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....