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If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40

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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40  [#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 pencils, is 40  [#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 pencils, is 40  [#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?
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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40  [#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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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40  [#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 pencils, is 40  [#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
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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40  [#permalink]

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New post 20 Jun 2018, 08:30
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.

Answer: B.




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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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40  [#permalink]

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New post 20 Jun 2018, 08:51
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.

Answer: B.




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 pencils, is 40  [#permalink]

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New post 17 Jan 2019, 11:53
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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.


Target question: Is 40 Swiss Francs enough to buy 12 notebooks and 12 pencils?
This is a great candidate for rephrasing the target question.

Let N = the cost of 1 notebook (in Swiss francs)
Let P = the cost of 1 pencil (in Swiss francs)

So, 12 notebooks cost 12N and 12 pencils cost 12P
So, we want to know whether 12N + 12P ≤ 40 (francs)
We can divide both sides by 12 to get: N + P ≤ 40/12
Simplify to get: N + P ≤ 10/3
We can now REPHRASE the target question....
REPHRASED target question: Is N + P ≤ 10/3?

Given: 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils
We can write: 9N + 3P ≤ 20

Statement 1: 20 Swiss Francs is enough to buy 7 notebooks and 5 pencils
We can write: 7N + 5P ≤ 20
We also know that 9N + 3P ≤ 20

Can we use these two inequalities to answer the REPHRASED target question?
It's hard to tell.
Let's add the inequalities to get: 16N + 8P ≤ 40
Divide both sides by 8 to get: 2N + P ≤ 5
In other words (N + P) + N ≤ 5

ASIDE: This inequality looks similar to our REPHRASED target question.
If we subtract N from both sides we get: (N + P) ≤ 5 - N
The REPHRASED target question asks Is N + P ≤ 10/3?
The answer to that question depends on the value of N.
So, let's test some (extreme) values that satisfy the given information:
Case a: N = 2 and P = 0 In this case, N + P = 2. So, the answer to the REPHRASED target question is YES, N + P ≤ 10/3
Case b: N = 0 and P = 4 In this case, N + P = 4. So, the answer to the REPHRASED target question is NO, N + P > 10/3
Since we cannot answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: 20 Swiss Francs is enough to buy 4 notebooks and 8 pencils
We can write: 4N + 8P ≤ 20
We also know that 9N + 3P ≤ 20
Let's add the inequalities to get: 13N + 11P ≤ 40
So close!!!
Too bad we don't have the same number of N's and P's!!
Wait. Perhaps we CAN have the same number of N's and P's if we create some EQUIVALENT inequalities.

First notice that 4N + 8P ≤ 20 is the same as 4(N + P) + 4P ≤ 20
And 9N + 3P ≤ 20 is the same as 3(N + P) + 6N ≤ 20

Now take 4(N + P) + 4P ≤ 20 and multiply both sides by 3 to get: 12(N + P) + 12P ≤ 60
And take 3(N + P) + 6N ≤ 20 and multiply both sides by 2 to get: 6(N + P) + 12N ≤ 40

ADD the two inequalities to get: 18(N + P) + 12P + 12N ≤ 100 [aha!! We now have the SAME number of P's and Q's]
Rewrite as: 18(N + P) + 12(P + N) ≤ 100
Simplify: 30(P + N) ≤ 100
Divide both sides by 30 to get: (P + N) ≤ 100/30
Simplify: (P + N) ≤ 10/3
Perfect! The answer to the REPHRASED target question is YES, N + P ≤ 10/3
Since we can answer the REPHRASED target question with certainty, statement 2 is SUFFICIENT

Answer: B

Cheers,
Brent
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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40  [#permalink]

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New post 26 Nov 2019, 01:29
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.

Can you explain the p<n or p>n part.. I am getting confused since that seems to be the one deciding which statement is suitable here..

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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40  [#permalink]

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New post 26 Nov 2019, 01:34
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.

Bunuel is my understanding correct here?
The question asks us whether we can substitute 3 notebooks for 3 pencils.
Statement 1 tells us we can substitute 2 notebooks for 2 pencils but since 2<3 hence there is no guarantee that 3 notebooks can replace 3 pencils.
Statement 2 states that we can substitute 5 notebooks for 5 pencils so since 5>3 hence 3 notebooks can replace 3 pencils??

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Re: If 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils, is 40   [#permalink] 26 Nov 2019, 01:34

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