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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
Then,
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)
SUFF.
Stat 2
Identical reasonning.
New system:
o 9*x + 3y = 20
o 4*x + 8*y = 20
Then,
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 .
By studying in the XY plan, it's easier to see the solution.
Why statment 1 is not correct?
The first inequations system give us a area in green (we do not represent the negative x or y as prices are always positive)
o 9*x + 3y <= 20
o 7*x + 5*y <= 20
So,
o y <= -3*x + 20/3
o y <= -7/5*x + 4
The figure 1 represents the area in green.
Then, the question tested is 12*x+12*y <= 40. That is equivalent to :
y <= -x + 10/3
The figure 2 represents where is situated the tested line.
Now, by drawing this line, we can see a green area that is ok. But, there is also the red area at top the left that verified the system but that is above the line -x + 10/3.
Thus, we cannot conclude.
Why statment 2 is correct?
Similarly, we obtain another system:
o y <= -3*x + 20/3
o y <= -x/2 + 5/2
The figure 3 shows this system of inequations.
Then, again, we test if the area in green of the figure 3 support also the inequation y <= -x + 10/3. The line y = -x + 10/3 must be above the green area.
The figure 4 shows the test.
Bingo, this time the line is above the green area.
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 ....
Re: DS: Swiss Francs [#permalink]
14 Jan 2007, 20:54
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.