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

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

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New post 02 Aug 2014, 02:12
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sri30kanth wrote:
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


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.

Hope it's clear.
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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 [#permalink]

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New post 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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New post 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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New post 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


Again please read carefully the whole discussion and the solutions provided.

(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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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 [#permalink]

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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
Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3   [#permalink] 22 Mar 2018, 22:00

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