BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

notebook and pencil-- inequality

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 48
Joined: Wed Aug 27, 2008 9:58 am

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.blocked
Top
Source: — Data Sufficiency |
Master | Next Rank: 500 Posts
Posts: 106
Joined: Wed Feb 04, 2009 6:09 am
Location: Louisville, KY
Thanked: 8 times

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.blocked
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.
Top
User avatar
Site Admin
Posts: 2567
Joined: Thu Jan 01, 2009 10:05 am
Thanked: 712 times
Followed by:550 members
GMAT Score:770

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.
Top
Senior | Next Rank: 100 Posts
Posts: 48
Joined: Wed Aug 27, 2008 9:58 am

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.
Top
User avatar
Site Admin
Posts: 2567
Joined: Thu Jan 01, 2009 10:05 am
Thanked: 712 times
Followed by:550 members
GMAT Score:770

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.
Top
Legendary Member
Posts: 1035
Joined: Wed Aug 27, 2008 10:56 pm
Thanked: 104 times
Followed by:1 members

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?
Top
User avatar
Site Admin
Posts: 2567
Joined: Thu Jan 01, 2009 10:05 am
Thanked: 712 times
Followed by:550 members
GMAT Score:770

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
Top
Legendary Member
Posts: 1035
Joined: Wed Aug 27, 2008 10:56 pm
Thanked: 104 times
Followed by:1 members

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.
Top
User avatar
Site Admin
Posts: 2567
Joined: Thu Jan 01, 2009 10:05 am
Thanked: 712 times
Followed by:550 members
GMAT Score:770

by DanaJ » Sat May 02, 2009 10:16 am
Thank you. Trust me: it's nothing particularly smart, it's just experience....
Top
Legendary Member
Posts: 1169
Joined: Sun Jul 06, 2008 2:34 am
Thanked: 25 times
Followed by:1 members

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?
Top
User avatar
Site Admin
Posts: 2567
Joined: Thu Jan 01, 2009 10:05 am
Thanked: 712 times
Followed by:550 members
GMAT Score:770

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
Top
Legendary Member
Posts: 1169
Joined: Sun Jul 06, 2008 2:34 am
Thanked: 25 times
Followed by:1 members

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.
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 286
Joined: Tue Sep 21, 2010 5:36 pm
Location: Kolkata, India
Thanked: 11 times
Followed by:5 members

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.
Top
User avatar
Senior | Next Rank: 100 Posts
Posts: 35
Joined: Sun Dec 02, 2007 1:06 pm

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?
Top
Junior | Next Rank: 30 Posts
Posts: 11
Joined: Sat Jan 15, 2011 11:30 am
Followed by:1 members

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
Top
Post new topic Post Reply