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by diebeatsthegmat » Wed Jul 13, 2011 11:35 pm
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.

the question i am asking here may sound silly to some people here but i dont understand why the setence " 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils" means 9n+3p<or= 20? why dont it be 9n+3p=20"?
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by Anurag@Gurome » Thu Jul 14, 2011 3:37 am
diebeatsthegmat wrote:the question i am asking here may sound silly to some people here but i dont understand why the setence " 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils" means 9n+3p<or= 20? why dont it be 9n+3p=20"?
Because if the price of 9 notebooks and 3 pencils is 5 or 12 or 15 or 19 or 20 or anything less than or equal to 20 Swiss Francs, then we say that 20 Swiss Francs is enough to buy them. And if the price of 9 notebooks and 3 pencils is something more than 20 Swiss francs, then 20 Swiss Francs is not enough to buy them.
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by Ozlemg » Thu Jul 14, 2011 5:49 am
what is the answer? [spoiler]D?[/spoiler]
The more you suffer before the test, the less you will do so in the test! :)
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by Ankittalwar » Wed Aug 03, 2011 10:40 pm
Anurag@Gurome wrote:
diebeatsthegmat wrote:the question i am asking here may sound silly to some people here but i dont understand why the setence " 20 Swiss Francs is enough to buy 9 notebooks and 3 pencils" means 9n+3p<or= 20? why dont it be 9n+3p=20"?
Because if the price of 9 notebooks and 3 pencils is 5 or 12 or 15 or 19 or 20 or anything less than or equal to 20 Swiss Francs, then we say that 20 Swiss Francs is enough to buy them. And if the price of 9 notebooks and 3 pencils is something more than 20 Swiss francs, then 20 Swiss Francs is not enough to buy them.
Hi Anurag need your help to validate the solution as currently on internet various versions of the solutions are there..

My 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
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