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Stat 1 As the question asks for 12*x and 12*y (symetrical contribution), we need to search the maximum of x and that of y and to verify that those maximums do not cross 40 once each is multiplied by 12.
This is reached when we intersect the 2 lines :
o 9*x + 3y = 20
o 7*x + 5*y = 20
o 45*x + 15*y = 100 (1)
o 21*x + 15*y = 60 (2)
(1) - (2)
<=> 24*x = 40
<=> x = 40/24 = 5/3
Also, y = (20 - 9*5/3)/3 = 5/3
With these maximums, we have 5/3*12 + 5/3*12 = 20 + 20 = 40 (=< 40)
o 9*x + 3y = 20
o 4*x + 8*y = 20
o 36*x + 12y = 80 (3)
o 36*x + 72*y = 180 (4)
From both the statements we get that price of one note book is equal to price of one pencil. By substituting pencil inplace of note book or note book in place of pencil, we can find out the answer. D it is .
I wish i had graph paper ! .... Kevin, is there an easier way to solve this question ?
To be able to say whether 12n+12p<=40 (n+p<10/3) all solutions of n and p proposed by the lines 9n+3p<=20 and (1) 7n+5p<=20, (2) 4n+8p<=20
By drawing lines for 9n + 3p <=20, 7n + 5p <= 20 & 4n + 8p <=20 we see that with just 1 or just 2 there are some regions of n,p specified by (1) or (2) that fall outside n+p<=10/3
Also if we use 1 & 2 we notice that we need to calculate the intersection points of all 4 eqs 1, 2, 9n+3p=20 and n+2p=10/3 (unless i have graph paper then we can just look and tell) to see if all regions of solution fall under the require n+2p<10/3 ....
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.