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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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15 Jan 2010, 10:01
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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.

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15 Jan 2010, 15:06
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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.

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04 Sep 2011, 21:40
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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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If 20 Swiss Francs is enough to buy 9 notebooks and 3  [#permalink]

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15 Jan 2010, 10:45
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1
B....

stem tells us 20>9n+3p... or 40>18n+6p...(i)
we are asked is 40>12n+12p?.. (ii)
So it is asked " Can we replace (18-12) or 6 notebooks with 12-6 or 6 pencils.

Logically..
In every equation we have 20 Swiss franc buying 12 items, with different combinations of notebooks and pencils
Initial equation gives more notebooks and lesser pencils, but the Q asks if 6 of each can be purchased in 20 Swiss franc.

1) 20>7n+5p
In Original equation 20>=9n+3p and 20>=7n+5p, number of notebooks is more than number of pencils. So n<p or n>p..
LOGICALLY in both cases number of notebooks is MORE than number of pencils
insuff

2). 20>4n+8p
from stem number of notebooks could be more and this eq tells us that pencils can be MORE..
so, we can say that any type of combinations of 12 items from 4 to 8 ( 4-8, 5-7,6-6,7-5,8-4) is possible
Hence 6 each is possible
suff

B
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15 Jan 2010, 21:51
2
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.

Thanks Bunuel!! OA is "B"

Kudos!!

Just wanna know that you solved this problem conceptually and no algebric approach is used. How did you know that this will be done without using algebra.
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17 Jan 2010, 02:09
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Hussain15 wrote:
Thanks Bunuel!! OA is "B"

Kudos!!

Just wanna know that you solved this problem conceptually and no algebric approach is used. How did you know that this will be done without using algebra.

This can be done with algebra or graphic approach as well. Choosing the way you solve depends which approach suits you personally the most, don't think that there is some ground rule for which way to choose.
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30 Jan 2010, 15:27
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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.

I like Bunuel's approach above. One can also do:

S1: Say notebooks are free, and pencils cost $4 each. Then the answer is 'no'. On the other hand, if pencils and notebooks are both free, the answer is 'yes'. Not sufficient. S2: * From the stem, we can buy 3 notebooks and 1 pencil for < 20/3 francs. * From S2 we can buy 2 notebooks and 4 pencils for < 10 francs. * Adding, we can buy 5 notebooks and 5 pencils for < 50/3 francs. * Thus, we can buy 12 notebooks and 12 pencils for < (12/5)(50/3) = 40 francs. The answer is B. _________________ GMAT Tutor in Toronto If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com Intern Joined: 14 May 2011 Posts: 6 Re: Tricky Inequality [#permalink] ### Show Tags 04 Sep 2011, 22:30 1 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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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.

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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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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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?

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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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]

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15 Jul 2012, 05:50
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sayak636 wrote:
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.

OA for this question is B and it's correct.

As for your solution, notice that:

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

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

So, we can not subtract 3n+p<=20/3 from 7n+5p<=20 since the signs are in the same direction.

Hope it helps.
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13 Sep 2013, 05:11
4
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.

Responding to a pm:

First of all, it is a tricky question. What makes it different from the run of the mill similar questions is the use of "is enough". Had the question said 9N and 3P cost 20 Swiss francs, life would have been much easier. Then your complain "I just don't understand that if we can replace 2 notebooks with 2 pencils, then why can't be substitute 3 notebooks with 3 pencils" would be totally justified.
The problem here is that we know that 9N and 3P cost less than or equal to 20 Swiss Francs.

So why does Bunuel say "We can substitute 2 notebooks with 2 pencils, but this not enough."? I can try to answer this question of yours using an example.

Say 1 N costs 1.5 SF and 1 P costs 1.85 SF
Then 9N and 3P cost 19.05 SF (which is less than 20 SF)

7N and 5P cost 19.75 SF. (So even though 1P costs more than 1N, 2P can substitute 2N because total cost is less than 20 SF. Obviously 1P can substitute 1N since there was enough leeway for even 2P in place of 2N)
But 6N and 6P cost 20.1 SF. (Now we see that 3P cannot substitute 3N. This time it crossed 20 SF)

I hope this answers why we can substitute 2P in place of 2N but not 3P in place of 3N.
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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3  [#permalink]

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16 Feb 2014, 16:57
He is my solution which I find more formal.

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.

What we have:
(A) 20 ≥ 9n+3p → 20 ≥ 3(n+p) + 6n

What we need to prove is:
(B) 40 ≥ 12n+12p → 40/12 ≥ n+p → n+p ≤ 10/3

Let's see the first statements:
(1) 20 ≥ 7n+5p → 20 ≥ 5(n+p)+2n
we can not combine (1) with (A) to prove or disprove (B)
Unsufficient

The second statement is:
(2) 20 ≥ 4n+8p → 20 ≥ 4(n+p)+4p
Let's combine it with (A) so we can find the solution:
20 ≥ 3(n+p)+6n x 2
20 ≥ 4(n+p)+4p x 3

40 ≥ 6(n+p)+12n
+
60 ≥ 12(n+p)+12p
=
100 ≥ 18(n+p)+12p+12n

100 ≥ 30(n+p)

n+p ≤ 10/3
Which is exactly what we need to prove (B).
(2) is Sufficient.

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13 Mar 2014, 05:42
1
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.

------------------------------------------------------------------------------

From given equation 9n + 3p < 20 i.e 3n + 1p < 20/3 --- eq 1

(1) 7n + 5p < 20 --- eq 2

Subtracting eq 1 from eq 2,
we get 4n+4p < 40/3 i.e 12n+12p < 40 (multiplying through out by 3). SUFFICIENT.

(2) 4n + 8p < 20 i.e 2n + 4p < 10 --- eq 3

Adding eq 1 with eq 3,
we get, 5n +5p < 50/3 i.e 12n + 12p < 40 (multiplying through out by 12/5). SUFFICIENT.

Ans : D

Kindly clarify if the approach is correct
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13 Mar 2014, 06:16
kuttu88 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.

------------------------------------------------------------------------------

From given equation 9n + 3p < 20 i.e 3n + 1p < 20/3 --- eq 1

(1) 7n + 5p < 20 --- eq 2

Subtracting eq 1 from eq 2,
we get 4n+4p < 40/3 i.e 12n+12p < 40 (multiplying through out by 3). SUFFICIENT.

(2) 4n + 8p < 20 i.e 2n + 4p < 10 --- eq 3

Adding eq 1 with eq 3,
we get, 5n +5p < 50/3 i.e 12n + 12p < 40 (multiplying through out by 12/5). SUFFICIENT.

Ans : D

Kindly clarify if the approach is correct

Actually it isn't. The correct answer is (B).

You cannot subtract inequalities like this.

e.g.
2 < 8 ... (I)
3 < 15 ...(II)

(I) - (II)
-1 < -7 not true

The best is to only add inequalities when they have the same inequality sign.

To see how to solve this question, check: 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 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  [#permalink]

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13 Apr 2014, 23:57
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  [#permalink]

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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
Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3   [#permalink] 01 Aug 2014, 10:20

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