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

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Math Expert
Joined: 02 Sep 2009
Posts: 52161
If 20 Swiss Francs is enough to buy 9 notebooks and 3  [#permalink]

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02 Aug 2014, 01:12
1
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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28 Jul 2015, 23:33
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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14 Sep 2017, 09:10
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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Joined: 02 Sep 2009
Posts: 52161
Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3  [#permalink]

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14 Sep 2017, 20:57
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, 05:31
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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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3  [#permalink]

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22 Mar 2018, 21:00
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  [#permalink]

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20 Jun 2018, 07:30
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.

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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Posts: 52161
Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3  [#permalink]

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20 Jun 2018, 07:51
1
MSGmat1 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.

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.

We cannot subtract the inequalities the way you did.

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 $$5>1$$ --> $$3-5<4-1$$.

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 $$2<5$$ --> $$3+2<4+5$$.

For more check Manipulating Inequalities.
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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3  [#permalink]

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19 Nov 2018, 04:19
@Gmatinsight:plz share ur approach to solve this sir??
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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3  [#permalink]

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14 Jan 2019, 21:10
If we use proper algebric Substitution Method to solve the equ. 1 9N+3P=20 and equ.2 7N+5P=20...we will get a value for "N" & "P".

Similarly for Option "B".
Hence Option "A" & Option "B" both are sufficient.

Why we can not use this process? When to use this kind of assumption process?

Math Expert
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Posts: 52161
Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3  [#permalink]

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14 Jan 2019, 21:15
tannumunu wrote:
If we use proper algebric Substitution Method to solve the equ. 1 9N+3P=20 and equ.2 7N+5P=20...we will get a value for "N" & "P".

Similarly for Option "B".
Hence Option "A" & Option "B" both are sufficient.

Why we can not use this process? When to use this kind of assumption process?

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

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15 Jan 2019, 11:27
Bunuel wrote:
tannumunu wrote:
If we use proper algebric Substitution Method to solve the equ. 1 9N+3P=20 and equ.2 7N+5P=20...we will get a value for "N" & "P".

Similarly for Option "B".
Hence Option "A" & Option "B" both are sufficient.

Why we can not use this process? When to use this kind of assumption process?

Thanks...
Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 &nbs [#permalink] 15 Jan 2019, 11:27

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