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

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krishna789
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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40 [#permalink] New post   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 pencils, is 40 [#permalink] New post   13 Apr 2014, 23:57
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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: If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40 [#permalink] New post   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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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40 [#permalink] New post   08 Nov 2017, 06:31
Algebra way: Prerequisite is to know that if a<b and c<d then (a+c)<(b+d)

Stem: 9n + 3p <= 20 (let's call it inequation A). Is 3n + 3p <= 10 ? (let's call it inequation B)
From the stem, note that n can not be more than 2.23 and p can not be more than 6.67.
This will help when testing values.

(1) - 7n + 5p <= 20 (INSUFFICIENT)
Adding inequation A and (1) gives 16n + 8p <= 40 , divide each side by 8, you get 2n + p <= 5
n=1 p=1 satisfy (1) and inequation A, and statisfy inequation B----- YES
n=0,5 p=3 satisfy (1) and inequation A, but does not statisfy inequation B----- NO

(2) - 4n + 8p <= 20 (SUFFICIENT)
divide each side by 4, you get n + 2p <= 5
Adding "9n + 3p <= 20" and "n + 2p <= 5" gives 10n + 5p <= 25 , divide each side by 5, you get 2n + p <= 5
Adding "2n + p <= 5" and "n + 2p <= 5" gives 3n + 3p <= 10

Answer is B
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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40 [#permalink] New post   22 Mar 2018, 22:00
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.

9N+3P<=20, Is 12N+12P<=40?
values for N and P(considering ints as its easy)-N=1, P=1 and N=1,P=2 and N=1,P=3

Option1 - 7N+5P<=20
Adding with 9N+3P<=20 we get => 16N+8P<=40

Values for N and P, N=1,P=1 and N=1, P=2 and N=1, P=3
now putting those values in 12N+12P<=40? - for N=1 , P=1 it gives val>40 and for other 2 it gives val<=40 - INSUFF

Option2 - 4N+8P<=20
Adding with 9N+3P<=20 we get => 13N+11P<=40

Values for N and P, N=1,P=1 and N=1, P=2
now putting those values in 12N+12P<=40, for both options we get val<=40 - SUFF
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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40 [#permalink] New post   20 Jun 2018, 08:30
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.




Hi Bunuel,

Can we do it in this way ? (Only to check that B alone is sufficient)
9n+3p ≤ 20 (given in question) -------(1)
4n + 8p ≤ 20 (from statement (b)) --------(2)

Add (1) and (2), we get
13n + 11p ≤ 40 -------(3)

Subtract (2) from (1), we get (is this step correct ?)
5n - 5p ≤0 which means
n-p ≤ 0 --------(4)

Now subtract (4) from (3), we get
12n + 12p ≤ 40

Therefore, B alone is sufficient.
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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40 [#permalink] New post   31 Jan 2024, 00:36
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.


I am not able to understand the meaning behind the logic of substituting number of notebooks with pencils. Can you explain in more simple language?
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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40 [#permalink] New post   31 Jan 2024, 00:59
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Rohan271 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.


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.


I am not able to understand the meaning behind the logic of substituting number of notebooks with pencils. Can you explain in more simple language?


The solution is elaborated in several posts on the previous two pages. Alternatively, there are other solutions that might be of help. Please consider reviewing those as well. Hope this helps.
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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40 [#permalink]
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