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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.

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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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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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.
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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.
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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.

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?
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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.

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.
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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
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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!
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