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

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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 ... -of-words/

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

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13 Sep 2016, 05:36
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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 [#permalink]

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14 Sep 2017, 07:54
i think the solution will be each alone.Please check and post the answer.I am getting each one as 40 .

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

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14 Sep 2017, 09:30
riccky wrote:
i think the solution will be each alone.Please check and post the answer.I am getting each one as 40 .

The OA is B. Please read the whole discussion before posting. Thank you.
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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 [#permalink]

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14 Sep 2017, 10:10
If i am wrong please correct me.

Let notebook=n and pencil =p

As per question 9n+3p costs 20
i.e. 9n+3p=20

Question : is 12n + 12p costs 40 swiss?

1.
given 9n+3p=20(take x as multiplier)
and also 7n+5p=20(take y as multiplier)

9x+7y=12
3x+5y=12

solving above two we get x as -1 and y as 3

so multiplying -1(9n+3p)+3(7n+5p)=20*-1+20*3=-20+60=40
i.e. 12n+12p=40(sufficient)

2.
given 9n+3p=20(take x as multiplier)
and also 4n+8p=20(take y as multiplier)

9x+4y=12
3x+8y=12

solving above two we get x as 4/5 and y as 6/5

so multiplying 4/5(9n+3p)+6/5(4n+8p)=20*4/5+20*6/5=16+24=40
i.e. 12n+12p=40(sufficient)

I think each of them are sufficient to answer 12n+12p=40 .

Please let me know if anything is wrong?
Thanks

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

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14 Sep 2017, 21:57
riccky wrote:
If i am wrong please correct me.

Let notebook=n and pencil =p

As per question 9n+3p costs 20
i.e. 9n+3p=20

Question : is 12n + 12p costs 40 swiss?

1.
given 9n+3p=20(take x as multiplier)
and also 7n+5p=20(take y as multiplier)

9x+7y=12
3x+5y=12

solving above two we get x as -1 and y as 3

so multiplying -1(9n+3p)+3(7n+5p)=20*-1+20*3=-20+60=40
i.e. 12n+12p=40(sufficient)

2.
given 9n+3p=20(take x as multiplier)
and also 4n+8p=20(take y as multiplier)

9x+4y=12
3x+8y=12

solving above two we get x as 4/5 and y as 6/5

so multiplying 4/5(9n+3p)+6/5(4n+8p)=20*4/5+20*6/5=16+24=40
i.e. 12n+12p=40(sufficient)

I think each of them are sufficient to answer 12n+12p=40 .

Please let me know if anything is wrong?
Thanks

(1) $$7n+5p\leq20$$. NOT 7n+5p=20.

(2) $$4n+8p\leq20$$. NOT 4n+8p=20.

This is explained many times on previous pages.
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If 20 Swiss Francs is enough to buy 9 notebooks and 3 [#permalink]

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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

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

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09 Nov 2017, 16:15
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.

Dear Bunuel

apparently, I am wrong, but I cannot find where
this is my approach

9n+3pn=20 ( the max price is 20)
1) 7n+5pn=20(it is enough with 20 Swiss Frances, so any price lowest this is acceptable)
from both we have:

9n+3pn=20 9n+3pn=20
-(7n+5pn)=20 ---->> -7n-5pn=-20

2n-2pn=0 ---> 2n=2pn ---->> n=pn---->> 12n(or pn)=20---->> n(or pn)=5/3

statement 1 is enough because it suggests 12* 5/3=20 whn max price is 20, so it is enough for purchasing 7n and 5 pn

where am I wrong?

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

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09 Nov 2017, 21:29
soodia 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.

Dear Bunuel

apparently, I am wrong, but I cannot find where
this is my approach

9n+3pn=20 ( the max price is 20)
1) 7n+5pn=20(it is enough with 20 Swiss Frances, so any price lowest this is acceptable)
from both we have:

9n+3pn=20 9n+3pn=20
-(7n+5pn)=20 ---->> -7n-5pn=-20

2n-2pn=0 ---> 2n=2pn ---->> n=pn---->> 12n(or pn)=20---->> n(or pn)=5/3

statement 1 is enough because it suggests 12* 5/3=20 whn max price is 20, so it is enough for purchasing 7n and 5 pn

where am I wrong?

Please read the whole discussion. This is answered many times on previous pages as well as in very post you are quoting.
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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3   [#permalink] 09 Nov 2017, 21:29

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