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

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Re: Tricky Inequality [#permalink]

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New post 21 May 2011, 05:13
I got a similar question wrong in GMATPrep today...

was thinking without algebra but did with enequality in opposite direction

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Re: Tricky Inequality [#permalink]

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New post 31 May 2011, 13:08
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.


Ok, If p<n then the answer is B.
Can any one show the calculation for better understanding?
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Re: Tricky Inequality [#permalink]

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New post 04 Jun 2011, 11:46
Bunuel you rock! I am still digesting why wrote :
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).

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Re: Tricky Inequality [#permalink]

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New post 04 Sep 2011, 20:53
I solved the problem using inequality but i took lot of time.

Please, Bunuel or anyone who understood bunuel's conceptual way to solve this problem,elaborate the below.

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.

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Re: Tricky Inequality [#permalink]

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Well, I tried in a different way and got B.

from stem and 1 9n+3p <=20 and 7n+5p <=20 => 16n+8p <=40 => 2n+p <=5
From this, we can get values for n,p (1,2) (1,3) and (2,1)
Subsituting these values in required inequality 12n+12p<=40
we can see that (1,3) does not satisfy the inequality. So A is insufficient

Coming to B, 9n+3p<=20 and 4n+8p<=20 which imply that 13n+11p <=40
possible values for n and p are (1,2) and (2,1)

Substituting these values in 12n+12p<=40, both the values satisfy the inequality..hence B is sufficient.
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Re: Tricky Inequality [#permalink]

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Hi,

I solved using the below method.

statement from the stem is 20>= 9(n) + 3(p)

The greatest value of n is 2.2 with p being very small. the greatest value of p is 6.33 with n being very small. If n = 2.2 and p is small then $40 is enough to purchase 12 notes and 12 pencils.However if p = 6 then the pencils alone will cost $72 and $40 will not be enough.

Statement 1 : 20>= 7n + 5p

Look at the greatest values for n and p again (keeping in mind the statements caps them at 2.2 and 6.3 respectively) N could be 2.8 in statement 1 but the cap is 2.2 by the original stem. p could be 3.9 from this statement. As stated above $40 will be enough when n = 2.2 and now check p= 3.9 to see that 12(3.9) is already over $40 - since you can get two answers this is insufficient.

Statement 2 says 20>= 4n + 8p

using the same logic - the greatest n = 2.2 (from the stem) but the greatest p can only be 2.5.

when n=2.2, $40(2.2 * 12 )will be enough.When p = 2.5,$40 (2.5 * 12) will be enough again.

Though I solved it.I'm interested in understanding bunuel's conceptual way to solve the problem.Can some one please elaborate bunuel's method.

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Re: Tricky Inequality [#permalink]

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raghavakumar85 wrote:
Well, I tried in a different way and got B.

from stem and 1 9n+3p <=20 and 7n+5p <=20 => 16n+8p <=40 => 2n+p <=5
From this, we can get values for n,p (1,2) (1,3) and (2,1)
Subsituting these values in required inequality 12n+12p<=40
we can see that (1,3) does not satisfy the inequality. So A is insufficient

Coming to B, 9n+3p<=20 and 4n+8p<=20 which imply that 13n+11p <=40
possible values for n and p are (1,2) and (2,1)

Substituting these values in 12n+12p<=40, both the values satisfy the inequality..hence B is sufficient.


Very smart approach. Kudos to you for this!
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Re: Tricky Inequality [#permalink]

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New post 01 Oct 2011, 01:46
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.

OA is:
[Reveal] Spoiler:
B


(1) 9n + 3p = 20
7n + 5p = 20
n - p = 0 => n = p
Not sufficient

(2) 9n + 3p = 20
4n + 8p = 20
solving these we get 12p + 12n = 40
Sufficient

(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 07 Jan 2012, 22:24
Hi Bunuel and all - I took different approach and got ans D. Here u go...
Satement : 9n+3p <=20 , is 12n+12p<=40...(1) ?
9n+3p<=20
or 3n+p <= 20/3
or 12n +4p<= 80/3
substituting in (1) - is 8p <=40-80/3 or is p<=5/3 ?

1>.
7n+5p <=20
from statement 15n+5p <= 100/3
from both - 8n<= 40/3
n<=5/3
substituting p<= 5/3

hence sufficient .
2> similarl approach and sufficient .

ANS is D .

Please check and correct me if this is not correct .

Thanks,
VCG

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

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New post 09 Jan 2012, 12:04
Wow, this is a 750+ level question. I tried to do this by equation solving and get fractions for number of notebooks. Wrong approach! Marked E.

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

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New post 15 Jan 2012, 11:57
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?

In this question, we are given a "market-line" of quotes. [y <= -3n + 20/3] or Line M.
All points in quadrant I that lie below Line M form a region of price quotes that can be filled given prior knowledge. Note that any quote above Line M does not satisfy the inequality and represents unknown information, implying that the order may or may not be filled at that quote.

Can 40 Swiss Francs buy 12 pencils and 12 notebooks? The concept of the question is whether, given Line M, one's order can be filled under a certain preference margin. The "preference line", Line P, is [y <= -n + 20/6]. Any "quote" for notebooks and pencils that is below both the market-line and the "preference line" is a definitive optimal quote.

We can plot the trajectory of any additional package of quotes to see such line's relation to Line M and Line P.

(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

(1) Statement 1's barter information is given by the line, y <= -7/5n + 4. Relative to Line M and Line P, this line borders a sub-region of quotes that are fillable, though they do not satisfy the preference. For example, the following bid, (7notebooks@1SF, 5pencils@2.6SF), can be filled under the market line (notebook@1SF, pencil@3.33SF), while the same bid is rejected as to the preference (notebook@1SF, pencil@2.33SF).

(2) Statement 2's bid-ask line is y <=-1/2n + 20/8. All quotes before point E (intersection) are under both Line M and Line P. These quotes are fillable and satisfy the preference line.

The fact that all of the lines intersect at point E (5/3, 5/3) implies a geometrically fair market.

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

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New post 28 Feb 2012, 23:53
Hi Bunuel - I have two doubts with your solution .
1.
"" 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).""

Why this is so ? If we can replace 2 notebooks with 2 pencils - why we wount be sure about replacing 3 notebooks with 3 pencils ?

2.
I find statement 1 is also suff.
Explanation :
Given, 9n+3p<=20 question is 12n+12p<=40 ?

(1) 7n+5p<=20
Multiplying both side by 2
14n+10p<= 40……..<2>

We can see from statement 1 and given info that 2 notebooks can be replaced by 2 pencils
Replacing 2 notebooks with 2 pencils in equation <2> we get 12n+12p<=40 ,
So statement 1 is suff.

Thanks ,
VCG.

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

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verycoolguy33 wrote:
Hi Bunuel - I have two doubts with your solution .
1.
"" 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).""

Why this is so ? If we can replace 2 notebooks with 2 pencils - why we wount be sure about replacing 3 notebooks with 3 pencils ?
2.
I find statement 1 is also suff.
Explanation :
Given, 9n+3p<=20 question is 12n+12p<=40 ?

(1) 7n+5p<=20
Multiplying both side by 2
14n+10p<= 40……..<2>

We can see from statement 1 and given info that 2 notebooks can be replaced by 2 pencils
Replacing 2 notebooks with 2 pencils in equation <2> we get 12n+12p<=40 ,
So statement 1 is suff.

Thanks ,
VCG.


The OA for this question is given under the spoiler in the initial post and it's B, not D.

As for the red part: if p>n then the value of 3 pencils will be higher than the value of 3 notebooks, and 20 Francs could not be enough to cover that (even if knew that 20 Francs could cover the price difference for 2 pencils and 2 notebooks).

Here is Ian's solution for this problem which might helps to clear you doubts: if-20-swiss-francs-is-enough-to-buy-9-notebooks-and-89226.html#p681382
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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 Feb 2012, 08:22
Hi Bunuel ,
I am not sure about the authenticity of the OA under the spoiler , so it would be great if you just verify my logic for stameent 1 . Thanks .

I find statement 1 is also suff.Explanation :
Given, 9n+3p<=20 question is 12n+12p<=40 ?
(1) 7n+5p<=20
Multiplying both side by 2
14n+10p<= 40……..<2>

We can see from statement 1 and given info that 2 notebooks can be replaced by 2 pencils
Replacing 2 notebooks with 2 pencils in equation <2> we get 12n+12p<=40 ,
So statement 1 is suff.

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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 Feb 2012, 08:27
verycoolguy33 wrote:
Hi Bunuel ,
I am not sure about the authenticity of the OA under the spoiler , so it would be great if you just verify my logic for stameent 1 . Thanks .

I find statement 1 is also suff.Explanation :
Given, 9n+3p<=20 question is 12n+12p<=40 ?
(1) 7n+5p<=20
Multiplying both side by 2
14n+10p<= 40……..<2>

We can see from statement 1 and given info that 2 notebooks can be replaced by 2 pencils
Replacing 2 notebooks with 2 pencils in equation <2> we get 12n+12p<=40 ,
So statement 1 is suff.


OA is correct, have no doubt about it. To check, you can see Ian's example with numbers proving that (1) is not sufficient. See my above post for the link.
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Re: Tricky Inequality [#permalink]

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New post 29 Feb 2012, 09:50
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.


After what you said all is clear but to figure out this tough question alone with your pencil in two minutes or less is quite difficult. I think as usual you have a really amazing approach , really flexible.

Basically we have to cover the difference between 9n and 6n = 3 AND 6p and 3p = 3

1) 9n - 7n = 2 and 5p - 3 p = 2

2) 9n - 4n = 5 AND 8p - 3p = 5

B it is. I hope this kind of approach is not prone of errors but seems 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 01 Mar 2012, 00:08
Hi Buneul,
Quote:
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.



For the second statement
9n+3P<=20 and 4n+8p<=20.
if i solve these two inequaities i get
n<=1.
N cant be negative or zero so n=1
with n=1 i can p<=11/3
p=1,2,3 and n-1
this stmt wud be insufficient then?

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

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New post 01 Mar 2012, 00:21
shankar245 wrote:
Hi Buneul,
Quote:
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.



For the second statement
9n+3P<=20 and 4n+8p<=20.
if i solve these two inequaities i get
n<=1.
N cant be negative or zero so n=1
with n=1 i can p<=11/3
p=1,2,3 and n-1
this stmt wud be insufficient then?


Please notice the following:
1. OA is B;
2. n and p can be zero, why not? We can have charity shop which gives either notebooks or pencils (or both) for free;
3. Values of n and p (prices of 1 notebook and 1 pencil respectively) are not necessary to be integers.

Reconsider your solution according to above.

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

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New post 21 Apr 2012, 20:04
Hey Bunuel,

Can you please verify if I did this correctly. I approached the problem in a different method. I just tried plugging different values.

\(N=Price of notebook\)
\(P=Price of Pencil\)
Given
\(20\geq9N+3P\)
We are asked if this statments is true:
\(40\geq12N+12P\) <=> \(20\geq6N+6P\)

1) \(20\geq7N+5P\)
Combine with the given statments
\(20\geq7N+5P\)
\(20\geq9N+3P\)
we get
\(40\geq16N+8P\) <=> \(20\geq8N+4P\)

Now we just plug numbers! If we pick a number that is greater than 5 for either N or P, this statement will be false. \(20\geq6N+6P\)
Well, we can pick N=0 and P=5 to \(20\geq8(0)+4(5)\) but it doesn't satisfy \(20\geq6(0)+6(5)=30\).
However we can easily pick an N and a P such that it is satified. N=2, P=1 is one that will satifiy both. Hence, Insufficient.

2) \(20\geq4N+8P\)
combine with given:
\(20\geq4N+8P\)
\(20\geq9N+3P\)
we get
\(40\geq13N+11P\)

Plug in numbers, note the highest possible value we can pick for N is N=3 and the highest possible value we can pick for P is 3:
N=3, P=0 satisfies both
N=2, P=1 satisfies both
N=0, P=3 satisfies both

We can never pick a P or an N that is greater than 4, otherwise this would cause the statment to be untrue. Hence sufficient.

So Bunuel, is my logic ok?The only problem that can arise with my method is for the first statment I picked N=0 (ie the price of notebooks is nothing, ie free) I would assume the only restriction is that the price cannot be negative but it can be free.

But then again, there are no restrictions that the price be an integer.hmm I could have also picked fractions for the prices. What do you think? Is plug method an alright way to tackle this problem that ask "X is sufficient to by Y Monkeys and Z Kangaroos"? Do you know any other problems i can try similar to this question.

Many thanks Bunuel!

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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 Jul 2012, 23:36
Bunuel... I have a doubt in regard to this question. Is it possible to add or subtract two equations, each with an inequality? Say, for example, if X + Y <= A and Z+ W<=B, then can we say that X-Z+Y-W<=A-B and X+Y+Z+W<=A+B?

If so, then S1 tells us that 7n+5p<=20 and we know that 3n+p<=20/3; if we subtract one from the other, we get 4n+4p<=40/3, or n +p<=10/3. Does this mean that S1 is sufficient?

Thanks.

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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3   [#permalink] 14 Jul 2012, 23:36

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