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# If 20 Swiss Francs is enough to buy 9 notebooks and 3

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Senior Manager
Joined: 24 Nov 2006
Posts: 351
Followers: 1

Kudos [?]: 16 [0], given: 0

I guess the approach for this sort of questions is
1st) translate the inequalities in the stem, (1), and (2) into graphs in the xy plane,
2nd) determine the relevant regions within the xy plane according to the <>= signs
3rd) find out whether the expression that is the object of the question has one definite answer (suff) or more than one answer (insuff) for each (1), (2) given.

And yes, the answer is B.
Senior Manager
Joined: 12 Mar 2006
Posts: 367
Schools: Kellogg School of Management
Followers: 2

Kudos [?]: 38 [0], given: 3

Re: DS: Swiss Francs [#permalink]  15 Jan 2007, 18:17
kevincan wrote:
kevincan 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.

First, note that the question can be simplified as follows: if 9n+3p<=20, is 6n+6p<=20 as well?

(1) 7n+5p<=20. Imagine that you are in a shop- you've picked up 9 notebooks and 3 pencils and the assistant tells you that the total is under 20 SF. Then you change your mind and exchange two of the notebooks for two pencils. The total is still less than 20 SF. It may well be, though, that the new total is closer to 20 SF than was the original:

For example n=0.5 and p=3: 9n+3p=13.5 and 7n+5p=18.5

In this case, 6n+6p=21. i.e. if you exhange another two notebooks for pencils, your purchase will be over 20 SF.

Of course, if n=p=0.5, then 6n+6p<20: NOT SUFF

(2) 4n+8p<=20: This means that if you exchange 5 of the original notebooks for 5 pencils, you will still be under the 20 SF limit. Surely, then, you could safely exchange only 2 of the notebooks for pencils.

Proof: 9n+3p<=20 implies that 18n+6p<=40, and 4n+8p<=20 implies that 12n+24p<=60.

Therefore 30n+30p<=100 and 6n+6p<=20.

Or, a proof by contradiction: suppose 20 SF is not enough to buy 6 of each: i.e. 9n+3p<=20 but 6n+6p>20.
Subtracting, we get that -3n+3p>0, i.e. p-n>0.

But 4n+8p=6n+6p+2(p-n)>20, which contradicts (2). Therefore 20 SF must be enough to buy 6 of each.

very cool kevin ... your q's are always very interesting ... thank you and keep posting....
Re: DS: Swiss Francs   [#permalink] 15 Jan 2007, 18:17

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