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

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If 20 Swiss Francs is enough to buy 9 notebooks and 3 [#permalink] New post 14 Jan 2007, 02:20
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A
B
C
D
E

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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.
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 [#permalink] New post 14 Jan 2007, 02:36
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.

9n+3p<= 20.....................(1)

is
12n+12p<=40 ie: is 3n+3p<= 10

from one

7n+3p<= 20 .......................(2) add 1 to 2

16n+6p<=40 ie 8n+3p<=20 ie 4n+1.5p<=10..............(4)........insuff

from two

4n+8p<=20 ie n+2p<=10...........(3)
add 1 to 3

10n+5p<=30 ie 2n+p<= 6,,,,,,,,,,(5).........insuff

both statments

add 4 to 5

6n+2.5p<=16............insuff

i say E
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 [#permalink] New post 14 Jan 2007, 02:47
(C) for me :)

12*n + 12*p =< 40

9*n + 3*p =< 20 (A)

Stat 1
7*n + 5*p =< 20 (B)

(A) + (B)
<=> 16*n + 8*p =< 40
<=> 2*n + p =< 5
<=> 2*n + p =< 5 (C)
<=> n + p =< 2*n + p =< 5
<=> n + p =< 5
<=> 8*n + 8*p =< 40

INSUFF.

Stat 2
4*n + 8*p =< 20
<=> n + 2*p =< 5 (D)
<=> n + p =< n + 2*p =< 5
<=> n + p =< 5

INSUFF.

Both (1) and (2)

(C) + (D)
<=> 3*n + 3*p =< 10
<=> 12*n + 12*p =< 40

SUFF.

Last edited by Fig on 14 Jan 2007, 05:00, edited 1 time in total.
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Re: DS: Swiss Francs [#permalink] New post 14 Jan 2007, 04:56
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.


9n+3p<=20------(1)
Q: 12n+12p<=40?

S1:7n+5p<=20----(2)
(1)&(2) gives 16n+8P<=40---(3)
(3)/2=>8n+4p<=20---(4)insuff

S2:4n+8p<=20---(5)--insuff

Together:
(4)+(5)=>12n+12p<=40
Hence C
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Re: DS: Swiss Francs [#permalink] New post 14 Jan 2007, 05:58
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.


Ans:B

The q is asking is 12x +12y <=40
or 3x+3y <= 10 or 6x+6y <=20

Given: 9x+3y <= 20, => y>x

1)7x+5y<=20
=> (6x+6y)+(x-y) <= 20. since x<y, doesnt say much

2) 4x+8y <=20
2x+4y<=10
(3x+3y)+(y-x) <=10

y>x so B is suff.
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Re: DS: Swiss Francs [#permalink] New post 14 Jan 2007, 07:00
hsampath 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.


Ans:B

The q is asking is 12x +12y <=40
or 3x+3y <= 10 or 6x+6y <=20

Given: 9x+3y <= 20, => y>x

1)7x+5y<=20
=> (6x+6y)+(x-y) <= 20. since x<y, doesnt say much

2) 4x+8y <=20
2x+4y<=10
(3x+3y)+(y-x) <=10

y>x so B is suff.


I'm not conviced on this part :)

Imagine:
x=y=1. We have 9 + 3 <= 20 and we have 4 + 8 <= 20. Then, 12 + 12 <= 40
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 [#permalink] New post 14 Jan 2007, 07:26
Ok another try :)

(D) :)

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)

(4) - (3)
<=> 60*y = 100
<=> y = 5/3

and, x = 5/3 one more time ;)

Finally, 5/3*12 + 5/3*12 = 20 + 20 = 40 (=< 40)

SUFF.
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 [#permalink] New post 14 Jan 2007, 07:31
kevincan, one more excellent question :)... a hard one and I'm not surprised of it ;)

Nice to see u back & Happy new year to u :D :)
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 [#permalink] New post 14 Jan 2007, 07:52
That's right. Now if I had written instead that 9x+3y <=20

implies y>= x, or y-x >=0, would everything else still be right?

Quote:
I'm not conviced on this part

Imagine:
x=y=1. We have 9 + 3 <= 20 and we have 4 + 8 <= 20. Then, 12 + 12 <= 40
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 [#permalink] New post 14 Jan 2007, 08:02
No :)... we cannot conclude y > = x :)

Take another example : x = 0,2 and y = 0,1. All is verified but x > y :)


hsampath wrote:
That's right. Now if I had written instead that 9x+3y <=20

implies y>= x, or y-x >=0, would everything else still be right?

Quote:
I'm not conviced on this part

Imagine:
x=y=1. We have 9 + 3 <= 20 and we have 4 + 8 <= 20. Then, 12 + 12 <= 40
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 [#permalink] New post 14 Jan 2007, 08:28
Agreed. I was just coming back to say I was wrong..:)


Fig wrote:
No :)... we cannot conclude y > = x :)

Take another example : x = 0,2 and y = 0,1. All is verified but x > y :)


hsampath wrote:
That's right. Now if I had written instead that 9x+3y <=20

implies y>= x, or y-x >=0, would everything else still be right?

Quote:
I'm not conviced on this part

Imagine:
x=y=1. We have 9 + 3 <= 20 and we have 4 + 8 <= 20. Then, 12 + 12 <= 40
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 [#permalink] New post 14 Jan 2007, 08:52
Ok, I also get D now.
Fig, can you confirm if my reasoning is right?

9x+3y <=20

to prove: 12x+12y <=40

1) 7x+5y <=20 => 9x+3y - 2x + 2y <=20
=> 2(y-x) <= 0 since 9x+3y <=20 ..Now this is the part I am unsure of.
=>y<=x

Adding this and 9x+3y<=20, we get: 16x+8y <=40
12x+12x+4(x-y) <=40.
Since x>=y, 12x+12y <= 40

2) 4x+8y <=20
=> (9x+3y) +5(y-x) <= 20
=>5(y-x)<=0 (same reasoning as for 1, 9x+3y<=20)
=>x>= y

Adding eq and eq from stem, 13x+11y <=40
=> 12x+12y + (x-y) <=40
=> 12x+12<=40 since x-y >= 0.
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 [#permalink] New post 14 Jan 2007, 09:03
No :) We are not allowed to do so :)

We can take an example :)

We know that 5 + 3 =< 20

Then,
8 + 6 =< 20
<=> (5 + 3) + (3+3) =< 20

But we cannot say that, as 5 + 3 =< 20, 3 + 3 =< 0 :)

hsampath wrote:
Ok, I also get D now.
Fig, can you confirm if my reasoning is right?

9x+3y <=20

to prove: 12x+12y <=40

1) 7x+5y <=20 => 9x+3y - 2x + 2y <=20
=> 2(y-x) <= 0 since 9x+3y <=20
..Now this is the part I am unsure of.
=>y<=x

Adding this and 9x+3y<=20, we get: 16x+8y <=40
12x+12x+4(x-y) <=40.
Since x>=y, 12x+12y <= 40

2) 4x+8y <=20
=> (9x+3y) +5(y-x) <= 20
=>5(y-x)<=0 (same reasoning as for 1, 9x+3y<=20)
=>x>= y

Adding eq and eq from stem, 13x+11y <=40
=> 12x+12y + (x-y) <=40
=> 12x+12<=40 since x-y >= 0.
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 [#permalink] New post 14 Jan 2007, 09:14
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 . :)
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 [#permalink] New post 14 Jan 2007, 09:54
kevincan wrote:
OA=B


Thanks Kevincan :).... would u give us the OE? :)
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 [#permalink] New post 14 Jan 2007, 13:28
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.
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 [#permalink] New post 14 Jan 2007, 13:30
The figure 4. :)
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 [#permalink] New post 14 Jan 2007, 16:28
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 ....

well...
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Re: DS: Swiss Francs [#permalink] New post 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.
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 [#permalink] New post 15 Jan 2007, 07:14
Thanks for the OE. Would appreciate if you could post OA/OE for some of your older Q's.
  [#permalink] 15 Jan 2007, 07:14
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