notebook and pencil-- inequality

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notebook and pencil-- inequality

by nh8404052006 » Fri May 01, 2009 1:45 am
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 : B
Please explain, thank
source: Gmat-maths.blogspot.com

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Re: notebook and pencil-- inequality

by marcusking » Fri May 01, 2009 4:51 am
nh8404052006 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 : B
Please explain, thank
source: Gmat-maths.blogspot.com
Just thinking out loud here. My initial guess would be E.

From the stem
20 >= 9n + 3p
Asking
40 >= 12n + 3p

i.) 20 >= 7n + 5p
We can't determine anything from this. It says that $20 is enough to buy but it doesn't say that the cost is $20. If a notebook cost $0.01 and a pencil cost $0.02 then all equations would be true. However a notebook might cost $0.10 and a pencil cost $0.30 and it would still be true for all equations. It is way to ambiguous.

ii) 20 >= 4n + 8p
again I don't see anything unique about this equation, or anything concrete we can derive from it.

The only answer I can think of is E.
IF it says that these items COST $20 then maybe we can do something but right now it might as well say $1,000,000 is enough to buy 9 notebooks and 3 pencils.

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by DanaJ » Fri May 01, 2009 5:17 am
I'll use "N" for notebooks and "P" for pencils, so it's easier for me to write the explanation.

You get that 9N + 3P <= 20 francs. You are interested to know if 12N + 12P < = 40 francs, i.e. 40 francs is enough to buy 12N and 12P or 12(N + P). In other words, the statements must yield info about N + P.

1. tells you that 7N + 5P <= 20. Couple that with 9N + 3P <= 20 to get that:
7N + 5P <= 20
9N + 3P <=20
-------------------
16N + 8P <=40 - as you can see, you can safely divide each side by 8 to get:
2N + P <= 5 - this is not what we're looking for (we are looking for N + P).
So 1 is insufficient.

2. tells you that 4N + 8P <=20. Use the same reasoning as above:
4N + 8P <= 20
9N + 3P <= 20
-----------------
13N + 12P <= 40

As you can see, 40 francs is enough to buy 13N and 12P, so it will surely be enough to buy 12N and 12P. This is why the answer is indeed the one provided.

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by nh8404052006 » Fri May 01, 2009 5:56 am
My attempt:
From question stem: 20 &#8805; 9N + 3 P &#8230;.... (1)
From Statement (1) 20 &#8805; 7N + 5 P &#8230;.... (2)
From Statement (2) 20 &#8805; 4N + 8 P &#8230;&#8230; (3)
(1) + (2) + (3) ==> 60 &#8805; 20N + 16 P &#8230; (4)
(4) / 2 ==> 30 &#8805; 10N + 8 P &#8230;&#8230;. (5)
(5) &#8211; (2) ==> 10 &#8805; 3N + 3 P &#8230;&#8230;..(6)
From question stem : 40 &#8805; 12N + 12 P &#8230;. (7)
(7) /4 ==> 10 &#8805; 3N + 3 P &#8230;&#8230;. (8)
Since equation (8) = equation (6) ==> Ans = C but OA = B

correct me if I make any mistake.

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by DanaJ » Fri May 01, 2009 7:41 am
Well, it's only natural that if stmt 2 is sufficient, then both stmts together are sufficient. It's the same as trying to solve this problem:

Is x greater than 4?

1. x is greater than 2
2. x is greater than 5

Of course, stmt 2 is sufficient, but it doesn't mean anything for the argument that you add the extra info that "x is greater than 2".

Plus, what you did here:


(5) – (2) ==> 10 &#8805; 3N + 3 P ……..(6)

is something I like to avoid. Subtracting inequalities is not the best method of solving a question and can sometimes backfire.

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by scoobydooby » Fri May 01, 2009 10:14 am
DanaJ wrote: 2. tells you that 4N + 8P <=20. Use the same reasoning as above:
4N + 8P <= 20
9N + 3P <= 20
-----------------
13N + 12P <= 40

As you can see, 40 francs is enough to buy 13N and 12P, so it will surely be enough to buy 12N and 12P. This is why the answer is indeed the one provided.
hey DanaJ,
dont we get 13N+11P<40 ? (8+3=11)
from here how do we get to infer 12N+12P <40?

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by DanaJ » Fri May 01, 2009 12:10 pm
OMG you are so right! Careless calculation mistake... No worries, it's easily fixed. Forget about that 13N + 11P <=40.

We can use the fact that 4N + 8P <=20 - divide by 4 on each side to get:

N + 2P <= 5

Use the other inequality:

N + 2P <= 5
9N + 3P <=20
--------------------
10N + 5P <= 25 - divide by 5 on each side to get:

2N + P <=5 -use this with N + 2P <= 5:

2N + P <=5
N + 2P <=5
------------
3N + 3P <=10 - multiply by 4 to get your final answer:

12N + 12P <=40

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by scoobydooby » Sat May 02, 2009 9:27 am
wow, thats a smart solution, DanaJ. i could never have thought of such manipulations with the given equations.

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by DanaJ » Sat May 02, 2009 10:16 am
Thank you. Trust me: it's nothing particularly smart, it's just experience....

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by aj5105 » Sat May 02, 2009 8:26 pm
DanaJ,

When we simplify 12P + 12N < 40, we get 1.5P + 1.5N < 5

From Statement (1), we have P + 2N < 5.

Can we make sense of these two equations?

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by DanaJ » Sat May 02, 2009 10:20 pm
I don't really understand what you mean by "making sense" here. The inequalities are consistent with each other, if that's what you're talking about.

Take N = 1 and P = 2 to prove that.

1.5N + 1.5P = 1.5 + 3 = 4.5 <=5

P + 2N = 2 + 2 = 4 <=5

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by aj5105 » Sun May 03, 2009 12:00 pm
Okay, What I meant was if we could find out values for N and P. Now, looking at the inequalities I do realize that I can get multiple values for N and P.

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by pesfunk » Sat Mar 05, 2011 12:12 am
Hello Dana - Your explanation is quite clear to me except that I do not understand why did you use the symbol <=
When the question says "is enough", shouldn't it me equal or more ?
DanaJ wrote:I'll use "N" for notebooks and "P" for pencils, so it's easier for me to write the explanation.

You get that 9N + 3P <= 20 francs. You are interested to know if 12N + 12P < = 40 francs, i.e. 40 francs is enough to buy 12N and 12P or 12(N + P). In other words, the statements must yield info about N + P.

1. tells you that 7N + 5P <= 20. Couple that with 9N + 3P <= 20 to get that:
7N + 5P <= 20
9N + 3P <=20
-------------------
16N + 8P <=40 - as you can see, you can safely divide each side by 8 to get:
2N + P <= 5 - this is not what we're looking for (we are looking for N + P).
So 1 is insufficient.

2. tells you that 4N + 8P <=20. Use the same reasoning as above:
4N + 8P <= 20
9N + 3P <= 20
-----------------
13N + 12P <= 40

As you can see, 40 francs is enough to buy 13N and 12P, so it will surely be enough to buy 12N and 12P. This is why the answer is indeed the one provided.

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by jaguar123 » Sat Jul 16, 2011 11:56 am
Though we have got this answer, we have not arrived in a logic manner. We have
adjusted our equations in many steps to get to the solution.
Any other straight forward method?

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by Ankittalwar » Wed Aug 03, 2011 10:36 pm
Hi Danaj .. Correct if i am wrong
*********************************************************************************************************
Solution

A) Statement 1 is sufficient
7N + 5P <= 20 - eqn 1
9N + 3P <=20 - eqn 2

Eqn 1 can be re written as - > 2N + 5(N+P) <=20 - eqn 3
Eqn 2 can be re written as - > 6N + 3(N+P) <=20 - eqn 4

Mulitiplying eqn 3 by 3 -> 6N + 15 (N+P) <=60 - eqn 5
Subtracting eqn 5 and eqn 4 -> 12(N+P)<=40

B) Statement 2 alone is sufficient
4N + 8P <= 20 - eqn 1
9N + 3P <= 20 - eqn 2

divinding eqn 1 by 4 we get -> N + 2P <= 5 - eqn 3 - > N + (N+P)<=5 - eqn 5
dividing eqn 2 by 3 we get -> 3N + P <= 20/3 - eqn 4 -> 2N + (N+P) <=20/3 - eqn 6

Multipltying eqn 5 by 2 and subtracting eqn 6 we get
(N+P)<= 10/3 - eqn 7
Multiplying eqn 7 by 12 we get 12(N+P)<=40