) 7n + 5p < 20  eq 2 Subtracting eq 1 from eq 2, we get 4n+4p < 40/3 i.e 12n+12p < 40 (multiplying through out by 3). SUFFICIENT. (2) 4n + 8p < 20 i.e 2n + 4p < 10  eq 3 Adding eq 1 with eq 3, we get, 5n +5p < 50/3 i.e 12n + 12p < 40 (multiplying through out by 12/5). SUFFICIENT. Ans : D Kindly clarify if the approach is correct



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Re: Tricky Inequality [#permalink]
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13 Mar 2014, 06:16
kuttu88 wrote: 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.  From given equation 9n + 3p < 20 i.e 3n + 1p < 20/3  eq 1 (1) 7n + 5p < 20  eq 2 Subtracting eq 1 from eq 2, we get 4n+4p < 40/3 i.e 12n+12p < 40 (multiplying through out by 3). SUFFICIENT. (2) 4n + 8p < 20 i.e 2n + 4p < 10  eq 3 Adding eq 1 with eq 3, we get, 5n +5p < 50/3 i.e 12n + 12p < 40 (multiplying through out by 12/5). SUFFICIENT. Ans : D Kindly clarify if the approach is correct Actually it isn't. The correct answer is (B). You cannot subtract inequalities like this. e.g. 2 < 8 ... (I) 3 < 15 ...(II) (I)  (II) 1 < 7 not true The best is to only add inequalities when they have the same inequality sign. To see how to solve this question, check: http://www.veritasprep.com/blog/2013/09 ... ofwords/
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Re: Tricky Inequality [#permalink]
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13 Mar 2014, 07:04
From given equation 9n + 3p < 20 i.e 3n + 1p < 20/3  eq 1 (1) 7n + 5p < 20  eq 2 Subtracting eq 1 from eq 2, we get 4n+4p < 40/3 i.e 12n+12p < 40 (multiplying through out by 3). SUFFICIENT. (2) 4n + 8p < 20 i.e 2n + 4p < 10  eq 3 Adding eq 1 with eq 3, we get, 5n +5p < 50/3 i.e 12n + 12p < 40 (multiplying through out by 12/5). SUFFICIENT. Ans : D Kindly clarify if the approach is correct [/quote] Actually it isn't. The correct answer is (B). You cannot subtract inequalities like this. e.g. 2 < 8 ... (I) 3 < 15 ...(II) (I)  (II) 1 < 7 not true The best is to only add inequalities when they have the same inequality sign.  But if we subtract eq I from eq II, we get 1 < 7 which is true. OR is there any rule for subtraction in such cases??



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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 [#permalink]
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13 Apr 2014, 22:24
Why should an inequality approach has to be followed for this qn?



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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 [#permalink]
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13 Apr 2014, 23:57
krishna789 wrote: Why should an inequality approach has to be followed for this qn? Because the question says that "20 Swiss Francs is enough to buy 9 notebooks and 3 pencils" This means that 9 notebooks and 3 pencils may cost anything less than or equal to 20 SF. They could cost 18 SF, 10 SF or 20 SF etc. Hence you need to use inequalities. Check out the link I gave above. On that, I have discussed how this question is different from other questions in which we make equations.
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Re: Tricky Inequality [#permalink]
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24 May 2014, 22:08
Understood, awesome answers everyone



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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 [#permalink]
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01 Aug 2014, 10:20
Bunuel,
Doesn't " 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils" come down to 9n + 3p =20? Why are we taking the equation as 9n+3P <= 20 ? Please explain. Thanks



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If 20 Swiss Francs is enough to buy 9 notebooks and 3 [#permalink]
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02 Aug 2014, 02:12



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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 [#permalink]
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23 Mar 2015, 12:25
Is this a realistic question to see on the GMAT?
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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 [#permalink]
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23 Mar 2015, 20:48
ak1802 wrote: Is this a realistic question to see on the GMAT?
Posted from my mobile device See this comment: http://gmatclub.com/forum/if20swissf ... ml#p924136
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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 [#permalink]
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29 Jul 2015, 00:33
Though didn't understand properly with the above explanations. However, later understood with detail explanations on.. http://www.veritasprep.com/blog/2013/09 ... ofwords/




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