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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 X-Z+Y-W<=A-B 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 \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

So, we can not subtract 3n+p<=20/3 from 7n+5p<=20 since the signs are in the same direction.

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. _________________

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

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12 Sep 2013, 15:37

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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.

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. _________________

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

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

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

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

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.

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. _________________

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

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

Consider this: if an apple costs $1 would it be correct to say that $100 is enough to buy one apple? Obviously the answer is yes. So, saying that $100 is enough to buy one apple means that p <= 100.

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